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Binomial transform of A002054.
2

%I #13 Jun 09 2019 06:38:13

%S 1,6,32,163,813,4013,19703,96477,471811,2306316,11274066,55128021,

%T 269688723,1320047550,6465047880,31682288305,155354522955,

%U 762235643450,3742013092520,18380748447345,90334417334095,444186621323945,2185193308168959,10755153057295603,52958560495897573

%N Binomial transform of A002054.

%H Vincenzo Librandi, <a href="/A034942/b034942.txt">Table of n, a(n) for n = 0..1000</a>

%H László Németh, <a href="https://arxiv.org/abs/1905.13475">Tetrahedron trinomial coefficient transform</a>, arXiv:1905.13475 [math.CO], 2019.

%F Recurrence: (n+3)*(3*n+1)*a(n) = 3*(6*n^2+15*n+11)*a(n-1) - 5*(n-1)*(3*n+4)*a(n-2). - _Vaclav Kotesovec_, Oct 08 2012

%F a(n) ~ 4*5^(n+1/2)/sqrt(Pi*n). - _Vaclav Kotesovec_, Oct 08 2012

%t Table[Sum[Binomial[n,k]*Binomial[2*k+3,k],{k,0,n}],{n,0,25}] (* _Vaclav Kotesovec_, Oct 08 2012 *)

%o (PARI) a(n)=sum(k=0,n,binomial(n,k)*binomial(2*k+3,k)); \\ _Joerg Arndt_, May 04 2013

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_.