A341873 Coefficients of the series whose 24th power equals E_2(x)^5/E_10(x), where E_2(x) and E_10(x) are the Eisenstein series A006352 and A013974.

1, 6, 7038, 2002644, 922569342, 380737463400, 175255606306116, 80315525064955440, 38028486993289854966, 18171889608389845598586, 8807723964899085718419480, 4305311468773791666900669828, 2122088430918938935321961736084

Offset: 0

Author

Peter Bala, Feb 23 2021

Keywords

• nonn
• easy

Comments

It is easy to see that E_2(x)^5/E_10(x) == 1 - 24*Sum_{k >= 1} (5*k - 11*k^9)*x^k/(1 - x^k) (mod 144), and also that the integer 5*k - 11*k^9 is always divisible by 6. Hence, E_2(x)^5/E_10(x) == 1 (mod 144). It follows from Heninger et al., p. 3, Corollary 2, that the series expansion of (E_2(x)^5/E_10(x))^(1/24) = 1 + 6*x + 7038*x^2 + 2002644*x^3 + 922569342*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, A013974, A341871 - A341875.

Programs

Maple

E(2,x)  := 1 -  24*add(k*x^k/(1-x^k),   k = 1..20):
E(10,x) := 1 - 264*add(k^9*x^k/(1-x^k), k = 1..20):
with(gfun): series((E(2,x)^5/E(10,x))^(1/24), x, 20):
seriestolist(%);