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Binomial transform of (2n)!.
6

%I #27 May 31 2019 17:27:33

%S 1,3,29,799,43353,3837851,501393973,90608944119,21633834338609,

%T 6593857931708083,2497877833687172301,1151118261673522046543,

%U 634098400947597342716809,411445662820653995008883019

%N Binomial transform of (2n)!.

%C Compare with A229464. - _Peter Bala_, Sep 25 2013

%H Vincenzo Librandi, <a href="/A064570/b064570.txt">Table of n, a(n) for n = 0..100</a>

%F In Maple notation: a(n)=hypergeom([1, 1/2, -n], [], -4), n=0, 1, ...

%F a(n) = Integral_{x>=0} ((x^4-1)/(x^2-1))^n*exp(-x) dx. - _Gerald McGarvey_, Oct 14 2006

%F From _Peter Bala_, Sep 25 2013: (Start)

%F a(n) = Sum_{k = 0..n} binomial(n,k)*(2*k)!.

%F Clearly a(n) is always odd; indeed, a(n) = 1 + 2*n*A229464(n-1) for n >= 1.

%F 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.

%F O.g.f. Sum_{k >= 0} (2*k)!*x^k/(1 - x)^(k + 1) = 1 + 3*x + 29*x^2 + 799*x^3 + .... (End)

%F 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

%F a(n) ~ sqrt(Pi) * 2^(2*n+1) * n^(2*n+1/2) / exp(2*n). - _Vaclav Kotesovec_, Sep 26 2013

%F From _Peter Bala_, Nov 26 2017: (Start)

%F E.g.f.: exp(x)*Sum_{n >= 0} A001813(n)*x^n.

%F a(k) = a(0) (mod k) for all k (by the inhomogeneous recurrence equation).

%F More generally a(n+k) = a(n) (mod k) for all n and k by an induction argument on n.

%F 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)

%t Table[Sum[Binomial[n,k]*(2*k)!,{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Sep 26 2013 *)

%Y Cf. A010050, A229464, A001813, A323280.

%K nonn,easy

%O 0,2

%A _Karol A. Penson_, Sep 20 2001