OFFSET
0,2
COMMENTS
Compare with A229464. - Peter Bala, Sep 25 2013
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..100
FORMULA
In Maple notation: a(n)=hypergeom([1, 1/2, -n], [], -4), n=0, 1, ...
a(n) = Integral_{x>=0} ((x^4-1)/(x^2-1))^n*exp(-x) dx. - Gerald McGarvey, Oct 14 2006
From Peter Bala, Sep 25 2013: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*(2*k)!.
Clearly a(n) is always odd; indeed, a(n) = 1 + 2*n*A229464(n-1) for n >= 1.
Recurrence equation: a(n) = 1 + 2*n*(2*n - 1)*a(n-1) - 2*n*(2*n - 2)*a(n-2) with a(0) = 1 and a(1) = 3.
O.g.f. Sum_{k >= 0} (2*k)!*x^k/(1 - x)^(k + 1) = 1 + 3*x + 29*x^2 + 799*x^3 + .... (End)
Recurrence (homogeneous): a(n) = (4*n^2 - 2*n + 1)*a(n-1) - 2*(n-1)*(4*n-3)*a(n-2) + 4*(n-2)*(n-1)*a(n-3). - Vaclav Kotesovec, Sep 26 2013
a(n) ~ sqrt(Pi) * 2^(2*n+1) * n^(2*n+1/2) / exp(2*n). - Vaclav Kotesovec, Sep 26 2013
From Peter Bala, Nov 26 2017: (Start)
E.g.f.: exp(x)*Sum_{n >= 0} A001813(n)*x^n.
a(k) = a(0) (mod k) for all k (by the inhomogeneous recurrence equation).
More generally a(n+k) = a(n) (mod k) for all n and k by an induction argument on n.
It follows that for each positive integer k, the sequence a(n) (mod k) is periodic, with the exact period dividing k. For example, modulo 10 the sequence becomes 1, 3, 9, 9, 3, 1, 3, 9, 9, 3, ... with exact period 5. (End)
MATHEMATICA
Table[Sum[Binomial[n, k]*(2*k)!, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Sep 26 2013 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Karol A. Penson, Sep 20 2001
STATUS
approved