OFFSET
0,3
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..449
FORMULA
a(n) = Sum_{k=0..floor(n/2)} n!/(n-2*k)!.
a(n) = n*(n-1)*a(n-2) + 1. - Vladeta Jovovic, Aug 24 2004
a(n) = Sum_{k=0..n} binomial(n, k)*(1+(-1)^k)k!/2. Binomial transform of A010050 (with interpolated zeros). - Paul Barry, Sep 14 2004
a(n) = Sum_{k=0..n} P(n, k)[1, 0, 1, 0, 1, 0, ...](k). - Ross La Haye, Aug 29 2005
a(n) = (1/(2*exp(1))) * (Integral_{t=0..2} t^n*exp(1-abs(1-t)) dt + Integral_{t=0..oo} ((2+t)^n + (-t)^n) * exp(-t) dt). - Groux Roland, Jan 15 2011
E.g.f.: 1/U(0) where U(k) = 1 - x^2/(1 - 1/(1 + x*(k+1)/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Oct 16 2012
If n is even then a(n) ~ n!*(e/2 + 1/(2*e)) = 1.543080634815243... * n!, if n is odd then a(n) ~ n!*(e/2 - 1/(2*e)) = 1.175201193643801... * n!. - Vaclav Kotesovec, Nov 20 2012
Conjecture: a(n) -a(n-1) -n*(n-1)*a(n-2) +(n-1)*(n-2)*a(n-3)=0. - R. J. Mathar, May 29 2013
From Peter Bala, Sep 05 2022: (Start)
The e.g.f. A(x) satisfies the differential equation (x^2 - 1)*A'(x) + (1 + 2*x - x^2)*A(x) = 0 with A(0) = 1. Mathar's recurrence above follows from this.
For k a positive integer, reducing the sequence modulo k produces a purely periodic sequence whose period divides k. For example, modulo 5 the sequence becomes [1, 1, 3, 2, 2, 1, 1, 3, 2, 2, ...] of period 5. (End)
MATHEMATICA
With[{nn=20}, CoefficientList[Series[Exp[x]/(1-x^2), {x, 0, nn}], x] Range[ 0, nn]!] (* Harvey P. Dale, Aug 11 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladeta Jovovic, Oct 19 2003
EXTENSIONS
Definition clarified by Harvey P. Dale, Aug 11 2017
STATUS
approved