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A377223
Coefficients of the series whose 12th power is 1/x * series_reversion(x*E_6(x)), where E_6(x) is the Eisenstein series of weight 6.
3
1, 42, 34020, 39770808, 54603156174, 82058923220904, 130685055490645992, 216707827984305135744, 370213729923354622242084, 647073665508052293475274898, 1151627718366568095339000345192, 2079918757332503030219456972007720, 3802403760868562402170776739039126584, 7022808067106759130277006634854345528104
OFFSET
0,2
COMMENTS
Let R = 1 + x*Z[[x]] denote the set of integral power series with constant term equal to 1. Let P_n = {g^n, g in R}. If f belongs to P_n then the power series 1/x * series_reversion(x*f(x)) is also in P_n. Apply Bala, Theorem 1, Corollary 2.
Here we take f to be the Eisenstein series E_6. See A013973. It is known that the 12th root f^(1/12) has integer coefficients (Heninger et al.). See A109817. It follows that the present sequence is integral.
LINKS
N. Heninger, E. M. Rains, and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
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.
FORMULA
G.f.: A(x) = the 12-fold iterate I^12( 1/E_6(x)^(1/12) ), where the operator I : R -> R is defined by I(f(x)) = 1/x * series_reversion(x/f(x)), showing that the g.f. A(x) is integral.
MAPLE
with(numtheory):
Order := 30:
E_6 := 1 - 504*add(sigma[5](n)*x^n, n = 1..30):
solve(series(x*E_6, x) = y, x):
seq(coeftayl(series((%/y)^(1/12), y), y = 0, n), n = 0..20);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 08 2024
STATUS
approved