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%I #55 Aug 17 2022 22:43:01
%S 1,1,-1,1,1,-19,151,-1091,7841,-56519,396271,-2442439,7701409,
%T 145269541,-4833158329,104056218421,-2002667085119,37109187217649,
%U -679877731030049,12440309297451121,-227773259993414719,4155839606711748061,-74724654677947488521,1293162252850914402221
%N E.g.f.: exp(x/(1+x)).
%C Row sums of triangle A111596.
%C With different signs see A066668.
%C From _Peter Bala_, Aug 15 2022: (Start)
%C The congruence a(n+k) == a(n) (mod k) holds for all n and k.
%C 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)
%H Seiichi Manyama, <a href="/A111884/b111884.txt">Table of n, a(n) for n = 0..450</a>
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Barry3/barry100r.html">The Restricted Toda Chain, Exponential Riordan Arrays, and Hankel Transforms</a>, J. Int. Seq. 13 (2010) # 10.8.4, example 4.
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Barry4/barry122.html">Exponential Riordan Arrays and Permutation Enumeration</a>, J. Int. Seq. 13 (2010) # 10.9.1, example 6.
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Barry1/barry97r2.html">Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms</a>, J. Int. Seq. 14 (2011) # 11.2.2, example 20.
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Barry5/barry112.html">Combinatorial Polynomials as Moments, Hankel Transforms, and Exponential Riordan Arrays</a>, J. Int. Seq. 14 (2011) 11.6.7, example 10.
%H A. Hennessy and Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Barry6/barry161.html">Generalized Stirling Numbers, Exponential Riordan Arrays, and Orthogonal Polynomials</a>, J. Int. Seq. 14 (2011) # 11.8.2
%F E.g.f.: exp(x/(1+x)).
%F From _Sergei N. Gladkovskii_, Jul 21 2012: (Start)
%F Let E(x) be the e.g.f., then
%F 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, 3-step).
%F 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, 1-step).
%F 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, 3-step).
%F (End)
%F E.g.f.: 1 + x*(E(0)-1)/(x+1) where E(k) = 1 + 1/(k+1)/(1+x)/(1-x/(x+1/E(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Jan 27 2013
%F 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
%F a(n) = sum(k=0..n, (-1)^(n-k)*L(n,k)); L(n,k) the unsigned Lah numbers. - _Peter Luschny_, Oct 18 2014
%F a(n) = hypergeom([-n+1,-n],[],-1). - _Peter Luschny_, Apr 08 2015
%F D-finite with recurrence a(n) +(2*n-3)*a(n-1) +(n-1)*(n-2)*a(n-2)=0. - _R. J. Mathar_, Jul 20 2017
%t nn=30; CoefficientList[Series[Exp[x/(1+x)],{x,0,nn}], x] Range[0,nn]! (* _Harvey P. Dale_, Jul 21 2011 *)
%o (Sage)
%o A111884 = lambda n: hypergeometric([-n+1,-n], [], -1)
%o [Integer(A111884(n).n(100)) for n in (0..23)] # _Peter Luschny_, Sep 23 2014
%Y Unsigned row sums of A111596: A000262.
%K sign,easy
%O 0,6
%A _Wolfdieter Lang_, Aug 23 2005
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