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A341874
Coefficients of the series whose 24th power equals E_2(x)^7/E_14(x), where E_2(x) and E_14(x) are the Eisenstein series A006352 and A058550.
2
1, -6, 8118, 1740636, 937783902, 364856395608, 172736345164500, 78278100914583312, 37268001893898954198, 17773741638825114790854, 8624927270409695050736952, 4214914849580580859932456300, 2078204723099375850950863499028
OFFSET
0,2
COMMENTS
It is easy to see that E_2(x)^7/E_14(x) == 1 - 24*Sum_{k >= 1} (7*k - 11*k^13)*x^k/(1 - x^k) (mod 144), and also that the integer 7*k - k^13 is always divisible by 6. Hence, E_2(x)^7/E_14(x) == 1 (mod 144). It follows from Heninger et al., p. 3, Corollary 2, that the series expansion of (E_2(x)^7/E_14(x))^(1/24) = 1 - 6*x + 8118*x^2 + 1740636*x^3 + 937783902*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.
MAPLE
E(2, x) := 1 - 24*add(k*x^k/(1-x^k), k = 1..20):
E(14, x) := 1 - 24*add(k^13*x^k/(1-x^k), k = 1..20):
with(gfun): series((E(2, x)^7/E(14, x))^(1/24), x, 20):
seriestolist(%);
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Peter Bala, Feb 23 2021
STATUS
approved