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.

1, 42, 34020, 39770808, 54603156174, 82058923220904, 130685055490645992, 216707827984305135744, 370213729923354622242084, 647073665508052293475274898, 1151627718366568095339000345192, 2079918757332503030219456972007720, 3802403760868562402170776739039126584, 7022808067106759130277006634854345528104

Offset: 0

Author

Peter Bala, Nov 08 2024

Keywords

• nonn
• easy

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

• Peter Bala, Fractional iteration of a series inversion operator
• 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.

Crossrefs

Cf. A013973, A109817, A377221, A377222.

Programs

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);

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.