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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.

%I A301921 #18 Aug 20 2025 00:30:37
%S A301921 1,1,3,19,159,1651,21303,324619,5653119,110909251,2424648903,
%T A301921 58430418619,1537673312079,43860906193651,1347852526593303,
%U A301921 44392923532503019,1560023977386027039,58259266750803410851,2303999137417453606503,96188099015599819297819,4227325636692027926037999
%N 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.
%C A301921 From _Peter Bala_, Aug 19 2025: (Start)
%C A301921 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)
%H A301921 Vaclav Kotesovec, <a href="/A301921/b301921.txt">Table of n, a(n) for n = 0..300</a>
%H A301921 N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%H A301921 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Rogers-RamanujanContinuedFraction.html">Rogers-Ramanujan Continued Fraction</a>
%F A301921 a(n) = Sum_{k=0..n} Stirling2(n,k)*A005169(k)*k!.
%F A301921 a(n) ~ c * d^n * n!, where d = 2.19787763261059933075080498218168228... and c = 0.250957960982243982921501085974065... - _Vaclav Kotesovec_, Dec 20 2018
%e A301921 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! + ...
%t A301921 nmax = 20; CoefficientList[Series[1/(1 + ContinuedFractionK[-(Exp[x] - 1)^k, 1, {k, 1, nmax}]), {x, 0, nmax}], x] Range[0, nmax]!
%t A301921 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}]
%Y A301921 Cf. A005169, A019538, A301923.
%K A301921 nonn
%O A301921 0,3
%A A301921 _Ilya Gutkovskiy_, Jun 19 2018