A301921 Expansion of e.g.f. 1/(1 - (exp(x) - 1)/(1 - (exp(x) - 1)^2/(1 - (exp(x) - 1)^3/(1 - ...)))), a continued fraction.

1, 1, 3, 19, 159, 1651, 21303, 324619, 5653119, 110909251, 2424648903, 58430418619, 1537673312079, 43860906193651, 1347852526593303, 44392923532503019, 1560023977386027039, 58259266750803410851, 2303999137417453606503, 96188099015599819297819, 4227325636692027926037999

Offset: 0

Author

Ilya Gutkovskiy, Jun 19 2018

Keywords

• nonn

Comments

From Peter Bala, Aug 19 2025: (Start)

Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 9 we obtain the sequence [0, 1, 3, 1, 6, 4, 0, 7, 3, 1, 6, 4, 0, 7, 3, 1, 6, 4, 0, 7, ...] with an apparent period of 6 = phi(9) beginning at n = 2. Cf. A004123. (End)

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 19*x^3/3! + 159*x^4/4! + 1651*x^5/5! + 21303*x^6/6! + ...
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..300
• N. J. A. Sloane, Transforms
• Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction

Crossrefs

Cf. A005169, A019538, A301923.

Programs

Mathematica

nmax = 20; CoefficientList[Series[1/(1 + ContinuedFractionK[-(Exp[x] - 1)^k, 1, {k, 1, nmax}]), {x, 0, nmax}], x] Range[0, nmax]!
b[n_] := b[n] = SeriesCoefficient[1/(1 + ContinuedFractionK[-x^k, 1, {k, 1, n}]), {x, 0, n}]; a[n_] := a[n] = Sum[StirlingS2[n, k] b[k] k!, {k, 0, n}]; Table[a[n], {n, 0, 20}]

Formula

a(n) = Sum_{k=0..n} Stirling2(n,k)*A005169(k)*k!.

a(n) ~ c * d^n * n!, where d = 2.19787763261059933075080498218168228... and c = 0.250957960982243982921501085974065... - Vaclav Kotesovec, Dec 20 2018