most recent 30 from http://mathoverflow.net 2013-05-19T02:46:11Z http://mathoverflow.net/feeds/user/3324 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130652/did-smith-correctly-state-the-mass-formula/130655#130655 Will Jagy 2013-05-15T01:16:52Z 2013-05-15T01:16:52Z

<p>Not entirely sure about Smith. As far as i know, Conway and Sloane's version is correct, I've used it, but they give no proof at all. This is one reason that Shimura got involved. He and his student Jonathan Hanke both published on proofs of the mass formula. Shimura wrote a book, I think that must be item 22 at <a href="http://www.ams.org/journals/bull/2006-43-03/S0273-0979-06-01107-4/" rel="nofollow">http://www.ams.org/journals/bull/2006-43-03/S0273-0979-06-01107-4/</a></p> <p>Let's see. All the difficulty lies in the 2-adic contribution. There have been attempts to make a canonical 2-adic representative for quadratic forms. See J. W. S. Cassels, <em>Rational Quadratic Forms.</em> On page 120 we read</p> <blockquote> <p>We do not attempt to specify a unique canonical form [see Jones(1944), Pall (1945), or Watson (1976a)]: that is more a job for a parliamentary draftsman than for a mathematician. </p> </blockquote> <p>Note that C+S use Watson's version. </p> <p>There is a fair amount of stuff at <a href="http://zakuski.math.utsa.edu/~kap/forms.html" rel="nofollow">http://zakuski.math.utsa.edu/~kap/forms.html</a> and <a href="http://zakuski.math.utsa.edu/~kap/more_than_this.html" rel="nofollow">http://zakuski.math.utsa.edu/~kap/more_than_this.html</a> which tends to the modern. </p> <p>Probably enough for now. </p>

http://mathoverflow.net/questions/130385/the-isoperimetric-problem-for-domains-constrained-to-lie-between-two-parallel-pla/130386#130386 Will Jagy 2013-05-12T01:37:52Z 2013-05-12T04:36:15Z

<p>You have not been very specific about boundary conditions. Still, the constant mean curvature surfaces (they are not called minimal if the mean curvature is nonzero) of revolution in $\mathbb R^3$ are the circular cylinder and the Delaunay surfaces, <a href="http://en.wikipedia.org/wiki/Constant-mean-curvature_surface" rel="nofollow">http://en.wikipedia.org/wiki/Constant-mean-curvature_surface</a> </p> <p>Well, I am not done, but let me at least suggest that the free boundary problem you appear to be describing ought to have the surface meeting the planes orthogonally. Furthermore, I think that the actual minimum surface area for a prescribed volume will simply be a circular cylinder of soap film. It is easy enough to prove that the minimizing surface cannot have a pinched waist and meet the planes obtusely, as would the catenoid. More work to be done. </p> <p>Anyway, see <a href="http://mathoverflow.net/questions/50300/is-there-a-complete-classification-of-constant-mean-curvature-surfaces" rel="nofollow">http://mathoverflow.net/questions/50300/is-there-a-complete-classification-of-constant-mean-curvature-surfaces</a> for a beginning.</p>

http://mathoverflow.net/questions/129879/filling-in-a-rational-orthogonal-matrix-given-one-row Will Jagy 2013-05-06T19:23:11Z 2013-05-08T14:36:02Z

<p>Quick version: given natural $n$ and a row of $n$ integers such that the sum of the squares is another square, call it $m^2.$ For $n=5,6,7$ is it always possible to fill in the rest of an $n$ by $n$ matrix of integers, call it $M,$ so that $M M^T = m^2 I? $ If so, $M/m$ is rational orthogonal.</p> <p>Notes: this is true for $n=1,2,3,4,8.$ 1 is trivial 2 uses complex numbers, 4 uses quaternions, 8 uses octonions. 3 uses quaternion stuff applied to ternary quadratic forms, papers of Jones and Pall mostly, the main one 1939. The naive adaptation of the Jones-Pall formalism to our $n=7$ does not work very well, see <a href="http://mathoverflow.net/questions/117212/octonions-and-the-dance-of-the-seven-veils" rel="nofollow">http://mathoverflow.net/questions/117212/octonions-and-the-dance-of-the-seven-veils</a></p> <p>This is false for $n = 9,17,25,33,\ldots.$ Indeed, take any odd $n = k^2,$ let the first row have all entries $1,$ no second row is possible that is orthogonal to the given first row, consists of integers, and has the same length. Problem mod 2, insofar as the dot product of the two rows is odd, therefore nonzero. Actually, for any $n >1, \; \; \; n \equiv 1 \pmod 8,$ one may specify any $n-3$ odd numbers, then find the final three (also odd) by Gauss three square theorem to get an odd square sum, no luck.</p> <p>Anyway, I did some computer checks, entirely successful for small entries for $n=5,6,7,$ and instinct tells me that it only gets easier with larger entries.</p> <p>So, that is the short version, does this work for any first row of integral length (sum of squares is another square) in dimension $n=5,6,7?$</p>

http://mathoverflow.net/questions/8326/is-there-an-approach-to-understanding-solution-counts-to-quadratic-forms-that-doe/128510#128510 Will Jagy 2013-04-23T16:54:05Z 2013-04-23T20:38:06Z

<p>The best short reference for this is <a href="http://zakuski.math.utsa.edu/~kap/Forms/Lehman_1992.pdf" rel="nofollow">Lehman Math Comp 1992 PDF</a>. A more elaborate discussion is in <a href="http://arxiv.org/abs/1101.2951" rel="nofollow">ME and ALEX</a> which appeared in the Journal of Number Theory, January 2012, volume 132, number 1, pages 258-275. I attempted to include this material in a final section of the JNT paper, that did not work out, see <a href="http://meta.mathoverflow.net/discussion/1060/attribution-on-mo-contact-undergraduate-project/" rel="nofollow">META</a> </p> <p>A (ternary) positive quadratic form is indicated by $\langle a,b,c,r,s,t \rangle$ which refers to $$ f(x,y,z) = a x^2 + b y^2 + c z^2 + r y z + s z x + t x y. $$ The discriminant is $$ \Delta = 4 a b c + r s t - a r^2 - b s^2 - c t^2. $$ Forms are gathered together into genera when they are equivalent locally. A fundamental result of Siegel is that we may calculate the weighted average of representations, over a genus, of a given target number. Siegel's result relates quadratic forms and modular forms. </p> <p>We have genera labelled $G_1, G_2, G_3.$ Given an odd prime $p,$ define useful integers $u,v$ such that $$ (-u | p) = -1, \; \; \; (-v | p) = +1. $$ The first one, $G_1,$ is the only genus of discriminant $p^2.$ Then we have two of the six genera of discriminant $4 p^2,$ these have level $4 p$ and are classically integral. Together $$ \begin{array}{lccccc} \mbox{Genus} &amp; \Delta &amp; \mbox{Level} &amp; \mbox{2-adic} &amp; \mbox{$p$-adic} &amp; \mbox{Mass} \\ G_1 &amp; p^2 &amp; 4 p &amp; y z - x^2 &amp; u x^2 + p(y^2 + u z^2) &amp; (p-1)/48 \\ G_2 &amp; 4p^2 &amp; 4 p &amp; 2 y z - x^2 &amp; u x^2 + p(y^2 + u z^2) &amp; (p-1)/32 \\ G_3 &amp; 4p^2 &amp; 4 p &amp; x^2 + y^2 + z^2 &amp; v x^2 + p(y^2 + v z^2) &amp; (p+1)/96 \end{array} $$ Note that $G_1$ and $G_2$ represent exactly the same numbers, but with different representation measures. Furthermore, when $p \equiv 3 \pmod 4,$ then $h(x,y,z) = x^2 + p y^2 + p z^2 \in G_2,$ but when $p \equiv 1 \pmod 4,$ then $h(x,y,z) = x^2 + p y^2 + p z^2 \in G_3.$</p> <p>Let $s(n)$ be the number of representations of $n$ as the sum of three squares. Then, taking one form $g$ per equivalence class in the specified genus, let $$ R_j(n) = \sum_{g \in G_j} \frac{r_g(n)}{|\mbox{Aut} g|} .$$</p> <p>The two new identities are $$ s(p^2 n) \; - \; p s(n) \; = \; 96 \; R_1(n)\; - \; 96 \; R_2(n), $$</p> <p>$$ (p+2) \; s( n) \; - \; s(p^2 n) \; = \; 96 \; R_3(n) .$$</p>

http://mathoverflow.net/questions/128115/mean-convex-embedded-sphere-and-convex-sphere/128126#128126 Will Jagy 2013-04-19T19:47:13Z 2013-04-19T19:47:13Z

<p>Huisken did not define the notion of mean convexity. A mean convex surface that is not convex could be, for example, the dumbbell shape made this way: draw an oval by making two parallel line segments of the same length, then gluing semicircles at both ends. Now gently bend the line segments a bit inward, so it is still a smooth continuous curve but not convex. Rotate this around an axis parallel to the original straight lines, the result is mean convex.</p> <p>Huisken did compare mean curvature flow on convex surfaces and then mean convex. i am pretty sure he is the one who settled the nature of pinching off; in finite time, the dumbbell shape described pinches off and becomes two surfaces. Just before that happens, is narrow tube that jojns them getting closer to a cylinder, or does it pinch off as a double cone? I don't remember the answer, but he did it. </p>

http://mathoverflow.net/questions/127160/numbers-integrally-represented-by-a-ternary-cubic-form Will Jagy 2013-04-11T00:19:00Z 2013-04-12T16:19:59Z

<p>Given integers $a,b,c,$ and cubic form $$ f(a,b,c) = a^3 + b^3 + c^3 + a^2 b - a b^2 + 3 a^2 c - a c^2 + b^2 c - b c^2 - 4 a b c $$ $$ f(a,b,c) = \det \left( \begin{array}{ccc} a &amp; b &amp; c \\ c &amp; a + c &amp; b + c \\ b + c &amp; b + 2 c &amp; a + b + 2 c \end{array} \right) . $$ what primes $p$ can be integrally represented as $$ p = f(a,b,c)? $$</p> <p><strong>(A):</strong> I think it is all primes $(p| 11) = -1 ,$ and all $p = u^2 + 11 v^2$ in integers, but not any $q = 3 u^2 + 2 u v + 4 v^2.$ Note that, if $-p$ is represented, so is $p.$ </p> <p><strong>(B):</strong> I also suspect that if prime $q = 3 u^2 + 2 u v + 4 v^2$ and $f(a,b,c) \equiv 0 \pmod q,$ then all three $a,b,c \equiv 0 \pmod q,$ and $f(a,b,c) \equiv 0 \pmod {q^3}.$ Checked correct for $q=3,5.$ Maybe I will do a few more. </p> <p>Note that if $f$ integrally represents both $m,n$ then it represents $mn.$ That is because $f(a,b,c) = \det(aI + b X + c X^2),$ where $$ X = \left( \begin{array}{ccc} 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 1 \end{array} \right) $$ Then $X^3 = X^2 + X + I$ and $X^4 = 2 X^2 + 2 X + I.$ </p> <p>If all suspicions are correct, we can correctly describe all numbers integrally represented by this polynomial: positive or negative are unimportant, most prime factors are unimportant, all that matters is that every <em>exponent</em> of a prime factor $q = 3 u^2 + 2 u v + 4 v^2$ must be divisible by 3.</p> <p>I should have done this last time: most of the class field part has already been done, by <a href="http://zakuski.utsa.edu/~jagy/Hudson_Williams_1991.pdf" rel="nofollow">Hudson and Williams (1991)</a>, Theorem 1 and Table 1 on page 134. You get my version of the polynomial by negating their variable $x.$</p> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= </p> <p><img src="http://i.stack.imgur.com/xnDUC.jpg" alt="enter image description here"></p> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p> <pre><code> p a b c 2 0 1 1 7 0 -11 6 11 0 -3 2 13 0 -1 2 17 -1 0 2 19 1 2 4 29 0 -7 4 41 0 3 2 43 0 4 -1 47 0 5 -2 53 0 1 4 61 0 46 -25 73 2 -36 19 79 0 3 4 83 0 24 -13 101 -1 12 -6 103 0 15 -8 107 1 -9 5 109 1 2 6 127 1 -2 4 131 1 7 -3 139 1 -6 4 149 -1 4 2 151 0 -20 11 163 0 5 2 167 -1 1 5 173 0 6 -1 193 1 -52 28 197 0 9 -4 199 -1 5 1 211 0 -12 7 227 -2 0 5 233 0 -16 9 239 0 -6 5 241 0 -4 5 257 0 -1 6 263 2 4 9 269 -1 0 6 271 2 8 -3 277 1 -7 5 281 0 2 7 283 -1 2 6 293 -1 -8 6 307 2 -1 6 311 0 5 6 337 -2 5 2 347 1 7 5 349 0 19 -10 359 -1 9 -3 373 2 5 10 397 1 -1 7 401 0 -68 37 409 3 -77 41 419 0 -7 6 421 0 7 2 431 1 -14 8 439 0 8 -1 457 0 1 8 461 0 -2 7 479 1 -8 6 491 0 7 4 499 0 13 -6 503 -1 -36 20 523 0 9 -2 541 2 -12 7 547 1 -11 7 557 -1 25 -13 563 -2 -11 8 569 0 8 1 571 1 -3 7 587 0 -29 16 593 3 -25 13 599 -1 0 8 601 0 7 6 607 0 11 -4 613 0 4 9 617 2 -1 8 659 0 8 3 673 0 -6 7 677 0 -17 10 683 -1 4 8 701 2 13 -6 733 1 10 -2 739 -1 14 -6 743 -2 1 8 757 0 81 -44 761 -1 8 2 769 0 -25 14 773 -1 7 5 787 2 5 12 809 -1 -10 8 811 -4 0 7 821 -1 3 9 827 2 10 7 853 0 -11 8 857 -2 3 8 863 0 9 2 877 -2 -15 10 883 0 -14 9 887 2 -3 8 907 0 -5 8 911 0 8 7 919 0 -2 9 929 1 7 11 937 3 8 14 941 3 -1 9 953 -1 6 8 967 1 13 -5 991 1 -35 19 997 -3 7 3 </code></pre> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p> <p>Primes represented by $x^2 + 11 y^2$ and then by $3 x^2 + 2 x y + 4 y^2,$ both up to $1000.$</p> <pre><code>jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primego Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 1 0 11 Discriminant -44 Modulus for arithmetic progressions? 11 Maximum number represented? 1000 p p mod 11 11 0 47 3 53 9 103 4 163 9 199 1 257 4 269 5 311 3 397 1 401 5 419 1 421 3 499 4 587 4 599 5 617 1 683 1 757 9 773 3 863 5 883 3 907 5 911 9 929 5 991 1 0 1 3 4 5 9 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primego Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 3 2 4 Discriminant -44 Modulus for arithmetic progressions? 11 Maximum number represented? 1000 p p mod 11 3 3 5 5 23 1 31 9 37 4 59 4 67 1 71 5 89 1 97 9 113 3 137 5 157 3 179 3 181 5 191 4 223 3 229 9 251 9 313 5 317 9 331 1 353 1 367 4 379 5 383 9 389 4 433 4 443 3 449 9 463 1 467 5 487 3 509 3 521 4 577 5 619 3 631 4 641 3 643 5 647 9 653 4 661 1 691 9 709 5 719 4 727 1 751 3 797 5 823 9 829 4 839 3 859 1 881 1 947 1 971 3 977 9 983 4 1 3 4 5 9 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ </code></pre> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p> <p>joro asked about high powers being represented primitively. It turns out that the polynomial is not divisible by 8 unless $a,b,c$ are all even. This, despite the fact that 2 is represented. I believe this happens for all the (unrepresented) primes $q = 3 u^2 + 2 u v + 4 v^2$ as well, in the strongest manner: the polynomial is not divisible by $q$ itself unless $a,b,c$ are. I thought there might be trouble with the prime 11, but no. Anyway, here are some prime powers represented primitively, where $47 = 36 + 11$ and $53 = 9 + 44:$</p> <pre><code> 7 0 1 2 49 1 -1 3 343 6 4 5 2401 -11 -3 9 16807 -11 30 -8 117649 -19 75 -29 823543 -2 -117 82 5764801 162 43 12 40353607 205 -64 186 11 -1 1 1 121 10 15 16 1331 -10 -2 7 14641 12 28 9 161051 1 25 59 1771561 53 -78 70 19487171 37 46 300 214358881 171 -210 460 13 1 3 3 169 10 17 18 2197 -4 3 10 28561 -15 -8 24 371293 8 71 34 4826809 -54 98 77 62748517 -257 125 167 47 1 3 5 2209 10 12 3 103823 108 181 202 4879681 104 32 153 229345007 -128 319 432 53 -1 1 3 2809 10 23 24 148877 100 163 170 7890481 100 18 187 418195493 342 -308 451 </code></pre> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p>

http://mathoverflow.net/questions/123306/kissing-number-of-spheres-in-non-euclidean-geometry/123316#123316 Will Jagy 2013-03-01T10:34:36Z 2013-03-01T10:34:36Z

<p>In one sense, which you are free to see as unimportant, the kissing number in the standard hyperbolic plane is unbounded, as it increases with increasing radii of the disks. One of the Laws of Cosines says $$ \cos \alpha = 1 - \frac{1}{1 + \cosh a}, $$ where $a$ is the side length of an equilateral triangle and $\alpha$ is the vertex angle. As $a$ increases without bound, $\alpha$ decreases without nonzero lower bound. So, with discs of large enough radius, we can place as many disjoint-interior disks as we like around a given central one. The equilateral triangle has vertices at the centers of three mutually tangent circles. </p> <p>I'm just saying.</p>

http://mathoverflow.net/questions/123061/brute-force-lattice-problems/123078#123078 Will Jagy 2013-02-27T06:15:06Z 2013-02-27T06:15:06Z

<p>I know this in the language of quadratic forms. Once you have expressed your form in any somewhat reduced form, you have $f(x) = (1/2) x^T A x$ where $A$ is a symmetric positive and integral matrix. To find all $f(x) \leq M$ what I do is find the largest possible value of each $x_i$ by Lagrange multipliers. This surrounds the ellipsoid by a rectangular shape. The diagonal entries of $A^{-1}$ are involved, as are some square roots. Then you just run a multiple loop, exhausting the rectangle thing. If this does not give enough short vectors, increase $M.$ The main point here is that you have small dimension. </p> <p>It is also possible to more precisely describe the ellipsoid, using the quadratic formula or the like, thus investigating fewer useless points. But for low dimension I would say it does not matter. </p>

http://mathoverflow.net/questions/120925/find-minimum-area-ellipse-which-encloses-two-ellipses/120927#120927 Will Jagy 2013-02-06T02:04:49Z 2013-02-06T02:31:19Z

<p>Yeah, there is a shear transformation that takes one of the ellipses to a circle. The least area ellipse enclosing the resulting figure is now evident by symmetry. Then use the inverse of the shear.</p> <p>Now that I think of it, you can just shrink along the major axis of one of the ellipses and expand on the minor axis to get the circle.</p> <p>All manipulations involved are with 2 by 2 matrices. The hypothesis of coincident centers is crucial.</p>

http://mathoverflow.net/questions/120436/representations-with-triangular-numbers/120615#120615 Will Jagy 2013-02-02T20:27:52Z 2013-02-02T20:27:52Z

<p>For $n &lt; 100,000,000$ the numbers with $a(n) > 3.5 n.$</p> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p> <pre><code> n a(n) k (n - trangular) a(n) / n 377 1620 4 26 4.297082228116711 944 3552 4 278 3.76271186440678 2409 9776 4 1083 4.058115400581154 2924 11856 4 1328 4.054719562243502 3134 11680 3 506 3.726866624122527 4457 19552 5 86 4.386807269463765 4551 18000 3 1776 3.955174686882004 4937 17848 3 281 3.615150901357099 5237 21216 5 86 4.051174336452167 5339 19836 4 1598 3.715302491103203 5774 23112 4 103 4.00277104260478 7007 24570 4 221 3.506493506493507 7743 29760 4 117 3.843471522665634 8119 28575 4 118 3.519522108634068 8643 31347 3 387 3.626865671641791 17951 63640 3 931 3.545206395186898 18695 78744 4 167 4.212035303557101 19280 71344 4 170 3.700414937759336 21695 84864 4 167 3.911684719981563 22322 86088 4 167 3.85664366992205 22364 82836 4 6788 3.703988553031658 23862 86328 4 209 3.617802363590646 28154 103132 3 424 3.66313845279534 29105 113760 4 1139 3.908606768596461 31106 115320 4 478 3.707323345978268 31354 110592 3 8134 3.52720546022836 34683 138075 4 230 3.981057001989447 36500 153900 5 185 4.216438356164383 37080 142788 3 13209 3.850809061488673 43301 154350 4 230 3.564582804092284 46545 164112 4 12092 3.525878182404125 48786 172224 5 270 3.530193088181035 53945 196800 5 317 3.648160163129113 71945 268560 4 2567 3.732851483772326 81771 327240 5 365 4.001907766812195 87050 312750 5 314 3.592762780011488 112478 420432 4 6448 3.737904301285585 113024 427500 5 449 3.782382502831257 119765 440100 5 449 3.674696280215422 122697 521235 4 432 4.248147876476198 122757 429660 4 492 3.500085534837117 146025 535680 5 495 3.668412942989214 153005 579420 4 17024 3.786935067481455 171965 611784 4 560 3.55760765272003 199989 776286 3 16674 3.881643490391971 210653 741312 4 1025 3.519114372926092 218744 768840 3 11054 3.514793548623048 306845 1074480 3 43670 3.501702814124395 319178 1170180 3 9100 3.666230128642952 339768 1341280 3 13740 3.947634856725766 358658 1372140 5 377 3.825761589034679 364041 1373664 4 120788 3.773377174549021 381404 1354792 3 19729 3.552117964153496 487529 2010666 4 10753 4.124197740031875 487577 1820280 4 2957 3.733318019512815 488474 1729000 4 896 3.539594737898025 488507 1835820 4 1916 3.758021891190914 497495 1838172 4 152630 3.694855224675625 499479 1760472 5 1976 3.524616650549873 530711 1975548 4 167333 3.722455347637415 562259 2238720 4 989 3.981652583595816 578204 2100612 4 167333 3.632994583226681 613193 2156700 4 2128 3.517163437938789 614294 2260830 4 4334 3.680371288015185 619904 2302944 4 11048 3.715001032417923 628836 2206344 3 162741 3.508615918935939 631124 2225040 3 32159 3.525519549248642 644321 2531088 5 1910 3.928302818005311 648078 2397933 3 2262 3.700068510271912 657074 2349540 3 2134 3.57576163415384 708534 3505172 5 2268 4.947076639935416 710774 2493556 4 183923 3.508226243503561 743525 3118144 3 54974 4.193731212803874 800605 2912328 4 18730 3.637659020365848 826139 3027460 4 1169 3.664589130884754 831102 3008580 4 83849 3.619988882231062 853241 3113928 4 250988 3.649529265471303 873014 3069730 3 125761 3.516243725759266 887654 3132864 5 1208 3.529375184475032 913079 3450600 5 2504 3.77908154716076 944576 3385536 4 1325 3.584185920455315 950129 3662880 3 172001 3.855139670507899 1023154 3583932 3 2848 3.502827531339368 1053054 4815972 3 185151 4.573338119412679 1069145 4198272 4 430130 3.926756426864457 1125620 4101264 5 2869 3.643559993603525 1194260 4374720 4 41339 3.663121933247367 1433870 5073000 4 4975 3.537977640929792 1437092 5030760 4 1427 3.500652706994403 1532012 5488000 4 1637 3.582217371665496 1540764 6844500 4 1629 4.44227668870768 1546109 5569360 3 236338 3.602178112927355 1574189 5747760 5 1538 3.651251533329225 1728807 6313800 3 5511 3.652113856549632 1738022 7705776 4 1706 4.433646984905829 1756593 6184200 5 1592 3.520565093906215 2101988 7656960 4 114967 3.642722984146437 2162096 8304000 3 8246 3.840717525956294 2392571 8396850 4 15281 3.50955102272827 2514050 9026292 5 1889 3.59033909429009 2566235 9526600 3 422750 3.712286676785252 2606609 9437922 4 1706 3.620766290609754 2666783 10844610 4 57313 4.066551346697501 2726700 9582840 4 1755 3.51444603366707 2943915 11066720 4 41070 3.759184623197341 2955869 10405836 4 7063 3.520398231450717 3051479 13031040 4 7201 4.27040133653222 3148760 11146122 3 7489 3.539844891322298 3199064 12361752 4 2408 3.864177771998309 3785324 14553280 3 681538 3.844659004090535 4148505 20141484 5 5624 4.855118651176749 4160028 16131140 3 765713 3.877651785036062 4278428 15179472 4 221452 3.547908717874883 4417170 15758184 4 5235 3.567484158409117 4555499 16685504 5 14908 3.66271708104864 4570599 17266730 3 57089 3.777782737011057 4713605 18242700 3 14960 3.870222473032848 4856214 22366476 4 80619 4.605743486592642 4918508 18635400 4 15362 3.788831897803155 5111928 19120640 4 296072 3.740396969597381 5237153 19183008 4 2923 3.662869501807566 5609519 20335128 4 3293 3.62511081609671 5791500 20667474 4 57309 3.568587412587413 5828400 22007024 4 5822 3.775825955665363 5921345 20997760 4 6265 3.546113256363208 5977013 21814375 3 944635 3.649711820937984 6060230 22278400 4 3290 3.676164105982776 6377159 24006312 4 20764 3.764421116048698 6539531 25291444 3 21676 3.867470618305808 6736259 24628800 4 21979 3.656153957263223 7278935 25518432 4 685039 3.505791987426732 7401413 26275872 4 7478 3.55011563332569 7590987 29577240 4 65727 3.896362884036029 7933785 28037850 5 27509 3.533981573738134 8750180 32834112 5 53645 3.752392750777698 8914607 32087200 4 4076 3.59939591279795 9297650 37763460 4 33290 4.061613418444446 9346322 36235648 4 8641 3.87699546409807 9526839 35421234 4 28578 3.718046877878381 10271700 37488704 4 4454 3.649707838040441 10293233 36452388 3 228892 3.541393457235448 10700699 41116075 3 877171 3.842372820691433 10830575 39205296 6 3044 3.619872075120665 10887764 38109696 4 236344 3.500231636174333 10892754 39281112 4 23301 3.606169018413525 11297264 41553600 5 150983 3.678200314695665 11570390 40535112 4 2990237 3.50334880673858 11607968 44046648 4 8632 3.794518385991415 12697697 49261264 5 4456 3.87954319590395 13381514 48341685 4 4136 3.612572164853693 13501710 50516180 4 10295 3.741465340316152 14685092 54864474 3 1595806 3.736066072994299 15072318 54310326 3 21477 3.603316092455056 15264369 54455592 3 9843 3.567497090773946 16014350 60271750 3 588115 3.763608888278325 17543358 65887452 3 5355 3.755692154261459 17768331 66752000 4 10511 3.756796291109165 18092546 65058240 5 5441 3.595858758629106 18594675 66302250 3 266190 3.565657910127496 18625892 67320000 4 628892 3.614323544880428 19048806 69817664 5 5100 3.665198963126613 21063119 77693616 4 148841 3.688609270070591 21256397 76865200 4 43556 3.616097309435837 22554933 79237704 4 493530 3.513098620155511 23093774 82862832 4 6438668 3.588102663514417 25116125 92868048 5 6884 3.69754681504412 25542609 90452432 5 6378 3.541236997363895 26052704 95841760 4 13768 3.678764400040779 26109692 94424640 4 214886 3.616459359229515 26516996 98940000 4 57821 3.73119187407201 27095031 101537634 4 6551 3.747463289486548 27486656 103165810 4 6665 3.753305240186365 27724950 97894820 4 7344030 3.530928640087719 27790965 102439944 3 8493387 3.686088050558878 27948929 98012200 3 14854 3.506832050702193 30642003 109327536 3 14952 3.567897829655588 31366983 110739296 3 15662 3.530441419884086 31660668 117679797 3 150777 3.716908215581554 32678023 117534903 3 14620 3.596756847866837 34051070 119893374 4 328492 3.520986976327029 34656740 123431424 3 107912 3.561541679915653 38501162 142436148 3 250531 3.699528549294175 40919563 146480400 4 45202 3.579715648478455 42757394 150734844 5 45491 3.525351521657283 43659675 160492544 4 9179 3.67599035036335 46971593 170623242 4 482690 3.632477229375636 54262017 219108240 4 145611 4.037967110584924 57969450 206383044 4 419094 3.560203589994385 61188365 218122920 3 21035 3.564777715501958 68917241 245462490 5 21050 3.561699314109223 69411164 244629060 4 56411 3.524347466640957 72437318 265728532 3 21770107 3.668392747506195 73247012 259686240 4 579472 3.545349262847746 73900616 262277748 4 19898588 3.549060375897273 84116339 300673704 5 1908086 3.574498219662175 84959663 343780800 4 179932 4.04640023113086 87521729 311169600 4 11894 3.555341097066307 n a(n) k (n - trangular) a(n) / n </code></pre> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p>

http://mathoverflow.net/questions/120436/representations-with-triangular-numbers/120502#120502 Will Jagy 2013-02-01T08:48:27Z 2013-02-02T19:02:09Z

<p>Up to 100,000,000, the evidence is consistent with $a(n) = o(n \log n).$ However, it is also consistent with the conjecture that $a(n) &lt; 5 n.$ Below are all $n$ such that $a(n) > 4 n,$ for $n &lt; 10^8$.</p> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p> <pre><code> n a(n) k n - triang a(n) / n a(n) / (n log n) 377 1620 4 26 4.297082228116711 0.7243601861246181 2409 9776 4 1083 4.058115400581154 0.521141979825834 2924 11856 4 1328 4.054719562243502 0.5080651557798358 4457 19552 5 86 4.386807269463765 0.5221002825521364 5237 21216 5 86 4.051174336452167 0.4730743737436168 5774 23112 4 103 4.00277104260478 0.4621539565467688 18695 78744 4 167 4.212035303557101 0.4282259482688222 36500 153900 5 185 4.216438356164383 0.4013718465190124 81771 327240 5 365 4.001907766812195 0.3537855116876313 122697 521235 4 432 4.248147876476198 0.3625481203181061 487529 2010666 4 10753 4.124197740031875 0.3148938427817576 708534 3505172 5 2268 4.947076639935416 0.3672402776801521 743525 3118144 3 54974 4.193731212803874 0.3102065465513547 1053054 4815972 3 185151 4.573338119412679 0.329795232400189 1540764 6844500 4 1629 4.44227668870768 0.3117870922046103 1738022 7705776 4 1706 4.433646984905829 0.308572334234016 2666783 10844610 4 57313 4.066551346697501 0.2748341420425469 3051479 13031040 4 7201 4.27040133653222 0.2860064408420606 4148505 20141484 5 5624 4.855118651176749 0.3186137460305866 4856214 22366476 4 80619 4.605743486592642 0.2991564294828833 9297650 37763460 4 33290 4.061613418444446 0.2531345905630797 54262017 219108240 4 145611 4.037967110584924 0.2267331768485826 84959663 343780800 4 179932 4.04640023113086 0.2216272082074552 jagy@phobeusjunior:~$ jagy@phobeusjunior:~$ date Sat Feb 2 2013 </code></pre> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p>

http://mathoverflow.net/questions/120436/representations-with-triangular-numbers/120469#120469 Will Jagy 2013-01-31T21:59:53Z 2013-02-01T02:18:37Z

<p>Up to 100,000, the program output is consistent with Barry's idea that a(n) is actually $o(n \log n),$ here each is the last occurrence of ratio at least...</p> <p>=-=-=-=-=-=-=-=-=-=-=</p> <pre><code> n a(n) a(n) / (n log n) 5 12 3 2 1.491203842943068 1.0 8 16 3 2 0.9617966939259756 0.9 33 96 4 5 0.8319990327350526 0.8 377 1620 4 26 0.7243601861246181 0.7 377 1620 4 26 0.7243601861246181 0.6 4457 19552 5 86 0.5221002825521364 0.5 36500 153900 5 185 0.4013718465190124 0.4 95643 334080 3 1248 0.3045757315691415 0.3 </code></pre> <p>=-=-=-=-=-=-=-=-=-=-=</p> <p>Using Barry's description, here are the numbers below 100,000 for which $a(n) > 4n.$ </p> <p>=-=-=-=-=-=-=-=-=-=-=</p> <pre><code> n a(n) k n-trianglular a(n) / n 377 1620 4 376 4.297082228116711 2409 9776 4 2408 4.058115400581154 2924 11856 4 2923 4.054719562243502 4457 19552 5 4456 4.386807269463765 5237 21216 5 5236 4.051174336452167 5774 23112 4 5773 4.00277104260478 18695 78744 4 18692 4.212035303557101 36500 153900 5 36499 4.216438356164383 81771 327240 5 81770 4.001907766812195 jagy@phobeusjunior:~$ </code></pre> <p>=-=-=-=-=-=-=-=-=-=-=</p> <p>Same list, but this time showing the largest triangular number used (as a difference)... </p> <p>=-=-=-=-=-=-=-=-=-=-=</p> <pre><code> n a(n) k n-trianglular a(n) / n 377 1620 4 26 4.297082228116711 2409 9776 4 1083 4.058115400581154 2924 11856 4 1328 4.054719562243502 4457 19552 5 86 4.386807269463765 5237 21216 5 86 4.051174336452167 5774 23112 4 103 4.00277104260478 18695 78744 4 167 4.212035303557101 36500 153900 5 185 4.216438356164383 81771 327240 5 365 4.001907766812195 jagy@phobeusjunior:~$ </code></pre> <p>=-=-=-=-=-=-=-=-=-=-=</p> <p>This is what I meant about experimenting. I allowed for as many as six triangular numbers, which may or may not really lead to optimum for all targets up to 55, but it seems likely. I told the computer that 1 choose 2 was defined to be 0, so both the product and the sum are unaffected if any of the $a_i$ are given as 1.</p> <p>Note: from Emil's comment about $n^{3/2},$ it is necessary to run this program up to 1000 or 10000 where $\sqrt n > \log n$ and most targets will have an optimal expression with, say, half a dozen triangular numbers, as opposed to the two or three typical here.</p> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p> <pre><code>best 4 6 best 5 12 best 6 4 best 7 8 best 8 16 best 9 12 best 10 5 best 11 10 best 12 16 best 13 15 best 14 30 best 15 6 best 16 12 best 17 24 best 18 18 best 19 36 best 20 25 best 21 7 best 22 14 best 23 28 best 24 21 best 25 30 best 26 60 best 27 28 best 28 8 best 29 16 best 30 32 best 31 24 best 32 48 best 33 96 best 34 32 best 35 64 best 36 9 best 37 18 best 38 36 best 39 27 best 40 54 best 41 108 best 42 36 best 43 48 best 44 96 best 45 10 best 46 20 best 47 40 best 48 30 best 49 56 best 50 112 best 51 40 best 52 80 best 53 160 best 54 120 best 55 11 </code></pre> <p>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=</p> <pre><code>16 2 2 2 2 1 1 4 6 3 2 1 1 1 1 4 best 4 6 32 2 2 2 2 2 1 5 12 3 2 2 1 1 1 5 best 5 12 64 2 2 2 2 2 2 6 24 3 2 2 2 1 1 6 9 3 3 1 1 1 1 6 4 4 1 1 1 1 1 6 best 6 4 48 3 2 2 2 2 1 7 18 3 3 2 1 1 1 7 8 4 2 1 1 1 1 7 best 7 8 96 3 2 2 2 2 2 8 36 3 3 2 2 1 1 8 16 4 2 2 1 1 1 8 best 8 16 72 3 3 2 2 2 1 9 27 3 3 3 1 1 1 9 32 4 2 2 2 1 1 9 12 4 3 1 1 1 1 9 best 9 12 144 3 3 2 2 2 2 10 54 3 3 3 2 1 1 10 64 4 2 2 2 2 1 10 24 4 3 2 1 1 1 10 5 5 1 1 1 1 1 10 best 10 5 108 3 3 3 2 2 1 11 128 4 2 2 2 2 2 11 48 4 3 2 2 1 1 11 10 5 2 1 1 1 1 11 best 11 10 216 3 3 3 2 2 2 12 81 3 3 3 3 1 1 12 96 4 3 2 2 2 1 12 36 4 3 3 1 1 1 12 16 4 4 1 1 1 1 12 20 5 2 2 1 1 1 12 best 12 16 162 3 3 3 3 2 1 13 192 4 3 2 2 2 2 13 72 4 3 3 2 1 1 13 32 4 4 2 1 1 1 13 40 5 2 2 2 1 1 13 15 5 3 1 1 1 1 13 best 13 15 324 3 3 3 3 2 2 14 144 4 3 3 2 2 1 14 64 4 4 2 2 1 1 14 80 5 2 2 2 2 1 14 30 5 3 2 1 1 1 14 best 14 30 243 3 3 3 3 3 1 15 288 4 3 3 2 2 2 15 108 4 3 3 3 1 1 15 128 4 4 2 2 2 1 15 48 4 4 3 1 1 1 15 160 5 2 2 2 2 2 15 60 5 3 2 2 1 1 15 6 6 1 1 1 1 1 15 best 15 6 486 3 3 3 3 3 2 16 216 4 3 3 3 2 1 16 256 4 4 2 2 2 2 16 96 4 4 3 2 1 1 16 120 5 3 2 2 2 1 16 45 5 3 3 1 1 1 16 20 5 4 1 1 1 1 16 12 6 2 1 1 1 1 16 best 16 12 432 4 3 3 3 2 2 17 192 4 4 3 2 2 1 17 240 5 3 2 2 2 2 17 90 5 3 3 2 1 1 17 40 5 4 2 1 1 1 17 24 6 2 2 1 1 1 17 best 17 24 729 3 3 3 3 3 3 18 324 4 3 3 3 3 1 18 384 4 4 3 2 2 2 18 144 4 4 3 3 1 1 18 64 4 4 4 1 1 1 18 180 5 3 3 2 2 1 18 80 5 4 2 2 1 1 18 48 6 2 2 2 1 1 18 18 6 3 1 1 1 1 18 best 18 18 648 4 3 3 3 3 2 19 288 4 4 3 3 2 1 19 128 4 4 4 2 1 1 19 360 5 3 3 2 2 2 19 135 5 3 3 3 1 1 19 160 5 4 2 2 2 1 19 60 5 4 3 1 1 1 19 96 6 2 2 2 2 1 19 36 6 3 2 1 1 1 19 best 19 36 576 4 4 3 3 2 2 20 256 4 4 4 2 2 1 20 270 5 3 3 3 2 1 20 320 5 4 2 2 2 2 20 120 5 4 3 2 1 1 20 25 5 5 1 1 1 1 20 192 6 2 2 2 2 2 20 72 6 3 2 2 1 1 20 best 20 25 972 4 3 3 3 3 3 21 432 4 4 3 3 3 1 21 512 4 4 4 2 2 2 21 192 4 4 4 3 1 1 21 540 5 3 3 3 2 2 21 240 5 4 3 2 2 1 21 50 5 5 2 1 1 1 21 144 6 3 2 2 2 1 21 54 6 3 3 1 1 1 21 24 6 4 1 1 1 1 21 7 7 1 1 1 1 1 21 best 21 7 864 4 4 3 3 3 2 22 384 4 4 4 3 2 1 22 405 5 3 3 3 3 1 22 480 5 4 3 2 2 2 22 180 5 4 3 3 1 1 22 80 5 4 4 1 1 1 22 100 5 5 2 2 1 1 22 288 6 3 2 2 2 2 22 108 6 3 3 2 1 1 22 48 6 4 2 1 1 1 22 14 7 2 1 1 1 1 22 best 22 14 768 4 4 4 3 2 2 23 810 5 3 3 3 3 2 23 360 5 4 3 3 2 1 23 160 5 4 4 2 1 1 23 200 5 5 2 2 2 1 23 75 5 5 3 1 1 1 23 216 6 3 3 2 2 1 23 96 6 4 2 2 1 1 23 28 7 2 2 1 1 1 23 best 23 28 1296 4 4 3 3 3 3 24 576 4 4 4 3 3 1 24 256 4 4 4 4 1 1 24 720 5 4 3 3 2 2 24 320 5 4 4 2 2 1 24 400 5 5 2 2 2 2 24 150 5 5 3 2 1 1 24 432 6 3 3 2 2 2 24 162 6 3 3 3 1 1 24 192 6 4 2 2 2 1 24 72 6 4 3 1 1 1 24 56 7 2 2 2 1 1 24 21 7 3 1 1 1 1 24 best 24 21 1152 4 4 4 3 3 2 25 512 4 4 4 4 2 1 25 1215 5 3 3 3 3 3 25 540 5 4 3 3 3 1 25 640 5 4 4 2 2 2 25 240 5 4 4 3 1 1 25 300 5 5 3 2 2 1 25 324 6 3 3 3 2 1 25 384 6 4 2 2 2 2 25 144 6 4 3 2 1 1 25 30 6 5 1 1 1 1 25 112 7 2 2 2 2 1 25 42 7 3 2 1 1 1 25 best 25 30 1024 4 4 4 4 2 2 26 1080 5 4 3 3 3 2 26 480 5 4 4 3 2 1 26 600 5 5 3 2 2 2 26 225 5 5 3 3 1 1 26 100 5 5 4 1 1 1 26 648 6 3 3 3 2 2 26 288 6 4 3 2 2 1 26 60 6 5 2 1 1 1 26 224 7 2 2 2 2 2 26 84 7 3 2 2 1 1 26 best 26 60 1728 4 4 4 3 3 3 27 768 4 4 4 4 3 1 27 960 5 4 4 3 2 2 27 450 5 5 3 3 2 1 27 200 5 5 4 2 1 1 27 486 6 3 3 3 3 1 27 576 6 4 3 2 2 2 27 216 6 4 3 3 1 1 27 96 6 4 4 1 1 1 27 120 6 5 2 2 1 1 27 168 7 3 2 2 2 1 27 63 7 3 3 1 1 1 27 28 7 4 1 1 1 1 27 best 27 28 1536 4 4 4 4 3 2 28 1620 5 4 3 3 3 3 28 720 5 4 4 3 3 1 28 320 5 4 4 4 1 1 28 900 5 5 3 3 2 2 28 400 5 5 4 2 2 1 28 972 6 3 3 3 3 2 28 432 6 4 3 3 2 1 28 192 6 4 4 2 1 1 28 240 6 5 2 2 2 1 28 90 6 5 3 1 1 1 28 336 7 3 2 2 2 2 28 126 7 3 3 2 1 1 28 56 7 4 2 1 1 1 28 8 8 1 1 1 1 1 28 best 28 8 1440 5 4 4 3 3 2 29 640 5 4 4 4 2 1 29 675 5 5 3 3 3 1 29 800 5 5 4 2 2 2 29 300 5 5 4 3 1 1 29 864 6 4 3 3 2 2 29 384 6 4 4 2 2 1 29 480 6 5 2 2 2 2 29 180 6 5 3 2 1 1 29 252 7 3 3 2 2 1 29 112 7 4 2 2 1 1 29 16 8 2 1 1 1 1 29 best 29 16 2304 4 4 4 4 3 3 30 1024 4 4 4 4 4 1 30 1280 5 4 4 4 2 2 30 1350 5 5 3 3 3 2 30 600 5 5 4 3 2 1 30 125 5 5 5 1 1 1 30 1458 6 3 3 3 3 3 30 648 6 4 3 3 3 1 30 768 6 4 4 2 2 2 30 288 6 4 4 3 1 1 30 360 6 5 3 2 2 1 30 36 6 6 1 1 1 1 30 504 7 3 3 2 2 2 30 189 7 3 3 3 1 1 30 224 7 4 2 2 2 1 30 84 7 4 3 1 1 1 30 32 8 2 2 1 1 1 30 best 30 32 2048 4 4 4 4 4 2 31 2160 5 4 4 3 3 3 31 960 5 4 4 4 3 1 31 1200 5 5 4 3 2 2 31 250 5 5 5 2 1 1 31 1296 6 4 3 3 3 2 31 576 6 4 4 3 2 1 31 720 6 5 3 2 2 2 31 270 6 5 3 3 1 1 31 120 6 5 4 1 1 1 31 72 6 6 2 1 1 1 31 378 7 3 3 3 2 1 31 448 7 4 2 2 2 2 31 168 7 4 3 2 1 1 31 35 7 5 1 1 1 1 31 64 8 2 2 2 1 1 31 24 8 3 1 1 1 1 31 best 31 24 1920 5 4 4 4 3 2 32 2025 5 5 3 3 3 3 32 900 5 5 4 3 3 1 32 400 5 5 4 4 1 1 32 500 5 5 5 2 2 1 32 1152 6 4 4 3 2 2 32 540 6 5 3 3 2 1 32 240 6 5 4 2 1 1 32 144 6 6 2 2 1 1 32 756 7 3 3 3 2 2 32 336 7 4 3 2 2 1 32 70 7 5 2 1 1 1 32 128 8 2 2 2 2 1 32 48 8 3 2 1 1 1 32 best 32 48 3072 4 4 4 4 4 3 33 1800 5 5 4 3 3 2 33 800 5 5 4 4 2 1 33 1000 5 5 5 2 2 2 33 375 5 5 5 3 1 1 33 1944 6 4 3 3 3 3 33 864 6 4 4 3 3 1 33 384 6 4 4 4 1 1 33 1080 6 5 3 3 2 2 33 480 6 5 4 2 2 1 33 288 6 6 2 2 2 1 33 108 6 6 3 1 1 1 33 567 7 3 3 3 3 1 33 672 7 4 3 2 2 2 33 252 7 4 3 3 1 1 33 112 7 4 4 1 1 1 33 140 7 5 2 2 1 1 33 256 8 2 2 2 2 2 33 96 8 3 2 2 1 1 33 best 33 96 2880 5 4 4 4 3 3 34 1280 5 4 4 4 4 1 34 1600 5 5 4 4 2 2 34 750 5 5 5 3 2 1 34 1728 6 4 4 3 3 2 34 768 6 4 4 4 2 1 34 810 6 5 3 3 3 1 34 960 6 5 4 2 2 2 34 360 6 5 4 3 1 1 34 576 6 6 2 2 2 2 34 216 6 6 3 2 1 1 34 1134 7 3 3 3 3 2 34 504 7 4 3 3 2 1 34 224 7 4 4 2 1 1 34 280 7 5 2 2 2 1 34 105 7 5 3 1 1 1 34 192 8 3 2 2 2 1 34 72 8 3 3 1 1 1 34 32 8 4 1 1 1 1 34 best 34 32 2560 5 4 4 4 4 2 35 2700 5 5 4 3 3 3 35 1200 5 5 4 4 3 1 35 1500 5 5 5 3 2 2 35 1536 6 4 4 4 2 2 35 1620 6 5 3 3 3 2 35 720 6 5 4 3 2 1 35 150 6 5 5 1 1 1 35 432 6 6 3 2 2 1 35 1008 7 4 3 3 2 2 35 448 7 4 4 2 2 1 35 560 7 5 2 2 2 2 35 210 7 5 3 2 1 1 35 384 8 3 2 2 2 2 35 144 8 3 3 2 1 1 35 64 8 4 2 1 1 1 35 best 35 64 4096 4 4 4 4 4 4 36 2400 5 5 4 4 3 2 36 1125 5 5 5 3 3 1 36 500 5 5 5 4 1 1 36 2592 6 4 4 3 3 3 36 1152 6 4 4 4 3 1 36 1440 6 5 4 3 2 2 36 300 6 5 5 2 1 1 36 864 6 6 3 2 2 2 36 324 6 6 3 3 1 1 36 144 6 6 4 1 1 1 36 1701 7 3 3 3 3 3 36 756 7 4 3 3 3 1 36 896 7 4 4 2 2 2 36 336 7 4 4 3 1 1 36 420 7 5 3 2 2 1 36 42 7 6 1 1 1 1 36 288 8 3 3 2 2 1 36 128 8 4 2 2 1 1 36 9 9 1 1 1 1 1 36 best 36 9 3840 5 4 4 4 4 3 37 2250 5 5 5 3 3 2 37 1000 5 5 5 4 2 1 37 2304 6 4 4 4 3 2 37 2430 6 5 3 3 3 3 37 1080 6 5 4 3 3 1 37 480 6 5 4 4 1 1 37 600 6 5 5 2 2 1 37 648 6 6 3 3 2 1 37 288 6 6 4 2 1 1 37 1512 7 4 3 3 3 2 37 672 7 4 4 3 2 1 37 840 7 5 3 2 2 2 37 315 7 5 3 3 1 1 37 140 7 5 4 1 1 1 37 84 7 6 2 1 1 1 37 576 8 3 3 2 2 2 37 216 8 3 3 3 1 1 37 256 8 4 2 2 2 1 37 96 8 4 3 1 1 1 37 18 9 2 1 1 1 1 37 best 37 18 3600 5 5 4 4 3 3 38 1600 5 5 4 4 4 1 38 2000 5 5 5 4 2 2 38 2160 6 5 4 3 3 2 38 960 6 5 4 4 2 1 38 1200 6 5 5 2 2 2 38 450 6 5 5 3 1 1 38 1296 6 6 3 3 2 2 38 576 6 6 4 2 2 1 38 1344 7 4 4 3 2 2 38 630 7 5 3 3 2 1 38 280 7 5 4 2 1 1 38 168 7 6 2 2 1 1 38 432 8 3 3 3 2 1 38 512 8 4 2 2 2 2 38 192 8 4 3 2 1 1 38 40 8 5 1 1 1 1 38 36 9 2 2 1 1 1 38 best 38 36 3200 5 5 4 4 4 2 39 3375 5 5 5 3 3 3 39 1500 5 5 5 4 3 1 39 3456 6 4 4 4 3 3 39 1536 6 4 4 4 4 1 39 1920 6 5 4 4 2 2 39 900 6 5 5 3 2 1 39 972 6 6 3 3 3 1 39 1152 6 6 4 2 2 2 39 432 6 6 4 3 1 1 39 2268 7 4 3 3 3 3 39 1008 7 4 4 3 3 1 39 448 7 4 4 4 1 1 39 1260 7 5 3 3 2 2 39 560 7 5 4 2 2 1 39 336 7 6 2 2 2 1 39 126 7 6 3 1 1 1 39 864 8 3 3 3 2 2 39 384 8 4 3 2 2 1 39 80 8 5 2 1 1 1 39 72 9 2 2 2 1 1 39 27 9 3 1 1 1 1 39 best 39 27 5120 5 4 4 4 4 4 40 3000 5 5 5 4 3 2 40 625 5 5 5 5 1 1 40 3072 6 4 4 4 4 2 40 3240 6 5 4 3 3 3 40 1440 6 5 4 4 3 1 40 1800 6 5 5 3 2 2 40 1944 6 6 3 3 3 2 40 864 6 6 4 3 2 1 40 180 6 6 5 1 1 1 40 2016 7 4 4 3 3 2 40 896 7 4 4 4 2 1 40 945 7 5 3 3 3 1 40 1120 7 5 4 2 2 2 40 420 7 5 4 3 1 1 40 672 7 6 2 2 2 2 40 252 7 6 3 2 1 1 40 648 8 3 3 3 3 1 40 768 8 4 3 2 2 2 40 288 8 4 3 3 1 1 40 128 8 4 4 1 1 1 40 160 8 5 2 2 1 1 40 144 9 2 2 2 2 1 40 54 9 3 2 1 1 1 40 best 40 54 4800 5 5 4 4 4 3 41 1250 5 5 5 5 2 1 41 2880 6 5 4 4 3 2 41 1350 6 5 5 3 3 1 41 600 6 5 5 4 1 1 41 1728 6 6 4 3 2 2 41 360 6 6 5 2 1 1 41 1792 7 4 4 4 2 2 41 1890 7 5 3 3 3 2 41 840 7 5 4 3 2 1 41 175 7 5 5 1 1 1 41 504 7 6 3 2 2 1 41 1296 8 3 3 3 3 2 41 576 8 4 3 3 2 1 41 256 8 4 4 2 1 1 41 320 8 5 2 2 2 1 41 120 8 5 3 1 1 1 41 288 9 2 2 2 2 2 41 108 9 3 2 2 1 1 41 best 41 108 4500 5 5 5 4 3 3 42 2000 5 5 5 4 4 1 42 2500 5 5 5 5 2 2 42 4608 6 4 4 4 4 3 42 2700 6 5 5 3 3 2 42 1200 6 5 5 4 2 1 42 2916 6 6 3 3 3 3 </code></pre>

http://mathoverflow.net/questions/118903/elementary-applications-of-linear-algebra-over-finite-fields/118922#118922 Will Jagy 2013-01-14T20:56:01Z 2013-01-14T21:11:20Z

<p>This one is pretty good. Kaplansky wanted squarefree numbers $x$ such that $$ \sigma ( x^3) = y^2, $$ where $\sigma$ is the sum of divisors function. Somewhere around here I have a short note of his. He referred to this (and some very similar problems) as Ozanam's problem, from page 56 of Dickson's History, volume 1. Let's see; for each prime $p$ up to some bound, I had the computer factor $ \sigma ( p^3),$ especially recording the exponents on the output primes $q.$ So, to the best of my memory (about 18 years ago), I wound up with a big matrix with entries in the field with two elements; each column meant a prime $p,$ each row was saying whether the exponent for the prime $q$ was even or odd. Then, a solution was a column vector, also of 0's and 1's, which my big matrix mapped to the zero vector. So, I did Gauss elimination over the field of two elements. And found hundreds of solutions. </p> <p>I will see if I can find something written about this. For that matter, the program or programs I wrote should still be there in my MSRI account. </p> <p>Note: i am having a little trouble remembering if it is the matrix i describe above or its transpose. So perhaps a little care is needed. It definitely worked, though, and quickly. I also built in some procedure where i could force some prime to be included, then see if i could find solutions with that restriction. As I recall, that needed more handholding for the computer, more attention by me.</p>

http://mathoverflow.net/questions/118138/cohen-and-selfridges-proof-about-odd-numbers-which-are-neither-the-sum-nor-diffe/118685#118685 Will Jagy 2013-01-11T23:59:51Z 2013-01-12T01:53:29Z

<p>John,</p> <p>I attended the talk by Carl Pomerance today, and I asked your question about the original Erdös paper. He said this was well-known as a minor gap in the Erdös paper, and if I email him he will send me the easy fill.</p> <p>William C. Jagy</p> <p>San Diego, CA.</p> <p>With a cold.</p>

http://mathoverflow.net/questions/117212/octonions-and-the-dance-of-the-seven-veils Will Jagy 2012-12-25T22:12:45Z 2012-12-26T22:44:01Z

<p>Let us take the octonions as having all integer coefficients and the multiplication table at <a href="http://math.ucr.edu/home/baez/octonions/node3.html" rel="nofollow">BAEZ</a></p> <p>We have a standard conjugation operator with $\bar{1} = 1$ and $\bar{e_i}= - e_i,$ extend by linearity. Then we get a norm, or squared Euclidean length, as $Nx = x \bar{x} = \bar{x} x,$ and the $N(xy) = Nx \; Ny.$ We get the real part $\mathcal Rx = \frac{1}{2} (x + \bar{x}). $ We will call an octonion $o$ pure if its real part is $0,$ that is $\bar{o} = -o.$ We also retrieve the ordinary dot product with $(x,y)= \mathcal R(x\bar{y}).$ Among a million other facts, multiplication on either the left or the right by a fixed octonion preserves angles, and multiplication by any of the $e_i$ rotates everything $90^\circ.$</p> <p>The main law we need is the "alternative algebra" law, $$ (xy)x = x(yx). $$ I like the quick summary of the laws in Robert Wilson 2008 <a href="http://www.maths.qmul.ac.uk/~raw/talks_files/octonions.pdf" rel="nofollow">SEMINAR PDF</a> </p> <p>LEMMA: $$ (\bar{t} x) t = \bar{t} (x t) $$</p> <p>Proof: Write $t = r + o $ with $r$ real and $o$ pure. The $\bar{t} = r - o.$ Calculation gives $$ (\bar{t} x) t = r^2 x + r x o - r o x - (ox)o, $$ while $$ \bar{t} (x t) = r^2 x + r x o - r o x - o(xo). $$ Since $ (ox)o = o (xo) $ the lemma follows.</p> <p>To the question of interest, which was done for the quaternions as the matrix in formula (10) on page 755 of Pall Automorphs (1940) and Theorem 3 on page 176 of Jones and Pall (1939), both available as pdfs at <a href="http://zakuski.math.utsa.edu/~kap/forms.html" rel="nofollow">TERNARY</a> </p> <p>Let $$ o = o_1 e_1 + o_2 e_2 + o_3 e_3 + o_4 e_4 + o_5 e_5 + o_6 e_6 + o_7 e_7 $$ be a "proper" pure quaternion, that is $\gcd(o_1,o_2,o_3,o_4,o_5,o_6,o_7) = 1,$ with square norm $$ No = m^2 = o_1^2 + o_2^2 + o_3^2 + o_4^2 + o_5^2 + o_6^2 + o_7^2 $$ for integer $m.$ </p> <p>QUESTION: is it necessarily the case that we can find some integral octonion $t,$ with $Nt = m,$ such that $$ o = \pm \bar{t} (e_j t) $$ for some $j =1,2,3,4,5,6,7?$</p> <p>EDDITTT: I got an email from a kind fellow who has a Maple package for playing with octonions, who says that with $o = e_4 + e_5 + e_6 + e_7$ of norm 4 has no expression as $ o = \bar{t} (e_1 t). $ I have asked him if checking all possible $\pm e_j$ makes any difference. It is a big help to be able to experiment, I don't have any computer programs myself this time. I will put in his name if he says that is alright. Oh, his Maple program may be using a slightly different multiplication table from the one I picked, i think there are 240 or 480 possible versions. </p> <p>Note that the proof by Jones and Pall uses associativity and, essentially, unique factorization (Theorem 2) for a proof by induction on the number of prime factors in out $m.$ That proof cannot work in quite such a pleasant manner. That is, their Theorem 2 is likely false, but perhaps this (Theorem 3) can be recovered as a very special case. Note that I can live with $m$ being odd, Jones and Pall needed to do that. For seven squares I do not think it should be necessary.</p> <p>Oh, automorphs. It is fairly likely that every rational automorph of the sum of seven squares can be written with the seven rows being $\bar{t} (e_i t),$ and then divide by $m.$ I'm just saying.</p>

http://mathoverflow.net/questions/117212/octonions-and-the-dance-of-the-seven-veils/117275#117275 Will Jagy 2012-12-26T20:28:12Z 2012-12-26T20:28:12Z

<p>Sometimes nobody knows a question because it is just nonsense.</p> <p>For <strong><em>distinct</em></strong> indices $i,j,k = 1,2,3,4,5,6,7,$ I have proved this little item by checking 35 triples in the Baez multiplication table: $$ (e_i e_j) e_k = (e_j e_k) e_i = (e_k e_i) e_j. $$ As a result, $$ (e_i e_j) e_k + (e_k e_j) e_i = 0. $$ I was pretty sure I had proved these with some calculations that forced the second version, but I also wanted to check, and it's true. </p> <p>W. Edwin Clark, emeritus at the University of South Florida, told me the example i edited into the question. A little more is true: if $Nt = 2,$ then the only possibilities I found for $\bar{t} e_i t$ are $\pm 2, \pm 2 e_j.$ </p> <p>Next I went ahead and checked $Nt = 3.$ This is also a complete bust. The possibilities for $\bar{t} e_i t$ are $\pm 3, \pm 3 e_j, \; \pm e_j \pm 2 e_k \pm 2 e_m.$ There is simply not enough variety to create anything with $9 = 4 + 1 + 1 + 1 + 1 + 1 + 0.$</p> <p>In conclusion, this was stupid, but now I know a bit more than I did.</p>

http://mathoverflow.net/questions/104889/what-numbers-are-integrally-represented-by-4-x2-2-x-y-7-y2-z3 Will Jagy 2012-08-17T03:31:48Z 2012-12-04T23:16:21Z

<p>This is related to my first MO question and Kevin Buzzard's conjecture at <a href="http://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-z" rel="nofollow">http://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-z</a></p> <p>In December 2010 my question appeared in the M.A.A. Monthly, show that $4 x^2 + 2 x y + 7 y^2 - z^3 \neq \pm 2 m^3, \; \pm 32 m^3$ when $m$ has certain prime factorizations. The answers were due by April 2011 so I feel willing to mention it, although the answers have not appeared yet, I just got the August-September issue. Sigh.</p> <p>A couple of days ago I thought I might check for identities, and found several good ones, showing that all odd numbers are represented for example. I believe there is no chance of completing this problem by identities owing to the non-represented numbers. So, that is the <strong>question</strong>, can anyone prove that $4 x^2 + 2 x y + 7 y^2 - z^3$ integrally represents everything else? </p> <p>For verisimilitude, we have:</p> <p>$$ \begin{array}{cc} x = 4 n^3 - 18 n^2 + 3 n - 21, &amp; y = -16 n^3 - 18 n + 1, \\ z = 12 n^2 + 12, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n+1.<br> \end{array} $$</p> <p>$$ \begin{array}{cc} x = 4 n^3 - 42 n^2 - 73 n - 359, &amp; y = -16 n^3 - 48 n^2 - 146n - 111, \\ z = 12 n^2 + 24n+ 88, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n-3.<br> \end{array} $$ </p> <p>$$ \begin{array}{cc} x = 4 n^3 + 42 n^2 - 65 n + 417, &amp; y = -16 n^3 + 48 n^2 - 166n + 137, \\ z = 12 n^2 - 24n+ 98, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n+5.<br> \end{array} $$ </p> <p>$$ \begin{array}{cc} x = 16 n^3 - 12 n^2 + 23 n + 6, &amp; y = 8 n^3 - 24 n^2 + 28n - 27, \\ z = 12 n^2 - 12 n+ 17, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n+10.<br> \end{array} $$</p> <p>$$ \begin{array}{cc} x = 16 n^3 - 12 n^2 + 3 n + 1, &amp; y = 8 n^3 - 24 n^2 + 18n - 7, \\ z = 12 n^2 - 12 n+ 7, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n-10.<br> \end{array} $$</p> <p>$$ \begin{array}{cc} x = 72 n^3 + 60 n^2 + 13 n, &amp; y = -72 n^3 - 24 n^2 + 2 n + 1, \\ z = 36 n^2 + 12 n+ 1, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n + 6.<br> \end{array} $$</p> <p>$$ \begin{array}{cc} x = 4 n^3 + 36 n^2 + 18 n + 135, &amp; y = -16 n^3 - 60 n + 4, \\ z = 12 n^2 + 42, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 24n + 4.<br> \end{array} $$</p> <p>$$ \begin{array}{cc} x = 9 n^3 - 30 n^2 + 29 n - 16, &amp; y = -9 n^3 + 12 n^2 - 8 n + 2, \\ z = 9 n^2 -12 n + 10, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 36n - 12.<br> \end{array} $$</p> <p>$$ \begin{array}{cc} x = 16 n^3 - 12 n^2 + 33 n + 7, &amp; y = 8 n^3 - 24 n^2 + 30 n - 37, \\ z = 12 n^2 -12 n + 21, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 162 n.<br> \end{array} $$</p> <p>Furthermore, if we have a prime $q = 4 u^2 + 2 u v + 7 v^2,$ the fact that $h(-108) = 3$ and $2^2 + 27 \cdot 1^2 = 31$ shows that $ 4 x^2 + 2 x y + 7 y^2$ represents $q^3, \; 31 q^3, \; 25 q^3.$ As a result $q = 4 u^2 + 2 u v + 7 v^2 - z^3$ represents $2 q^3 = q^3 + q^3, \; 32 q^3 = 31 q^3 + q^3, \; -2 q^3 = 25 q^3 - 27 q^3. $ I'm not sure how to do $-32 q^3.$ </p> <p>P.S. Not that it really increases the difficulty, but representing $\pm 2 q^3, \pm 32 q^3$ is not actually enough... if we can represent some $n,$ for any $k$ we know we can also represent $n k^6,$ but not necessarily $n k^3.$ I'm just saying. </p> <p>P.P.S. Komputer Kalkulation:</p> <pre><code> Targets between -1,000,000 and 1,000,000 that appear to have no integer expression as 4 x^2 + 2 x y + 7 y^2 + z^3 : -953312 = -1 * 2^5 * 31^3 -780448 = -1 * 2^5 * 29^3 -715822 = -1 * 2 * 71^3 -500000 = -1 * 2^5 * 5^6 -410758 = -1 * 2 * 59^3 -389344 = -1 * 2^5 * 23^3 -332750 = -1 * 2 * 5^3 * 11^3 -297754 = -1 * 2 * 53^3 -207646 = -1 * 2 * 47^3 -159014 = -1 * 2 * 43^3 -157216 = -1 * 2^5 * 17^3 -137842 = -1 * 2 * 41^3 -59582 = -1 * 2 * 31^3 -48778 = -1 * 2 * 29^3 -42592 = -1 * 2^5 * 11^3 -31250 = -1 * 2 * 5^6 -24334 = -1 * 2 * 23^3 -9826 = -1 * 2 * 17^3 -4000 = -1 * 2^5 * 5^3 -2662 = -1 * 2 * 11^3 -250 = -1 * 2 * 5^3 -32 = -1 * 2^5 -2 = -1 * 2 2 = 2 32 = 2^5 250 = 2 * 5^3 2662 = 2 * 11^3 4000 = 2^5 * 5^3 9826 = 2 * 17^3 24334 = 2 * 23^3 31250 = 2 * 5^6 42592 = 2^5 * 11^3 48778 = 2 * 29^3 59582 = 2 * 31^3 137842 = 2 * 41^3 157216 = 2^5 * 17^3 159014 = 2 * 43^3 207646 = 2 * 47^3 297754 = 2 * 53^3 332750 = 2 * 5^3 * 11^3 389344 = 2^5 * 23^3 410758 = 2 * 59^3 500000 = 2^5 * 5^6 715822 = 2 * 71^3 780448 = 2^5 * 29^3 953312 = 2^5 * 31^3 phoebus:~/Cplusplus&gt; </code></pre> <p>Monday, August 20: A student of Kevin Buzzard, in what would be a Master's thesis in the U.S., proved that for any integers $A,B,$ both the inhomogeneous polynomials $$ x^2 + x y + 6 y^2 + z^3 + A z^2 + B z $$ and $$ x^2 + x y + 8 y^2 + z^3 + A z^2 + B z $$ are universal, they integrally represent all integers. He also did a fixed one, $$ 2x^2 + x y + 2 y^2 + z^3 + z. $$ So the hard case really is these non-universal ones. </p>

http://mathoverflow.net/questions/115193/integers-represented-by-the-polynomial-a2b3c6/115205#115205 Will Jagy 2012-12-02T21:29:44Z 2012-12-03T02:47:37Z

<p>This is question 3 on page 146 of the second edition of <em>The Hardy-Littlewood Method</em> by R. C. Vaughan, with $b \geq 0.$ <br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <img src="http://cs.smith.edu/~orourke/MathOverflow/Vaughan.jpg" alt="Vaughan" /> <br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;_{(Image added by J.O'Rourke)}<br /></p> <p>The answer is <em>No</em>, by a simple volume argument. It is not even necessary to know the exact constant, just that the number of lattice points with $x,y,z \geq 0 $ and $$ x^2 + y^3 + z^6 \leq N$$ is less than $CN,$ with a constant $0 &lt; C &lt; 1.$</p> <p>I'm in question 5 on the same page. </p> <p>Image hosting did not work. Please see: <a href="http://zakuski.utsa.edu/~jagy/Vaughan.pdf" rel="nofollow">VAUGHAN PDF</a> </p> <p>Oh, well. The relevant calculation is the sum of the reciprocals of the exponents, in case the polynomial is the sum of distinct monomials in the different variables. You might think that $$ x^2 + y^2 + z^9$$ ought to represent all large numbers with $z \geq 0.$ There are no local obstructions. But this is not the case. We can think of the exponent in this volume calcultaion as $10/9.$</p> <p>It is reasonable to ask, how large an exponent reciprocal sum can we get and still fail to represent large integers? The best I have is $4/3,$ and in the simplest form we require coefficients, as in $$ x^2 + 27 y^2 + 7 z^3. $$ This is a version of the $ 4 x^2 + 2 x y + 7 y^2 + z^3, $ which is more natural but looks less diagonal.</p>

http://mathoverflow.net/questions/114995/the-mean-curvature-of-a-hypersurface/115023#115023 Will Jagy 2012-11-30T20:45:53Z 2012-11-30T20:59:32Z

<p>I was unable to post an image without a big song and dance. If you go to <a href="http://zakuski.utsa.edu/~jagy/bib.html" rel="nofollow">MY STUFF</a> and open the Michigan Math J. 1991, page 256 has the formula when the hypersurface is given as the level set of a smooth function. It is just a rotated version of the graph formula, as it must be. Meanwhile, the mean curvature is just a dimension constant times the divergence of the (oriented) unit normal, where it does not matter whther you take the divergence on the manifold or extend the unit normal field off the manifold and use the ambient divergence. All the same. I do not see how you are going to separate out first and second order derivatives. </p> <p>Wait, I can try to typeset:</p> <p>$$ n H = \frac{1}{|\nabla F|} \; \sum_{i=1}^{n+1} \sum_{j=1}^{n+1} \left( \delta_{ij} - \frac{F_i F_j}{|\nabla F|^2} \right) F_{ij} $$ That actually looks correct. Good for me. Level set of the function $F$ and $H$ is the mean curvature with one of the choices of unit normal field.</p>

http://mathoverflow.net/questions/114707/the-diophantine-equation-x2-y2-z2-1/114712#114712 Will Jagy 2012-11-27T22:38:51Z 2012-11-27T22:49:24Z

<p>It is also true that the automorphism group of the quadratic form is known. See <a href="http://mathoverflow.net/questions/110956/is-there-a-topograph-for-pythagorean-triples" rel="nofollow">http://mathoverflow.net/questions/110956/is-there-a-topograph-for-pythagorean-triples</a> and the three matrices. If you have any particular $x^2 + y^2 - z^2 = n,$ write $(x,y,z)$ as a column vector. Multiply by any of the three square matrices or its inverse and you get another solution for $n.$ Multiply again you get another, and so on for any combination of group elements. </p> <p>I see, for your ordering you need to switch first and last elements to use these three matrices.</p>

http://mathoverflow.net/questions/114018/fastest-way-to-factor-integers-260/114097#114097 Will Jagy 2012-11-21T20:30:18Z 2012-11-21T20:30:18Z

<p>After your edit: another ad for <a href="http://en.wikipedia.org/wiki/Standard_Template_Library" rel="nofollow">C++ STL</a>. The "set" template type is implemented as a <a href="http://en.wikipedia.org/wiki/Red%25E2%2580%2593black_tree" rel="nofollow">red/black tree</a> which sorts at each insertion step. It is also dynamic memory allocation. Meanwhile, the user creates a "class," each instance will be a node in the tree, and the user defines the two relations == for equality and &lt; for strictly less than. </p> <p>As I said in a comment, each node could have some ID number for the curve, the number you are factoring, perhaps a complete factorization (it appears you wish to continue with a curve only if you have a complete factorization), and the set of divisor differences, perhaps implemented as another "set" if you like. </p> <p>This is largely opposite to difficult factorizations: the number of divisors $d(n)$ of some positive integer $n$ satisfies $$ d(n) \leq n^{\left( \frac{\log 2}{\log \log n} \right) \left( 1.5379396\ldots \right)} $$ or roughly $ n^{0.28596} $ up to $2^{60}.$ So a superior highly composite number of that size could have about 146214 divisors. Oh, dear. Well, those are rare and instantly factored. </p>

http://mathoverflow.net/questions/114018/fastest-way-to-factor-integers-260/114028#114028 Will Jagy 2012-11-21T06:42:21Z 2012-11-21T06:42:21Z

<p>In case of relevance: I am assuming that you are sifting through your 280,000 curves and throwing out those that are not suitable. The question is, do you need a complete factorization of your numbers to rule out a curve? For example, if I want to discard any number that is not the sum of two squares, I can very rapidly get rid of a high percentage by trial division by primes up to 1000. That is, if there is a prime $q &lt; 1000$ with $q \equiv 3 \mod 4$ and my test subject number $n \equiv 0 \pmod q$ but $n \neq 0 \pmod {q^2},$ then i know that $n$ is not the sum of two squares. And, for such numbers, I have not needed to perform a full factorization. In fact, any occurrence of an odd power of such $q$ will decide the question.</p> <p>Well, that has generally been how I have gotten things done, trial division up to a small bound for initial screening, up to a larger bound for those that remain, finally appeal to elliptic curve for the stubborn ones. </p> <p>Let's see, $2^{60} \approx 10^{18}.$ That is really not so bad. First screening by primes up to $1024 = 2^{10}.$ Second screening by primes up to $1,048,576 = 2^{20}.$ If any number is not completely factored by now, it is the product of two primes only. </p>

http://mathoverflow.net/questions/111984/gram-matrix-modulo-4/112022#112022 Will Jagy 2012-11-10T21:02:41Z 2012-11-10T21:02:41Z

<p>I use the 2-adic decompositions for various tasks. I can't say I know what would be useful for you, but let me call your attention to page 141, Lemma 4.3. This refers back to Lemma 5.2 on page 123, (paraphrase) if $d(g) = u^2 d(f)$ for some unit $u$ and $$ g_{ij} \equiv f_{ij} \pmod {2^{ v_2(d(f)) + 2}}, $$ then $f,g$ are $\mathbb Z_2$-equivalent. Here $v_2$ is the 2-adic valuation, the highest power of 2 dividing $d(f)$ in this case. </p> <p>Meanwhile, I can point you to A Canonical quadratic form for the ring of 2-adic integers, Burton W. Jones, Duke Math. J. 11 (1944), 715-727. This and two others are at the top of page 120 in Cassels, look for ``parliamentary draftsman.'' Note that Watson (1976), Mathematika 23 94-106, The 2-adic density of a quadratic form, is the source for the 2-adic terms in Conway and Sloane on the Mass Formula. In turn Watson's own 1960 book is the main source for the 1976 article.</p>

http://mathoverflow.net/questions/111489/the-quadratic-form-x2ny2-via-prime-factors/111518#111518 Will Jagy 2012-11-05T03:53:41Z 2012-11-05T03:53:41Z

<p>Why not. My answer at <a href="http://math.stackexchange.com/questions/229201/the-quadratic-form-x2-ny2-via-prime-factors/229270#229270" rel="nofollow">http://math.stackexchange.com/questions/229201/the-quadratic-form-x2-ny2-via-prime-factors/229270#229270</a> </p>

http://mathoverflow.net/questions/110239/is-there-an-algorithm-for-writing-a-number-as-a-sum-of-three-squares/110266#110266 Will Jagy 2012-10-21T20:10:24Z 2012-10-21T20:10:24Z

<p>One point that I do not see in the answer to which Igor links is size. Your target number is some $k \equiv 3 \pmod 8.$ So we take some odd $z$ and find out whether $k - z^2$ is the sum of two squares by factoring. My advice is to take $z$ as large as possible to begin, the decrease $z$ by 2 at each failure. There are two reasons for this. </p> <p>First, the numbers $j \equiv 2 \pmod 8$ that actually are the sum of two squares are more frequent the smaller the approximate size of $j.$ Combining all congruence classes $\pmod 8,$ the number of integers up to some real positive $x$ is about $$ \frac{0.7642 \; x}{\sqrt{\log x}}, $$ so they get less frequent near $x$ as $x$ gets bigger.</p> <p>Second, deciding whether $k-z^2$ is the sum of two squares is just factoring, and factoring is quicker for smaller numbers: powers of $2$ are irrelevant, any positive integer $j$ is the sum of two squares if and only if, when factoring $j,$ the exponent of any prime divisor $q \equiv 3 \pmod 4$ is even. Indeed, if there are any such, what you actually do is divide out all the appropriate $q^{2a}$ to arrive at a smaller number $j_0,$ write that as $x_0^2 + y_0^2 = j_0$ by solving that for each remaining prime power $p^w$ with $p \equiv 1 \pmod 4,$ which involves finding a square root of $-1 \pmod p$ and then screwing around. Combining pieces comes from $$ (a^2 + b^2)(c^2 + d^2) = (ad-bc)^2 + (ac + bd)^2. $$ Oh, when yopu are done with $x_0^2 + y_0^2 = j_0,$ you put back each $q \equiv 3 \pmod 4$ with $(q^a x_0)^2 + (q^a y_0)^2 = q^{2a} j_0.$</p> <p>Well, there is more to it, as you can see. But start with large $z.$ Size Matters.</p>

http://mathoverflow.net/questions/109164/synthetic-approach-to-hyperbolic-geometry/109184#109184 Will Jagy 2012-10-08T21:14:33Z 2012-10-08T21:14:33Z

<p>Download the article at <a href="http://mathdl.maa.org/mathDL/22/?pa=content&amp;sa=viewDocument&amp;nodeId=3729&amp;pf=1" rel="nofollow">MARVIN</a> and look at the references.</p>

http://mathoverflow.net/questions/107479/solving-znaib-using-only-radicals-of-positive-real-numbers/107486#107486 Will Jagy 2012-09-18T16:49:56Z 2012-09-18T16:49:56Z

<p>As soon as you get to general $$ z^3 = a + b i $$ this may be impossible. In solving a cubic $z^3 + p z + q = 0$ with, say, rational coefficients, one may use Cardano's formula. If there is only one real (irrational) root and two complex conjugate roots, then this works in the sense of being able to separately calculate real and imaginary parts using real square and cube roots. However, if there are three real irrational roots, a situation called <a href="http://en.wikipedia.org/wiki/Casus_irreducibilis" rel="nofollow">CASUS IRREDUCIBILIS</a>, then Cardano's formula is just a sum of cube roots of complex numbers $a+bi,$ with no way to separate real and imaginary parts. The terms are summed in a way that guarantees real answers, but that is not satisfying.</p> <p>Now, if you begin with general $ z^3 = a + b i, $ say with $a,b \in \mathbb Q,$ and carefully write out equations for the real and imaginary parts of this $z,$ you get cubics with three real roots, where you cannot separate parts for those...it is all pretty circular, and hopeless. </p>

http://mathoverflow.net/questions/107210/what-fraction-of-n-x-n-invertible-integer-matrices-contain-at-least-one-unit/107225#107225 Will Jagy 2012-09-15T02:16:58Z 2012-09-15T04:19:32Z

<p>I did the 2 by 2 case for $m$ up to $100.$ For very small entry bound $m,$ a $\pm 1$ is always required in order to get determinant $\pm 1.$ I think the limit of $r_2(m)$ is $0$ as $m$ goes to $\infty.$ </p> <p>Edit, 9:17 pm. I did 3 by, i killed it after it finished $m=8.$</p> <pre><code>jagy@phobeusjunior:~$ ./units_3 m H_3(m) G_3(m) r_3(m) 1 6960 6960 1 2 135408 135408 1 3 1279344 1281648 0.9982023145200555 4 5094192 5194416 0.9807054344511491 5 19593840 20852976 0.939618402668281 6 43474800 47054640 0.9239216366334967 7 113376432 131283120 0.8636025103608141 8 214735152 256950192 0.8357073031492422 ^C jagy@phobeusjunior:~$ jagy@phobeusjunior:~$ </code></pre> <p>I think for $r_3(m)$ you also get limit $0.$ And so on. It should not be difficult switching to $3$ by $3,$ it will just execute even more slowly. 7:16 pm.</p> <pre><code>jagy@phobeusjunior:~$ ./units m H_2(m) G_2(m) r_2(m) 1 40 40 1 2 104 104 1 3 232 232 1 4 328 360 0.9111111111111111 5 520 616 0.8441558441558441 6 616 744 0.8279569892473119 7 840 1128 0.7446808510638298 8 968 1384 0.6994219653179191 9 1192 1768 0.6742081447963801 10 1320 2024 0.6521739130434783 11 1608 2664 0.6036036036036037 12 1704 2920 0.5835616438356165 13 1992 3688 0.5401301518438177 14 2152 4072 0.5284872298624754 15 2408 4584 0.525305410122164 16 2568 5096 0.5039246467817896 17 2888 6120 0.4718954248366013 18 2984 6504 0.4587945879458795 19 3336 7656 0.4357366771159875 20 3496 8168 0.4280117531831538 21 3784 8936 0.423455684870188 22 3944 9576 0.4118629908103592 23 4296 10984 0.3911143481427531 24 4424 11496 0.3848295059151009 25 4776 12776 0.3738259236067627 26 4968 13544 0.3668044890726521 27 5256 14696 0.3576483396842678 28 5416 15464 0.3502327987584066 29 5832 17256 0.3379694019471488 30 5928 17768 0.3336334984241333 31 6344 19688 0.3222267370987403 32 6504 20712 0.3140208574739282 33 6792 21992 0.308839578028374 34 7016 23016 0.3048314216197428 35 7400 24552 0.3014011078527207 36 7560 25320 0.2985781990521327 37 7944 27624 0.2875760208514335 38 8104 28776 0.2816235752015568 39 8456 30312 0.2789654262338348 40 8616 31336 0.2749553229512382 41 9096 33896 0.2683502478168516 42 9192 34664 0.2651742441726286 43 9608 37352 0.2572285285928465 44 9832 38632 0.2545040381031269 45 10120 40168 0.2519418442541326 46 10344 41576 0.2487973831056379 47 10760 44520 0.2416891284815813 48 10888 45544 0.2390655190584929 49 11368 48232 0.2356941449659977 50 11560 49512 0.2334787526256261 51 11912 51560 0.2310318076027928 52 12072 53096 0.2273617598312491 53 12488 56424 0.2213242591804906 54 12648 57576 0.2196748645268862 55 13128 60136 0.2183051749368099 56 13352 61672 0.2165001945777663 57 13704 63976 0.2142053269976241 58 13864 65768 0.2108016056440822 59 14344 69480 0.206447898675878 60 14440 70504 0.2048110745489618 61 14920 74344 0.2006886904121382 62 15144 76264 0.1985733766914927 63 15464 78568 0.1968231341003971 64 15752 80616 0.1953954549965267 65 16200 83688 0.1935761399483797 66 16360 84968 0.1925430750400151 67 16776 89192 0.1880886178132568 68 16936 91240 0.1856203419552828 69 17352 94056 0.1844858382240367 70 17512 95592 0.1831952464641393 71 18120 100072 0.1810696298664961 72 18216 101608 0.1792772222659633 73 18696 106216 0.1760186789184304 74 18920 108520 0.1743457427202359 75 19208 111080 0.1729204177169608 76 19496 113384 0.1719466591406195 77 19912 117224 0.1698628267249027 78 20072 118760 0.169013135735938 79 20616 123752 0.1665912470101493 80 20808 125800 0.1654054054054054 81 21224 129256 0.1642012749891688 82 21416 131816 0.1624688960368999 83 21896 137064 0.1597501896924065 84 22056 138600 0.1591341991341991 85 22536 142696 0.1579301452037899 86 22760 145384 0.1565509271996919 87 23112 148968 0.1551474142097632 88 23272 151528 0.1535821762314556 89 23880 157160 0.1519470603206923 90 24040 158696 0.1514845994858093 91 24584 163304 0.1505413217067555 92 24808 166120 0.1493378280760896 93 25096 169960 0.1476582725347141 94 25320 172904 0.1464396428075695 95 25800 177512 0.1453422867186444 96 25960 179560 0.144575629316106 97 26504 185704 0.1427217507431181 98 26728 188392 0.1418743895706824 99 27176 192232 0.1413708435640268 100 27400 194792 0.1406628608977781 jagy@phobeusjunior:~$ </code></pre>

http://mathoverflow.net/questions/106217/last-term-of-repeating-continued-fraction-expansion/106220#106220 Will Jagy 2012-09-03T05:44:37Z 2012-09-03T06:00:04Z

<p>EDIT: I got your fact right here.</p> <p>ORIGINAL: It seems Henry got it. Meanwhile, let me point out how things appear from the Lagrange viewpoint of right-adjacent reduced forms: given odd numbers $n$ and $1 \leq m \leq n,$ the cycle for the form $\langle -1, n, m \rangle $ has penultimate form $\langle m, n, -1 \rangle, $ then "digit" $\delta = -n,$ then the end of the cycle is again $\langle -1, n, m \rangle .$ Well, see the method in my answer at <a href="http://mathoverflow.net/questions/22811/upper-bound-of-period-length-of-continued-fraction-representation-of-very-composi" rel="nofollow">http://mathoverflow.net/questions/22811/upper-bound-of-period-length-of-continued-fraction-representation-of-very-composi</a> where the fact you need, the final $\delta = -n,$ follows from the definition of the $\delta$'s. </p> <p>Examples: </p> <pre><code>========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 -1 5 1 0 form -1 5 1 1 0 0 1 To Return 1 0 0 1 0 form -1 5 1 delta 5 1 form 1 5 -1 delta -5 2 form -1 5 1 minimum was 1rep 1 0 disc 29 dSqrt 5.3851648071 M_Ratio 29 Automorph, written on right of Gram matrix: -1 5 5 -26 Trace: -27 gcd(a21, a22 - a11, a12) : 5 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 -1 5 3 0 form -1 5 3 1 0 0 1 To Return 1 0 0 1 0 form -1 5 3 delta 1 1 form 3 1 -3 delta -1 2 form -3 5 1 delta 5 3 form 1 5 -3 delta -1 4 form -3 1 3 delta 1 5 form 3 5 -1 delta -5 6 form -1 5 3 minimum was 1rep 1 0 disc 37 dSqrt 6.0827625303 M_Ratio 37 Automorph, written on right of Gram matrix: -13 72 24 -133 Trace: -146 gcd(a21, a22 - a11, a12) : 24 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 -1 5 5 0 form -1 5 5 1 0 0 1 To Return 1 0 0 1 0 form -1 5 5 delta 1 1 form 5 5 -1 delta -5 2 form -1 5 5 minimum was 1rep 1 0 disc 45 dSqrt 6.7082039325 M_Ratio 45 Automorph, written on right of Gram matrix: -1 5 1 -6 Trace: -7 gcd(a21, a22 - a11, a12) : 1 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 -1 7 1 0 form -1 7 1 1 0 0 1 To Return 1 0 0 1 0 form -1 7 1 delta 7 1 form 1 7 -1 delta -7 2 form -1 7 1 minimum was 1rep 1 0 disc 53 dSqrt 7.2801098893 M_Ratio 53 Automorph, written on right of Gram matrix: -1 7 7 -50 Trace: -51 gcd(a21, a22 - a11, a12) : 7 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 -1 7 3 0 form -1 7 3 1 0 0 1 To Return 1 0 0 1 0 form -1 7 3 delta 2 1 form 3 5 -3 delta -2 2 form -3 7 1 delta 7 3 form 1 7 -3 delta -2 4 form -3 5 3 delta 2 5 form 3 7 -1 delta -7 6 form -1 7 3 minimum was 1rep 1 0 disc 61 dSqrt 7.8102496759 M_Ratio 61 Automorph, written on right of Gram matrix: -79 585 195 -1444 Trace: -1523 gcd(a21, a22 - a11, a12) : 195 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 -1 7 5 0 form -1 7 5 1 0 0 1 To Return 1 0 0 1 0 form -1 7 5 delta 1 1 form 5 3 -3 delta -1 2 form -3 3 5 delta 1 3 form 5 7 -1 delta -7 4 form -1 7 5 minimum was 1rep 1 0 disc 69 dSqrt 8.3066238629 M_Ratio 69 Automorph, written on right of Gram matrix: 2 -15 -3 23 Trace: 25 gcd(a21, a22 - a11, a12) : 3 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 -1 7 7 0 form -1 7 7 1 0 0 1 To Return 1 0 0 1 0 form -1 7 7 delta 1 1 form 7 7 -1 delta -7 2 form -1 7 7 minimum was 1rep 1 0 disc 77 dSqrt 8.7749643874 M_Ratio 77 Automorph, written on right of Gram matrix: -1 7 1 -8 Trace: -9 gcd(a21, a22 - a11, a12) : 1 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ </code></pre>

http://mathoverflow.net/questions/105644/cancellation-theorem-for-lattices/105649#105649 Will Jagy 2012-08-27T18:27:03Z 2012-08-27T18:44:56Z

<p>Right. If $L = U$ is the lattice of the quadratic form $u(x,y) = 2 xy,$ and $M,N$ are positive definite, the conclusion is that $M,N$ are in the same genus. That is, they are rationally equivalent "without essential denominator." There is no complete proof printed in one place. I first saw this on page 378 of SPLAG by Conway and Sloane, first edition. The observation may be due to Conway. This is a small part of finding certain automorphism groups, and is first apparent in the articles on the automorphism group of the Leech Lattice. Anyway, click on my name and just go through my question with promising titles. In a minute I will find the one with a sketch of a proof, put a link here. </p> <p>Found it, <a href="http://mathoverflow.net/questions/70666/lorentzian-characterization-of-genus" rel="nofollow">http://mathoverflow.net/questions/70666/lorentzian-characterization-of-genus</a> </p> <p>I also checked with Wai Kiu Chan about the case of "odd" lattices such as the sum of squares, it turns out it does not matter, same outcome. </p> <p>Meanwhile, it is exactly this observation that allows one to conclude, given a positive "even" lattice with covering radius strictly below $\sqrt 2,$ such as $\mathbb E_8,$ that there is only one class in the genus, i.e. that your integral cancellation holds. See <a href="http://mathoverflow.net/questions/69444/a-priori-proof-that-covering-radius-strictly-less-than-sqrt-2-implies-class-nu" rel="nofollow">http://mathoverflow.net/questions/69444/a-priori-proof-that-covering-radius-strictly-less-than-sqrt-2-implies-class-nu</a> </p>

http://mathoverflow.net/questions/130513/another-colored-balls-puzzle-part-ii Will Jagy 2013-05-13T19:21:07Z 2013-05-13T19:21:07Z

You really need to choose your friends more carefully.

http://mathoverflow.net/questions/130385/the-isoperimetric-problem-for-domains-constrained-to-lie-between-two-parallel-pla/130427#130427 Will Jagy 2013-05-12T21:24:39Z 2013-05-12T21:24:39Z

Oh, including contact pieces with the planes. That explains the oval pictures.

http://mathoverflow.net/questions/130392/xy-z2-where-x-and-y-are-squares Will Jagy 2013-05-12T05:58:39Z 2013-05-12T05:58:39Z

try <a href="http://math.stackexchange.com/questions" rel="nofollow">math.stackexchange.com/questions</a>

http://mathoverflow.net/questions/107073/translation-of-kahlers-ber-eine-bemerkenswerte-hermitesche-metrik Will Jagy 2013-05-09T05:16:09Z 2013-05-09T05:16:09Z

The unusual thing is that there was an early translation into Mongolian. You don't see that every day.

http://mathoverflow.net/questions/107073/translation-of-kahlers-ber-eine-bemerkenswerte-hermitesche-metrik/130136#130136 Will Jagy 2013-05-09T05:15:09Z 2013-05-09T05:15:09Z

Did you do this yourself? If so, at least put a line in the document about translated by <i>_</i> and perhaps a date, maybe the original title and full reference. My friend Dmitry does this sometimes from Russian, he did all that just for a three page thing that noone but I would ever see.

http://mathoverflow.net/questions/129879/filling-in-a-rational-orthogonal-matrix-given-one-row/130082#130082 Will Jagy 2013-05-08T20:30:47Z 2013-05-08T20:30:47Z

Thank you. I had that article at one point.

http://mathoverflow.net/questions/130036/localized-l2-norm-of-quasimode-for-laplacian Will Jagy 2013-05-08T00:14:45Z 2013-05-08T00:14:45Z

crosspost <a href="http://math.stackexchange.com/questions/384999/localized-l2-norm-of-quasimode-for-laplacian" rel="nofollow" title="localized l2 norm of quasimode for laplacian">math.stackexchange.com/questions/384999/&hellip;</a>

http://mathoverflow.net/questions/129879/filling-in-a-rational-orthogonal-matrix-given-one-row Will Jagy 2013-05-06T20:37:02Z 2013-05-06T20:37:02Z

@Gerhard, I used to have a very nice bathroom scale, based on strain gauge. Later I dripped a bunch of water on it and it died. <a href="http://en.wikipedia.org/wiki/Weighing_matrix" rel="nofollow">en.wikipedia.org/wiki/Weighing_matrix</a> Will Orrick is an MO regular, not sure about the other name. It appears the important case for most people has entries $0,1,-1.$

http://mathoverflow.net/questions/129879/filling-in-a-rational-orthogonal-matrix-given-one-row Will Jagy 2013-05-06T19:54:44Z 2013-05-06T19:54:44Z

@Tony, yes, I pay attention to that. If I click in the middle, where it currently says &quot;edited 13 mins ago&quot; it shows me the revision list and the official edit count. So that is how I know when to start an answer of my own, for example.

http://mathoverflow.net/questions/129879/filling-in-a-rational-orthogonal-matrix-given-one-row Will Jagy 2013-05-06T19:42:54Z 2013-05-06T19:42:54Z

That's nice, if you do a large number of edits, but they are all in a few minutes, it clumps them together and counts only one edit.

http://mathoverflow.net/questions/129804/galois-group-of-constructible-numbers Will Jagy 2013-05-06T06:53:33Z 2013-05-06T06:53:33Z

In case it helps, adjoining $i$ to your field gives all constructible points in the plane, regarded as $\mathbb C.$ I found it useful.

http://mathoverflow.net/questions/127889/is-rigour-just-a-ritual-that-most-mathematicians-wish-to-get-rid-of-if-they-could/129799#129799 Will Jagy 2013-05-06T05:28:55Z 2013-05-06T05:28:55Z

fedja, there are courses <a href="http://en.wikipedia.org/wiki/Assertiveness#Training" rel="nofollow">en.wikipedia.org/wiki/Assertiveness#Training</a> where you can learn to deal with your crippling inhibitions, then speak as you truly feel.

http://mathoverflow.net/questions/129743/a-question-about-prime-factorization-of-n Will Jagy 2013-05-05T17:48:13Z 2013-05-05T17:48:13Z

try <a href="http://math.stackexchange.com/questions" rel="nofollow">math.stackexchange.com/questions</a>

http://mathoverflow.net/questions/89054/when-is-the-set-of-numbers-represented-by-certain-quaternary-quadratic-forms-comp Will Jagy 2013-05-04T20:44:58Z 2013-05-04T20:44:58Z

@Yazdegerd, It would be nice if you could elaborate on that a little. I now know something of Cassels-Davenport owing to communication with Pete Clark. Meanwhile, a simple conjecture I settled on was $ f(x,y) + f(z,t) $ where $f$ is a positive binary form of order three in the class group. Maybe I can find it, Noam Elkies proved it for $f(x,y) = 3 x^2 + 2 x y + 4 y^2.$ It is true if $f$ is the principal form, roughly by quaternion multiplication.

http://mathoverflow.net/questions/128939/3sat-to-quadratic-diophantine-equation Will Jagy 2013-04-27T19:14:24Z 2013-04-27T19:14:24Z

<a href="http://mathoverflow.net/questions/36420/is-the-solution-bounded-diophantine-problem-np-complete" rel="nofollow" title="is the solution bounded diophantine problem np complete">mathoverflow.net/questions/36420/&hellip;</a>