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a(n) = binomial(2+2*n, 2+n) * binomial(7+2*n, n).
1

%I #6 Nov 08 2019 20:52:03

%S 1,36,825,16016,286650,4900896,81477396,1330243200,21455160012,

%T 343138081000,5455289950110,86359817849760,1362899694292500,

%U 21460589553110400,337374701839395000,5297540558850547200,83114164698623615700,1303247055281641470000,20427480491760087405660

%N a(n) = binomial(2+2*n, 2+n) * binomial(7+2*n, n).

%H Andrew Howroyd, <a href="/A114238/b114238.txt">Table of n, a(n) for n = 0..200</a>

%e if n=0 then C(2+2*0,2+0)*C(7+2*0,0+0) = C(2,2)*C(7,0) = 1*1 = 1.

%e if n=8 then C(2+2*8,2+8)*C(7+2*8,0+8) = C(18,10)*C(23,8) = 43758*490314 = 21455160012.

%o (PARI) a(n)={binomial(2+2*n, 2+n) * binomial(7+2*n, 0+n)} \\ _Andrew Howroyd_, Nov 08 2019

%Y Cf. A062190.

%K easy,nonn

%O 0,2

%A _Zerinvary Lajos_, Feb 04 2006

%E Terms a(13) and beyond from _Andrew Howroyd_, Nov 08 2019