

A111884


E.g.f.: exp(x/(1+x)).


23



1, 1, 1, 1, 1, 19, 151, 1091, 7841, 56519, 396271, 2442439, 7701409, 145269541, 4833158329, 104056218421, 2002667085119, 37109187217649, 679877731030049, 12440309297451121, 227773259993414719, 4155839606711748061, 74724654677947488521, 1293162252850914402221
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OFFSET

0,6


COMMENTS

The congruence a(n+k) == a(n) (mod k) holds for all n and k.
It follows that the sequence obtained by taking a(n) modulo a fixed positive integer k is periodic with period dividing k. For example, taken modulo 10 the sequence becomes [1, 1, 9, 1, 1, 1, 1, 9, 1, 1, ...], a purely periodic sequence with period 5. More generally, the same property holds for any sequence with an e.g.f. of the form F(x)*exp(x*G(x)), where F(x) and G(x) are power series with integer coefficients and G(0) = 1. (End)


LINKS



FORMULA

E.g.f.: exp(x/(1+x)).
Let E(x) be the e.g.f., then
E(x) = 1/G(0) where G(k)= 1  x/((1+x)*(2*k+1)  x*(1+x)*(2*k+1)/(x  (1+x)*(2*k+2)/G(k+1))); (continued fraction, 3rd kind, 3step).
E(x) = 1 + x/(G(0)x) where G(k)= 1 + 2*x + (1+x)*k  x*(1+x)*(k+1)/G(k+1); (continued fraction, Euler's 1st kind, 1step).
E(x) = G(0) where G(k)= 1 + x/((1+x)*(2*k+1)  x*(1+x)*(2*k+1)/(x + 2*(1+x)*(k+1)/G(k+1))); (continued fraction, 3rd kind, 3step).
(End)
E.g.f.: 1 + x*(E(0)1)/(x+1) where E(k) = 1 + 1/(k+1)/(1+x)/(1x/(x+1/E(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Jan 27 2013
E.g.f.: E(0)/2, where E(k)= 1 + 1/(1  x/(x + (k+1)*(1+x)/E(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Jul 31 2013
a(n) = sum(k=0..n, (1)^(nk)*L(n,k)); L(n,k) the unsigned Lah numbers.  Peter Luschny, Oct 18 2014
Dfinite with recurrence a(n) +(2*n3)*a(n1) +(n1)*(n2)*a(n2)=0.  R. J. Mathar, Jul 20 2017


MATHEMATICA

nn=30; CoefficientList[Series[Exp[x/(1+x)], {x, 0, nn}], x] Range[0, nn]! (* Harvey P. Dale, Jul 21 2011 *)


PROG

(Sage)
A111884 = lambda n: hypergeometric([n+1, n], [], 1)


CROSSREFS



KEYWORD

sign,easy


AUTHOR



STATUS

approved



