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%I #20 Dec 04 2017 02:59:31
%S 1,7,133,5419,383785,41782831,6472067437,1352114646163,
%T 366325440650449,124893891684358615,52323557348796456661,
%U 26420766706149889279867,15824833185409769038803193,11092546337733020334329204479,8995627147680234199615065312445
%N Binomial transform of (2*n + 1)!.
%C Companion sequence to A064570.
%F a(n) = Sum_{k = 0..n} binomial(n,k)*(2*k + 1)!.
%F Clearly a(n) is always odd; indeed, for n >= 1, a(n) = 1 + 6*n*b(n-1), where b(n) = [1, 11, 301, 15991, 1392761, ...] is the binomial transform of A051618.
%F a(n) = Integral_{x >= 0} x*(1 + x^2)^n*exp(-x) dx.
%F a(n) = (2*n + 1)*A064570(n) - 2*n*A064570(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) = 7.
%F O.g.f.: Sum_{k >= 0} (2*k + 1)!*x^k/(1 - x)^(k + 1) = 1 + 7*x + 133*x^2 + 5419*x^3 + ....
%F a(n) ~ sqrt(Pi) * 2^(2*n + 2) * n^(2*n + 3/2) / exp(2*n). - _Vaclav Kotesovec_, Oct 30 2017
%F From _Peter Bala_, Nov 26 2017: (Start)
%F E.g.f.: exp(x)*Sum_{n >= 0} A000407(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, 7, 3, 9, 5, 1, 7, 3, 9, 5, ... with exact period 5. (End)
%e a(3) = 1*1! + 3*3! + 3*5! + 1*7! = 5419.
%t Table[Sum[Binomial[n, k] * (2*k+1)!, {k, 0, n}], {n, 0, 15}] (* _Vaclav Kotesovec_, Oct 30 2017 *)
%Y Cf. A000522, A051618, A064570, A294352, A000407.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Sep 25 2013
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