OFFSET
0,2
COMMENTS
Since E_2(x)*E_4(x)/E_6(x) == 1 - 24*Sum_{k >= 1} (k - 10*k^3 - 21*k^5)*x^k/(1 - x^k) (mod 144), and since the integer k - 10*k^3 - 21*k^5 is always divisible by 6 it follows that E_2(x)*E_4(x)/E_6(x) == 1 (mod 144). It follows from Heninger et al., p. 3, Corollary 2, that the series expansion of (E_2(x)*E_4(x)/E_6(x))^(1/24) = 1 + 30*x + 5310*x^2 + 2453220*x^3 + 910100190*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
FORMULA
a(n) ~ c * exp(2*Pi*n) / n^(23/24), where c = 0.0431061156115657949750305669836959595841497962033916083447436... - Vaclav Kotesovec, Mar 08 2021
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):
E(6, x) := 1 - 504*add(k^5*x^k/(1-x^k), k = 1..20):
with(gfun): series((E(2, x)*E(4, x)/E(6, x))^(1/24), x, 20):
seriestolist(%);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Feb 23 2021
STATUS
approved