OFFSET
0,2
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
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(%);
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Peter Bala, Feb 22 2021
STATUS
approved