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%I #53 Aug 15 2022 10:30:02
%S 1,1,9,211,9285,658171,68504709,9837380491,1863598406805,
%T 450247033371451,135111441590583909,49300373690091496171,
%U 21495577955682021043125,11037123350952586270549531,6591700149366720366704735109
%N a(n) = Sum_{k=0..n} k^n * k! * Stirling2(n,k).
%C Conjecture: Let p be prime. The sequence obtained by reducing a(n) modulo p for n >= 1 is purely periodic with period p - 1. For example, modulo 7 the sequence becomes [1, 2, 1, 3, 3, 0, 1, 2, 1, 3, 3, 0, ...], with an apparent period of 6. Cf. A338040. - _Peter Bala_, May 31 2022
%H Vincenzo Librandi, <a href="/A122399/b122399.txt">Table of n, a(n) for n = 0..200</a>
%F E.g.f.: Sum_{n >= 0} (exp(n*x) - 1)^n. - _Vladeta Jovovic_, Sep 03 2006
%F E.g.f.: Sum_{n>=0} exp(n^2*x) / (1 + exp(n*x))^(n+1). - _Paul D. Hanna_, Oct 26 2014
%F E.g.f.: Sum_{n>=0} exp(-n*x) / (1 + exp(-n*x))^(n+1). - _Paul D. Hanna_, Oct 30 2014
%F O.g.f.: Sum_{n>=0} n^n * n! * x^n / Product_{k=1..n} (1 - n*k*x). - _Paul D. Hanna_, Jan 05 2013
%F Limit n->infinity (a(n)/n!)^(1/n)/n = ((1+exp(1/r))*r^2)/exp(1) = A317855/exp(1) = 1.162899527477400818845..., where r = 0.87370243323966833... is the root of the equation 1/(1+exp(-1/r)) = -r*LambertW(-exp(-1/r)/r). - _Vaclav Kotesovec_, Jun 21 2013
%F a(n) ~ c * A317855^n * (n!)^2 / sqrt(n), where c = 0.327628285569869481442286492410507030710253054522608... - _Vaclav Kotesovec_, Aug 09 2018
%F Let A(x) = 1 + x + 9*x^2/2! + 211*x^3/3! + ... denote the e.g.f. of the sequence. Let F(x) denote the series reversion of A(x) - 1 = x - 9*x^2/2 + 16*x^3/3 - 205*x^4/4 - 2714*x^5/5 - .... Then both dF/dx = 1 - 9*x + 16*x^2 - 205*x^3 - 2714*x^4 - ... and exp(F(x)) = 1 + x - 4*x^2 + x^3 - 38*x^4 - 606*x^5 - ... have integer coefficients. Note that 1 + series reversion(exp(F(x)) - 1) is the o.g.f. for A122400. - _Peter Bala_, Aug 09 2022
%e E.g.f.: A(x) = 1 + x + 9*x^2/2! + 211*x^3/3! + 9285*x^4/4! + 658171*x^5/5! + ...
%e such that
%e A(x) = 1 + (exp(x)-1) + (exp(2*x)-1)^2 + (exp(3*x)-1)^3 + (exp(4*x)-1)^4 + ...
%e The e.g.f. is also given by the series:
%e A(x) = 1/2 + exp(x)/(1+exp(x))^2 + exp(4*x)/(1+exp(2*x))^3 + exp(9*x)/(1+exp(3*x))^4 + exp(16*x)/(1+exp(4*x))^5 + exp(25*x)/(1+exp(5*x))^6 + ...
%e or, equivalently,
%e A(x) = 1/2 + exp(-x)/(1+exp(-x))^2 + exp(-2*x)/(1+exp(-2*x))^3 + exp(-3*x)/(1+exp(-3*x))^4 + exp(-4*x)/(1+exp(-4*x))^5 + exp(-5*x)/(1+exp(-5*x))^6 + ...
%p a := n -> add(k^n*k!*combinat[stirling2](n,k),k=0..n); # _Max Alekseyev_, Feb 01 2007
%t Flatten[{1,Table[Sum[k^n*k!*StirlingS2[n,k],{k,0,n}],{n,1,20}]}] (* _Vaclav Kotesovec_, Jun 21 2013 *)
%o (PARI) {a(n)=polcoeff(sum(m=0, n, m^m*m!*x^m/prod(k=1, m, 1-m*k*x+x*O(x^n))), n)}
%o for(n=0, 20, print1(a(n), ", ")) \\ _Paul D. Hanna_, Jan 05 2013
%o (PARI) {a(n)=n!*polcoeff(sum(k=0, n, (exp(k*x +x*O(x^n)) - 1)^k), n)}
%o for(n=0,25,print1(a(n),", ")) \\ _Paul D. Hanna_, Oct 26 2014
%o (PARI) /* From e.g.f. infinite series: */
%o \p100 \\ set precision
%o {A=Vec(serlaplace(sum(n=0, 500, 1.*exp(n^2*x +O(x^26))/(1 + exp(n*x +O(x^26)))^(n+1)) ))}
%o for(n=0, #A-1, print1(round(A[n+1]), ", ")) \\ _Paul D. Hanna_, Oct 30 2014
%Y Cf. A195415, A122400, A220181, A224899, A245322, A320083, A338040.
%K nonn,easy
%O 0,3
%A _Vladeta Jovovic_, Aug 31 2006
%E More terms from _Max Alekseyev_, Feb 01 2007
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