|
%I #29 Feb 23 2022 23:05:34
%S 13,19,43,61,97,103,109,127,157,163,181,193,241,277,283,331,349,373,
%T 379,409,433,463,487,499,523,601,607,619,631,661,673,691,727,733,757,
%U 769,787,811,859,883,937,967,991
%N Bisection of primes congruent to 1 modulo 3 (A002476), depending on the corresponding A001479 entry being congruent to 1 modulo 3 or not. Here the first case.
%C The other primes congruent to 1 modulo 3 are given in A272201.
%C Each prime == 1 (mod 3) has a unique representation A002476(m) = A(m)^2 + 3*B(m)^2 with positive A(m) = A001479(m+1) and B(m) = A001480(m+1), m >= 1 (see also A001479). The present sequence gives such primes corresponding to A(m+1) == 1 (mod 3). The ones corresponding to A(m+1) not == 1 (mod 3) (the complement) are given in A272201.
%C This bisection of the primes from A002476 is needed in the formula for the coefficients of the q-expansion (q = exp(2*Pi*i*z), Im(z) > 0) of the modular cusp form (eta(6*z))^4|_{z=z(q)} = Eta64(q) with Dedekind's eta function. See A000727 which gives the coefficients of the q-expansion of F(q) = Eta64(q^(1/6))/q^(1/6) = (Product_{m>=0} (1 - q^m))^4. The coefficients F(q) = Sum_{n>=0} f(6*n+1)*q^n are given in the Finch link on p. 5, using multiplicativity. For primes congruent to 1 modulo 6 the formula involves x_p and y_p which are the present A and B for prime p == 1 (mod 3).
%C See also the p-defects of the elliptic curve y^2 = x^3 + 1, related to (eta(6*z))^4, given in A272198 with another (equivalent) way to find the coefficients of the Eta64(q) expansion, hence those of F(q).
%H Robert Israel, <a href="/A272200/b272200.txt">Table of n, a(n) for n = 1..10000</a>
%H S. R. Finch, <a href="https://arxiv.org/abs/math/0701251">Powers of Euler's q-Series</a>, arXiv:math/0701251 [math.NT], 2007.
%F This sequence collects the 1 (mod 3) primes p(m) = A002476(m) = A(m)^2 + 3*B(m)^2 with positive A(m) == 1 (mod 3), for m >= 1. A(m) = A001479(m+1).
%p filter:= proc(n) local S,x,y;
%p if not isprime(n) then return false fi;
%p S:= remove(hastype,[isolve(x^2+3*y^2=n)],negative);
%p subs(S[1],x) mod 3 = 1
%p end proc:
%p select(filter, [seq(i,i=7..1000,6)]); # _Robert Israel_, Apr 29 2019
%t filterQ[n_] := Module[{S, x, y}, If[!PrimeQ[n], Return[False]]; S = Solve[x > 0 && y > 0 && x^2 + 3 y^2 == n, Integers]; Mod[x /. S[[1]], 3] == 1];
%t Select[Range[7, 1000, 6], filterQ] (* _Jean-François Alcover_, Apr 21 2020, after _Robert Israel_ *)
%Y Cf. A000727, A001479, A002476, A001480, A272198, A272201 (complement relative to A002476).
%K nonn,easy
%O 1,1
%A _Wolfdieter Lang_, Apr 28 2016
|
