A341872 Coefficients of the series whose 72nd power equals E_2(x)^3/E_6(x), where E_2(x) and E_6(x) are the Eisenstein series A006352 and A013973.

1, 6, 1998, 722484, 291762942, 125454173544, 56146411655460, 25832836404319152, 12128921727745915062, 5783583949613172902394, 2791762868052719757442008, 1360988846025232489401029220, 668925190887642335984231235348, 331039288912491308442251418152952

Offset: 0

Author

Peter Bala, Feb 22 2021

Keywords

• nonn
• easy

Comments

It is easy to see that E_2(x)^3/E_6(x) == 1 - 72*Sum_{k >= 1} (k - 7*k^5)*x^k/(1 - x^k) (mod 432), and also that the integer k - 7*k^5 is always divisible by 6. Hence, E_2(x)^3/E_6(x) == 1 (mod 432). It follows from Heninger et al., p. 3, Corollary 2, that the series expansion of (E_2(x)^2/E_6(x))^(1/72) = 1 + 6*x + 1998*x^2 + 722484*x^3 + 291762942* x^4 + ... has integer coefficients.

Links

• N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
• Wikipedia, Eisenstein series

Crossrefs

Cf. A006352, A013973, A341871 - A341875.

Programs

Maple

E(2,x) := 1 -  24*add(k*x^k/(1-x^k),   k = 1..20):
E(6,x) := 1 - 504*add(k^5*x^k/(1-x^k), k = 1..20):
with(gfun): series((E(2,x)^3/E(6,x))^(1/72), x, 20):
seriestolist(%);

Formula

a(n) ~ c * exp(2*Pi*n) / n^(71/72), where c = 0.013960369132490470055158573616810629626490780934389076244815126342923645628... - Vaclav Kotesovec, Mar 08 2021