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a(n) = Sum_{k=0..floor(n/3)} binomial(2*n-3*k, n).
11

%I #29 Apr 08 2024 09:12:57

%S 1,2,6,21,75,273,1009,3770,14202,53846,205216,785460,3017106,11624580,

%T 44905518,173863965,674506059,2621371005,10203609597,39773263035,

%U 155231706951,606554343495,2372544034143,9289131196485,36401388236461

%N a(n) = Sum_{k=0..floor(n/3)} binomial(2*n-3*k, n).

%H Seiichi Manyama, <a href="/A105872/b105872.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: 2/(4*x^2+sqrt(1-4*x)*(3*x+1)-5*x+1). - _Vladimir Kruchinin_, May 24 2014

%F Conjecture: -3*(n+1)*(7*n-2)*a(n) +6*(7*n+5)*(2*n-1)*a(n-1) -(n+1)*(7*n-2)*a(n-2) +2*(7*n+5)*(2*n-1)*a(n-3)=0. - _R. J. Mathar_, Nov 28 2014

%F a(n) ~ 2^(2*n+3) / (7*sqrt(Pi*n)). - _Vaclav Kotesovec_, Jan 28 2023

%F a(n) = [x^n] 1/((1-x^3) * (1-x)^(n+1)). - _Seiichi Manyama_, Apr 08 2024

%t Table[Sum[Binomial[2n-3k,n],{k,0,Floor[n/2]}],{n,0,30}] (* _Harvey P. Dale_, Jan 13 2015 *)

%o (PARI) a(n) = sum(k=0, n\3, binomial(2*n-3*k, n)); \\ _Seiichi Manyama_, Jan 28 2023

%Y Cf. A144904, A360150, A360151, A360152, A360153.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Apr 23 2005

%E Erroneous title changed by _Paul Barry_, Apr 14 2010

%E Name corrected by _Seiichi Manyama_, Jan 28 2023