A229464 Binomial transform of (2*n + 1)!.

1, 7, 133, 5419, 383785, 41782831, 6472067437, 1352114646163, 366325440650449, 124893891684358615, 52323557348796456661, 26420766706149889279867, 15824833185409769038803193, 11092546337733020334329204479, 8995627147680234199615065312445

Offset

0,2

Comments

Companion sequence to A064570.

Links

Table of n, a(n) for n=0..14.

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

Cf. A000522, A051618, A064570, A294352, A000407.

Sequence in context: A274258 A251577 A082164 * A317216 A119670 A003374

Adjacent sequences: A229461 A229462 A229463 * A229465 A229466 A229467

Keyword

nonn,easy

Author

Peter Bala, Sep 25 2013

Status

approved