A275651 a(n) = (2*n)!*Sum_{k = 0..n} (-1)^k/(2*k)!.

1, 1, 13, 389, 21785, 1960649, 258805669, 47102631757, 11304631621681, 3459217276234385, 1314502564969066301, 607300185015708631061, 335229702128671164345673

Offset: 0

Author

Peter Bala, Sep 02 2016

Keywords

• nonn
• easy

Comments

Compare with the derangement numbers A000166(n) := n!*sum_{k = 0..n} (-1)^k/k! and also A074790.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..225

Crossrefs

Cf. A000166, A049470 (cos(1)), A051396, A073743, A074790.

Programs

Maple

A275651 := proc(n) option remember; if (n = 0) then 1 else 2*n*(2*n - 1)*A275651(n-1)+(-1)^n end if; end proc:
seq(A275651(n), n = 0..20);

Mathematica

Table[(2 n)!*Sum[(-1)^k/(2 k)!, {k, 0, n}], {n, 12}] (* Michael De Vlieger, Sep 04 2016 *)

Formula

a(n) ~ (2*n)!*cos(1).

E.g.f. for the aerated sequence: cos(x)/(1 - x^2) = 1 + x^2/2! + 13*x^4/4! + 389*x^6/6! + ....

Recurrence equations:

a(n) = 2*n*(2*n - 1)*a(n-1) + (-1)^n with a(0) = 1.

a(n) = (4*n^2 - 2*n - 1)*a(n - 1) + (2*n - 2)*(2*n - 3)*a(n - 2) with a(0) = 1, a(1) = 1.

The latter recurrence is also satisfied by the sequence b(n) := (2*n)! with b(0) = 1, b(1) = 2. This leads to the continued fraction representation a(n) = (2*n )!*( 1/(1 + 1/(1 + 2/(11 + 12/(29 + ... + (2*n - 2)*(2*n - 3)/(4*n^2 - 2*n - 1) )))) ) for n >= 3. Taking the limit gives the continued fraction representation cos(1) = A049470 = 1/(1 + 1/(1 + 2/(11 + 12/(29 + ... + (2*n - 2)*(2*n - 3)/((4*n^2 - 2*n - 1) + ... ))))). Cf. A073743.