A165433 A transform of the double factorial numbers A001147.
1, 1, 2, 3, 7, 14, 39, 97, 308, 897, 3139, 10304, 38997, 140893, 570002, 2230599, 9567979, 40091222, 181203603, 805962157, 3819522284, 17912075229, 88646095447, 435959031488, 2245454002137, 11530035000169, 61627679281154
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..800
Programs
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Magma
[(&+[Binomial(n-k,k)*Factorial(2*k)/(Factorial(k)*2^k): k in [0.. Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Oct 20 2018
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Maple
a:=proc(n) add(binomial(n-k,k)*factorial(2*k)/(factorial(k)*2^k),k=0..floor(n/2)) end proc: seq(a(n),n=0..30); # Muniru A Asiru, Oct 20 2018
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Mathematica
Table[Sum[Binomial[n-k, k]*(2*k)!/(k!*2^k), {k, 0, Floor[n/2]}], {n,0, 30}] (* G. C. Greubel, Oct 20 2018 *) -
PARI
vector(30, n, n--; sum(k=0, floor(n/2), binomial(n-k,k)*(2*k)!/(k!*2^k))) \\ G. C. Greubel, Oct 20 2018
Formula
G.f.: 1/(1-x-x^2-2x^4/(1-x-5x^2-12x^4/(1-x-9x^2-30x^4/(1-x-13x^2-56x^4/(1-.... (continued fraction);
a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*(2k)!/(k!*2^k).
Conjecture: 2*a(n) -3*a(n-1) +(3-2*n)*a(n-2) +(2*n-3)*a(n-3)=0. - R. J. Mathar, Nov 14 2011
G.f.: T(0)/(1-x), where T(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 29 2013
a(n) ~ 2^(-1/2) * exp(sqrt(n)/2 - n/2 + 1/16) * n^(n/2) * (1 + 121/(192*sqrt(n))). - Vaclav Kotesovec, Apr 18 2024
Comments