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
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..224
Programs
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Mathematica
Table[(2n+1)!Sum[(-1)^k/(2k+1)!,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Sep 14 2019 *) -
PARI
a(n) = (2*n+1)!*sum(k=0,n,(-1)^k/(2*k+1)!); \\ Michel Marcus, Sep 09 2016
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Sage
[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
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)