A377977 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.

1, 6, -5028, 5704188, -7284893010, 9926715853068, -14092613175928308, 20580782244716567592, -30684764269418402550900, 46478269075227117026711730, -71284154421570122590465786956, 110437754516732491586466670733772, -172528135408494997625486967978486588, 271418933884659782820559630827037837908

Offset: 0

Author

Peter Bala, Nov 15 2024

Keywords

• sign
• easy

Comments

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.).

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

Proof.

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

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

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.

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

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.

Links

• N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006; J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.

Crossrefs

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

Programs

Maple

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

Mathematica

terms = 20; E4[x_] = 1 + 240*Sum[k^3*x^k/(1 - x^k), {k, 1, terms}]; E6[x_] = 1 - 504*Sum[k^5*x^k/(1 - x^k), {k, 1, terms}]; CoefficientList[Series[(3*E4[x] - 2*E6[x])^(1/288), {x, 0, terms}], x] (* Vaclav Kotesovec, Aug 03 2025 *)

Formula

a(n) ~ (-1)^(n+1) * c * d^n / n^(289/288), where d = 1704.7780406875645261102091212390097973945883014209828800432529862899259963... and c = 0.0034650713031853295969588514070741337333119867976967661391075146399616... - Vaclav Kotesovec, Aug 03 2025