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a(n) = binomial(n, floor((n-4)/2)).
6

%I #19 Jun 21 2022 05:07:49

%S 0,0,0,0,1,1,6,7,28,36,120,165,495,715,2002,3003,8008,12376,31824,

%T 50388,125970,203490,497420,817190,1961256,3268760,7726160,13037895,

%U 30421755,51895935,119759850,206253075

%N a(n) = binomial(n, floor((n-4)/2)).

%H G. C. Greubel, <a href="/A037956/b037956.txt">Table of n, a(n) for n = 0..1000</a>

%F E.g.f.: Bessel_I(4,2x) + Bessel_I(5,2x). - _Paul Barry_, Feb 28 2006

%F (n+5)*(n-4)*a(n) = -(n^2-3*n-20)*a(n-1) - (n^2-13*n-88)*a(n-2) + 2*(2*n+3)*(n-2)*a(n-3) +20*(n-2)*(n-3)*a(n-4). - _R. J. Mathar_, Nov 24 2012

%F From _Robert Israel_, Oct 28 2019: (Start)

%F G.f.: 16*x^4*(1+2*x+sqrt(1-4*x^2))/(sqrt(1-4*x^2)*(1+sqrt(1-4*x^2))^5).

%F Mathar's recurrence verified using the D.E. (4*x^4-x^2)*y'' + (16*x^3+2*x^2-2*x)*y' + (8*x^2+2*x+20)*y = 0 satisfied by the G.f. (End)

%p seq(binomial(n,floor((n-4)/2)),n=0..50); # _Robert Israel_, Oct 28 2019

%t Table[Binomial[n,Floor[(n-4)/2]],{n,0,40}] (* _Harvey P. Dale_, Mar 02 2015 *)

%o (Magma) [Binomial(n, Floor((n-4)/2)): n in [0..40]]; // _G. C. Greubel_, Jun 20 2022

%o (SageMath) [binomial(n, (n-4)//2) for n in (0..40)] # _G. C. Greubel_, Jun 20 2022

%Y Cf. A035951, A035952, A035953, A035954, A035955, A035957.

%Y Cf. A089940, A101491.

%K nonn

%O 0,7

%A _N. J. A. Sloane_