A074790 a(n) = (2*n+1)!*Sum_{k=0..n} (-1)^k/(2*k+1)!.

1, 5, 101, 4241, 305353, 33588829, 5239857325, 1100370038249, 299300650403729, 102360822438075317, 42991545423991633141, 21753721984539766369345, 13052233190723859821607001, 9162667699888149594768114701

Offset

0,2

Links

Seiichi Manyama, Table of n, a(n) for n = 0..224

Formula

a(n) = round(sin(1)*(2*n+1)!).

a(n) = A009551(2*n+1).

From Peter Bala, Jan 30 2015: (Start)

G.f.: sin(x)/(1 - x^2) = x + 5*x^3/3! + 101*x^5/5! + 4241*x^7/7! + ....

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-1)*(2*n-2)*a(n-2) with a(0) = 1, a(1) = 5.

The sequence b(n) := (2*n + 1)! also satisfies the second recurrence but with b(0) = 1, b(1) = 6. This leads to the continued fraction representation a(n) = (2*n + 1)!*(1 - 1/(6 + 6/(19 + 20/(41 + ... + (2*n - 1)*(2*n - 2)/(4*n^2 + 2*n - 1) )))) for n >= 2. Taking the limit gives the continued fraction representation sin(1) = 1 - 1/(6 + 6/(19 + 20/(41 + ... + (2*n - 1)*(2*n - 2)/((4*n^2 + 2*n - 1) + ... )))). (End)

Mathematica

Table[(2n+1)!Sum[(-1)^k/(2k+1)!, {k, 0, n}], {n, 0, 20}] (* Harvey P. Dale, Sep 14 2019 *)

Prog

(PARI) a(n) = (2*n+1)!*sum(k=0, n, (-1)^k/(2*k+1)!); \\ Michel Marcus, Sep 09 2016
(SageMath) [factorial(2*n+1)*sum((-1)^j/factorial(2*j+1) for j in (0..n)) for n in (0..20)] # G. C. Greubel, Jul 09 2021

Crossrefs

Cf. A000166, A009551, A049469, A051397.

Sequence in context: A266235 A266452 A354332 * A009757 A119556 A394683

Adjacent sequences: A074787 A074788 A074789 * A074791 A074792 A074793

Keyword

nonn,easy

Author

Benoit Cloitre, Sep 07 2002

Status

approved