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A111884
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E.g.f.: exp(x/(1+x)).
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23
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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
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0,6
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COMMENTS
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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)
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LINKS
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FORMULA
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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, 3-step).
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).
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).
(End)
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
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)^(n-k)*L(n,k)); L(n,k) the unsigned Lah numbers. - Peter Luschny, Oct 18 2014
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
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MATHEMATICA
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nn=30; CoefficientList[Series[Exp[x/(1+x)], {x, 0, nn}], x] Range[0, nn]! (* Harvey P. Dale, Jul 21 2011 *)
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PROG
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(Sage)
A111884 = lambda n: hypergeometric([-n+1, -n], [], -1)
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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