login
a(n) = (2*n)!*Sum_{k = 0..n} (-1)^k/(2*k)!.
4

%I #14 Sep 09 2016 12:15:36

%S 1,1,13,389,21785,1960649,258805669,47102631757,11304631621681,

%T 3459217276234385,1314502564969066301,607300185015708631061,

%U 335229702128671164345673

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

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

%H Seiichi Manyama, <a href="/A275651/b275651.txt">Table of n, a(n) for n = 0..225</a>

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

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

%F Recurrence equations:

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

%F 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.

%F 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.

%p 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:

%p seq(A275651(n), n = 0..20);

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

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

%K nonn,easy

%O 0,3

%A _Peter Bala_, Sep 02 2016