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2d
comment Are the nontrivial zeros of the Riemann zeta simple?
Not that this answers your question, but a multiple zero is in some sense even less likely than one would expect if the zeros were "randomly distributed" (given their average density): they seem to repel each other, so if you plot the zero spacings scaled to average 1 then their density approaches $0$ as the distance approaches $0$. [I hedge with "in some sense" because even random spacing would make the probability of a coincidence $0$. NB I'm trying to minimize confusion by using "zero" for a root of $\zeta(s)$ and "$0$" for the smallest nonnegative real number.]
2d
comment Are the nontrivial zeros of the Riemann zeta simple?
@DavidHansen what about elliptic curves of rank $3$ (or for that matter even $2$) and higher?
Apr
24
comment On cluster points of a particular sequence
Even more simply: 3, 18, 105, 612, 3567, 20790, 121173, 706248 are divisible by 3; Dividing by 3 yields 1, 6, 35, 204, 1189, 6930, 40391, 235416 = OEIS A001109 = solutions of $8x^2+1=\Box$, and square roots of square-triangular numbers. (Naturally the A106328 comments include this connection with A00109.)
Apr
20
comment magma generators for unit group/ sage totally positive
(and anything that's in gp is automatically accessible by Sage even if there's no native Sage way to do it)
Apr
20
comment magma generators for unit group/ sage totally positive
In gp you could do K=bnfinit(x^3+x^2-2*x-1); U=K.fu; S=bnfsignunit(K) to get a vector U of fundamental units and a matrix S of their real embeddings' signs. Then it just takes a bit of linear algebra mod 2 (possibly using matsolvemod) to construct a basis of totally positive units.
Apr
19
comment Smooth, irreducible surface with real part containing two projective planes
I noticed this error and was correcting it as you wrote.
Apr
19
revised Smooth, irreducible surface with real part containing two projective planes
added 608 characters in body
Apr
19
answered Smooth, irreducible surface with real part containing two projective planes
Apr
19
comment Equations for Elliptic Curves
(Inevitably that's not far from the usual Riemann-Roch argument, but is what people actually did in the old days and sometimes still do nowadays.)
Apr
19
comment Equations for Elliptic Curves
Would you prefer something along these lines? We know the curve has a model $Y^2 = P(X)$ with $P$ of degree $3$ or $4$ and some rational point. Changing projective coordinates, we may assume this point has $X = \infty$. If $\deg P = 3$ then we're done. Else $\deg P = 4$ and the leading coefficient of $P$ is a square, so at infinity $P = Q^2 + R$ for some $Q,R$ with rational coefficients such that $\deg Q = 2$ and $\deg Q \leq 1$. Now write $Y=Q+x$ and get an equation quadratic in $X$ whose discriminant is cubic in $x$. This gives a birational map from $Y^2 = P(X)$ to $y^2 = cubic(x)$.
Apr
18
comment What does “game theory” cover and how should it be called?
Combinatorial game theory can also be summarized as an extension of the Sprague-Grundy theory en.wikipedia.org/wiki/Sprague-Grundy_theorem of impartial games ( = generalized Nim ) to games that may be "partial" in the sense that the two opponents needn't always have the same move sets.
Apr
17
comment On the parity of $[x^n]$
...(as well as Max Alekseyev's $(3+\sqrt{17})/2$ example).
Apr
17
comment On the parity of $[x^n]$
It was a guess when I wrote it (and the comment above), but meanwhile I thought about it a bit more and I think we can construct $x$ by intersecting an infinite sequence of nested intervals as long as the first interval $[x_1, x_1+1)$ has $x_1 \geq 4$ (so $4$ or $5$ will do). The $n$-th interval has the form $[x_n, (x_n^n+1)^{1/n})$ for some integer $x_n$, and that puts $x^{n+1}$ in an interval of length at least 4 so its intersection with $\{X\in{\bf R}:\lfloor X\rfloor\equiv b_{n+1} \bmod 2\}$ must contain a semiopen interval of length $1$. I should update my answer to incorporate this...
Apr
16
comment Is there a formula for the number of elements in $S_n$ having length $k$ with respect to the generators taken to be the transpositions?
I see that "Mahonian" = in honor of MacMahon; is that a common formation? (Cf. Descartes → cartesian, but Desargues → Desarguesian [though "Arguesian" is occasionally seen too].)
Apr
10
comment A long-lasting conjecture of Pólya & Szegő
Is the regular $n$-gon even proved to be a local minimum for $\lambda_1$? The survey you cite doesn't cite such a result in the "case of polygons" sections (3.2, page 5).
Apr
1
comment Examples of math hoaxes/interesting jokes published on April Fool's day?
That's actually not far (even in many subsidiary details) from yet another item in Martin Gardner's famous April column that already accounts for two answers here.
Mar
30
awarded  Enlightened
Mar
30
awarded  Nice Answer
Mar
30
revised On the parity of $[x^n]$
Add the explicit example with a (totally real, discriminant 148) cubic irrationality.
Mar
30
comment On the parity of $[x^n]$
@ARupinski Yes. To get a prime between $N$ and $N+H$, we need $H$ to be some power of $N$ (it should really be a power of $\log N$ but we can't prove anything like that), which is why Mills needs double exponentials. For an integer of given parity, $H=2$ is enough, so plain $[x^n]$ probably suffices.