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Sum of cubes of trinomial coefficients: a(n) = Sum_{k=0..2n} trinomial(n,k)^3 where trinomial(n,k) = [x^k] (1 + x + x^2)^n.
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%I #2 Mar 30 2012 18:37:04

%S 1,3,45,831,17181,375903,8530929,198643455,4714491357,113550338127,

%T 2767105469745,68077260387315,1688160321677025,42142679453321307,

%U 1058050429855640217,26695057057648257231,676431705046728704733

%N Sum of cubes of trinomial coefficients: a(n) = Sum_{k=0..2n} trinomial(n,k)^3 where trinomial(n,k) = [x^k] (1 + x + x^2)^n.

%o (PARI) a(n)=sum(k=0,2*n,polcoeff((1+x+x^2)^n,k)^3)

%Y Cf. A132304, A132305.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Aug 18 2007