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Expansion of the 288th root of the series 3*E_4(x) - 2*E_6(x), where E_4(x) and E_6(x) are the Eisenstein series of weight 4 and 6.
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%I #8 Nov 26 2024 16:28:02

%S 1,6,-5028,5704188,-7284893010,9926715853068,-14092613175928308,

%T 20580782244716567592,-30684764269418402550900,

%U 46478269075227117026711730,-71284154421570122590465786956,110437754516732491586466670733772,-172528135408494997625486967978486588,271418933884659782820559630827037837908

%N Expansion of the 288th root of the series 3*E_4(x) - 2*E_6(x), where E_4(x) and E_6(x) are the Eisenstein series of weight 4 and 6.

%C Let R = 1 + x*Z[[x]] denote the set of integer power series with constant term equal to 1. Let P(n) = {g^n, g in R}. The Eisenstein series E_4(x) lies in P(8) and E_6(x) lies in P(12) (Heninger et al.).

%C We claim that the series 3*E_4(x) - 2*E_6(x) belongs to P(288).

%C Proof.

%C E_4(x) = 1 + 240*Sum_{n >= 1} sigma_3(n)*x^n.

%C E_6(x) = 1 - 504*Sum_{n >= 1} sigma_5(n)*x^n.

%C Hence, 3*E_4(x) - 2*E_6(x) = 1 - (12^3)*Sum_{n >= 1} (1/12)*(5*sigma_3(n) + 7*sigma_5(n))*x^n belong to R, since the polynomial (1/12)*(5*k^3 + 7*k^5) is integral for integer values of k. See A245380.

%C Hence, 3*E_4(x) - 2*E_6(x) == 1 (mod 12^3) == 1 (mod (2^6)*(3^3)).

%C It follows from Heninger et al., Theorem 1, Corollary 2, that the series 3*E_4(x) - 2*E_6(x) belongs to P((2^5)*(3^2)) = P(288). End Proof.

%H N. Heninger, E. M. Rains and N. J. A. Sloane, <a href="https://arxiv.org/abs/math/0509316">On the Integrality of n-th Roots of Generating Functions</a>, arXiv:math/0509316 [math.NT], 2005-2006; J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.

%p with(numtheory):

%p E := proc (k) local n, t1; t1 := 1 - 2*k*add(sigma[k-1](n)*q^n, n = 1..30)/bernoulli(k); series(t1, q, 30) end:

%p seq(coeftayl((3*E(4) - 2*E(6))^(1/288), q = 0, n), n = 0..20);

%Y Cf. , A004009 (E_4), A013973 (E_6), A108091 ((E_4)^1/8), A109817 ((E_6)^1/12), A245380, A341871 - A341875, A377973, A377974, A377975.

%K sign,easy

%O 0,2

%A _Peter Bala_, Nov 15 2024