# Tag Info

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The only CAS's that have built-in support for modular and automorphic forms, as far as I know, are Sage and Magma. [Edit: I had forgotten Pari/GP, which will introduce substantial modular forms functionality as of version 2.10 which is currently in alpha testing; see Aurel's comment below]. Both Sage and Magma offer roughly comparable functionality. In Sage ...

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I think that there is indeed some possibility to lower the bound, and this is something I've looked at seriously a few times. I spent a semester (in 2019) with the Computational Number Theory Group here at BYU trying to do it, without success. Let me outline some of the difficulties we found. Issue 1. The main technique used in establishing the current ...

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As Jan Grabowski notes, all the primes that you mention have been discovered by GIMPS. GIMPS draws a distinction between testing and double-checking. When a Lucas–Lehmer test is performed on a Mersenne number and delivers a verdict that the number is prime, the computation is immediately verified. Only after the verification checks out is it ...

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Using $$\frac{10^c-1}{9}=\sum_{m=0}^{c-1} 10^m,$$ the product in question equals $$\sum_{n=0}^{2a+2b+c-1}r(n)10^n,$$ where $r(n)$ is the number of times $n$ occurs among the numbers (counted with multiplicity) $$\begin{matrix} m+2a+2b,&m+2a+b,&m+2a,\\ m+a+2b,&m+a+b,&m+a,\\ m+2b,&m+b,&m,\\ \end{matrix}$$ for some $m\in\{0,1,\dots,c-1\}$...

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You are looking at a computable field (if your focus is on the field), or a computable presentation of a field (if your focus is on the details of how elements and operations are coded). These objects are studied in computable structure theory.

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Answer to Question 1. Yes, $E(\mathbb{Z})=F$. The inclusion $F\subseteq E(\mathbb{Z})$ follows from $E\subseteq E(\mathbb{Z})$ and Dickson's result $E=F$. It remains to show $E(\mathbb{Z})\subseteq F$. Pick $n\in E(\mathbb{Z})$. By definition there exist $u,v\in\mathbb{Z}^3$ such that $\lVert u\rVert=\lVert v\rVert$ and $n=\lVert u\rVert^2+\lvert u\cdot v\... 15 I did the computation in Sage, and there is no such form$f$. There are 21 Galois orbits of newforms of level$\Gamma_0(1776)$and trivial character, of which the largest has size 3, and none of the reductions modulo any of the primes above 7 in the coefficient fields is congruent to the form associated to$\bar\rho$. Looking again at your question, the ... 15 It seems what you are asking is "If we have a precise asymptotic for the number of elements of a set, can we solve binary additive problems involving that set?" The answer in general seems to be `no'. Let's consider Goldbach's conjecture that every large integer$n$is the sum of two primes. It is not hard to see from pigeonholing that the typical$...

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This is an elaboration of the answer that Noam Elkies provided in the comments. Suppose that $p=x^2 + xy + y^2$. Then note that $x$ and $y$ are small relative to $p$ (at most half as many digits). Note also that if $\zeta \not\equiv 1\pmod p$ satisfies $\zeta^3 \equiv 1\pmod p$ then $\zeta^2 + \zeta + 1 \equiv 0 \pmod p$, so $$(x - \zeta y)(x - \zeta^2 y) =... 14 The answer to question 1 is yes. Pick any field element x_1 outside the rationals. Let x_2 and x_3 be its conjugates under the Galois group. Then the cross ratio$$w=\frac{(x_1-1)(x_2-x_3)}{(x_1-x_2)(1-x_3)}$$does the trick. (This is the same argument as in this answer.) Note that these w usually won't be algebraic integers, and they cannot be ... 14 [EDITED] It is likely that there are no solutions at all for n \ge 4. For n \ge 5 a solution would be a counterexample to the Lander, Parkin, and Selfridge conjecture. The best FLT "near miss" that I know of is 13^5 + 16^5 = 17^5 + 12. 13 Google provides an answer to this question. The first deterministic polynomial time algorithm for this is due to H. W. Lenstra, Jr., in his paper "Finding isomorphisms between finite fields" (Mathematics of Computation, v. 56 (1991), 329-347). Another algorithm appears in the paper by B. Allombert, "Explicit computation of isomorphisms between finite ... 13 In Sage you can do something like: x=polygen(QQbar) f=x^5 - x +1 roots = f.roots(QQbar) a1 = roots[1][0] a2 = roots[-1][0] (a1 - a2).minpoly() x^20 - 10*x^16 - 95*x^12 + 625*x^10 - 40*x^8 + 3750*x^6 + 400*x^4 + 5000*x^2 + 2869 13 The problem asks for the least number N such that the number of divisors of N is at least n+2. Since all numbers below N must have fewer divisors, clearly d(N) > d(m) for all 1\le m < N. Such a champion value N for the divisor function was termed by Ramanujan as a highly composite number, and he determined the prime factorization of ... 13 This is just to present a numerical curiosity. I suspect that it is coincidental, but I feel compelled to share it in case anyone wants to look further. First some disclaimers: Consider a random walk which starts at a value x_0 and, for constants c_1,c_2 moves at step k with probability \frac{c_1}{\log^2 k} and, if so, increases or decreases by \... 12 More specifically, e_k(j)=c(j+1,j+1-k), where c(j+1,j+1-k) is a signless Stirling number of the first kind. For a discussion of this polynomial see http://math.mit.edu/~rstan/pubs/pubfiles/29.pdf. In particular, Theorem 2.1 gives a combinatorial interpretation of the coefficients of the polynomial (1-x)^{2k+1}\sum_j e_k(j)x^j, but I am unaware of a ... 12 You can get a conjectural lower bound for |a^x-b^y| using the ABC-conjecture. I'll do the case a^x > b^y for simplicity. Taking A=a^x, B=-b^y, and C=a^x-b^y, we get for every \epsilon >0 that there is a K=K_\epsilon>0 so that$$ a^x \le K\prod_{p\mid ab(a^x-b^y)} p^{1+\epsilon} \le K(ab(a^x-b^y))^{1+\epsilon}. $$Replacing K with a ... 12 Computation of ranks of elliptic curves relies on descent. The first step of descent is the computation of a finite Selmer group, which in turn uses the computation of the class group of a potentially large number field. This is the step where GRH is used: it allows you to assume that the class group is generated by the set of prime ideals up to a relatively ... 10 One shortcut you could use for computing the level 17 form you link to would be the following. There are exactly 8 Eisenstein series of weight 1 for \Gamma_1(17) and they are all given by completely explicit q-expansion formulae (involving sums of divisors, etc). The pairwise products of these series span the space of forms of weight 2. So if F is the ... 10 Deuring proved that for every a, |a| < 2\sqrt{p}, there exists an elliptic curve with p+1-a points over \mathbb{F}_p. M Deuring, Die Typen der Multiplikatorenringe elliptischer Funktionenkörper, Abh. Math. Sem. Univ. Hamburg 14 (1941), 197-272. 10 Yes. What you ask about are Stirling numbers of the first kind s(j,j-k). Formula (21) http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html is an explicit expression for fixed k. 10 I want to leave a few elementary comments, maybe they will be helpful. The question asks about recurrence relation$$ u_{n+1}= \lfloor u_n \rfloor (u_n − \lfloor u_n \rfloor + 1) $$Suppose you write rational u in terms of natural numbers \frac{pq+r}{p}. Then \lfloor u \rfloor = q. The recurrence relation is now$$ u_{n+1} = q_n (\frac{pq_n+r_n}{p} − ...

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As discussed in the comments, I don't see how to extract the desired matrix from the original question (about spanning the vector space). However, the matrix being nonzero IS equivalent to the following: The polynomials $\sum_{i = 0}^{p-2} (i+1)^{-1} t^{i+n}$ for $0\leq n\leq p-2$, along with the polynomial $t^{p-1}$, span the vector space $\mathbb{F}_p[t]/(... 9 The expression converges to$0$, even when$\phi(n)$is replaced by the larger$n$. The contribution from$n\le\sqrt N/dcan be given by ignoring the logarithm in the denominator: \begin{align*} \frac1{N^2} \sum_{d=1}^N \log d \sum_{n=1}^{\sqrt N/d} \frac{\phi(n)}{\log dn} &\le \frac1{N^2} \sum_{d=1}^N \log d \sum_{n=1}^{\sqrt N/d} n \\\ &\lesssim \... 9 Using resultants can greatly help. For example, knowing that\sqrt[5]{2}$is a zero of$P(x)=x^5-2$and$1-\exp\frac{2\pi i}{5}$is a zero of$Q(x)=(x-1)^5+1$, we conclude$\sqrt[5]{2}\cdot (1-\exp\frac{2\pi i}{5})$is a zero of$\mathrm{Res}_y(P(y),y^5Q(\frac{x}{y}))$. In PARI/GP: ? polresultant(y^5-2,y^5*((x/y-1)^5+1),y) %1 = x^25 + 2500*x^15 + 50000*x^5 ... 9 A standard way to measure the size (= complexity) of vectors with coordinates in$\overline{\mathbb Q}$is through the theory of Weil height functions. You can find the definition and properties of the Weil height $$H : \mathbb P^n(\overline{\mathbb Q}) \longrightarrow [1,\infty)$$ in many places, such as [1] and [2]. Since you are asking about vectors, ... 9 You can check the current status (updated hourly). Currently among the different stats are: All exponents below 46 251 389 have been tested and verified. All exponents below 82 038 613 have been tested at least once. 9 given your interest: the list of all$A x^2 + B y^2 + C z^2$with ordered positive coefficients, such that the represented numbers can be described by congruences Ummm. Allowing mixed terms, all 913 (probably) regular positive forms 9 Seems that here is a proof that the graph is connected Let$C$be a component.$C$contains an even number, as an odd number$2k+1$is adjacent to$2k(k+1)$.$C$contains a number divisible by$6$: if$2k\in C$with$3\nmid k$, then$k^2-1\in C$as well. If$3k\in C$and$p$is a prime, then$3kp\in C$. Induction on$p$. The base cases$p=2,3$follow ... 8 I recently implemented an algorithm to determine these number fields by computing the$j$-polynomial: Let$\varphi: X_0(N) \to E$be a fixed modular parametrization and$P \in E(\mathbb{Q})$. By j-polynomial I mean the polynomial$F_P(x) = \prod_{z : \varphi(z) = P}(x - j(z))$. There's a Laurent series$x(q)\$ with integer coefficients, which is the modular ...

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