A341871 Coefficients of the series whose 48th power equals E_2(x)^2/E_4(x), where E_2(x) and E_4(x) are the Eisenstein series A006352 and A004009.

1, -6, 558, -88884, 15433662, -2864048616, 552921962724, -109731286565040, 22220439670517814, -4569456313225317114, 951159953810624453208, -199945837161334089352548, 42373766861587365894611604

Offset: 0

Author

Peter Bala, Feb 22 2021

Keywords

• sign
• easy

Comments

It is easy to see that E_2(x)^2/E_4(x) == 1 - 48*Sum_{k >= 1} (k + 5*k^3)*x^k/(1 - x^k) (mod 288), and also that the integer k + 5*k^3 is always divisible by 6. Hence, E_2(x)^2/E_4(x) == 1 (mod 288). It follows from Heninger et al., p. 3, Corollary 2, that the series expansion of (E_2(x)^2/E_4(x))^(1/48) = 1 - 6*x + 558*x^2 - 88884*x^3 + 15433662*x^4 - ... has integer coefficients.

Note that (E_2(x)^2/E_4(x))^(1/48) = (E_2(x)^4/E_8(x))^(1/96).

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. A004009, A006352, A108091, A289392, A341872, A341873, A341874, A341875.

Programs

Maple

E(2,x) := 1 -  24*add(k*x^k/(1-x^k),   k = 1..20):
E(4,x) := 1 + 240*add(k^3*x^k/(1-x^k), k = 1..20):
with(gfun): series((E(2,x)^2/E(4,x))^(1/48), x, 20):
seriestolist(%);