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A229464
Binomial transform of (2*n + 1)!.
2
1, 7, 133, 5419, 383785, 41782831, 6472067437, 1352114646163, 366325440650449, 124893891684358615, 52323557348796456661, 26420766706149889279867, 15824833185409769038803193, 11092546337733020334329204479, 8995627147680234199615065312445
OFFSET
0,2
COMMENTS
Companion sequence to A064570.
FORMULA
a(n) = Sum_{k = 0..n} binomial(n,k)*(2*k + 1)!.
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.
a(n) = Integral_{x >= 0} x*(1 + x^2)^n*exp(-x) dx.
a(n) = (2*n + 1)*A064570(n) - 2*n*A064570(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) = 7.
O.g.f.: Sum_{k >= 0} (2*k + 1)!*x^k/(1 - x)^(k + 1) = 1 + 7*x + 133*x^2 + 5419*x^3 + ....
a(n) ~ sqrt(Pi) * 2^(2*n + 2) * n^(2*n + 3/2) / exp(2*n). - Vaclav Kotesovec, Oct 30 2017
From Peter Bala, Nov 26 2017: (Start)
E.g.f.: exp(x)*Sum_{n >= 0} A000407(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, 7, 3, 9, 5, 1, 7, 3, 9, 5, ... with exact period 5. (End)
EXAMPLE
a(3) = 1*1! + 3*3! + 3*5! + 1*7! = 5419.
MATHEMATICA
Table[Sum[Binomial[n, k] * (2*k+1)!, {k, 0, n}], {n, 0, 15}] (* Vaclav Kotesovec, Oct 30 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Sep 25 2013
STATUS
approved