login
Self-convolution cube of A120590, such that a(n) = 4*A120590(n) for n>=2.
2

%I #3 Mar 30 2012 18:36:58

%S 1,3,12,76,600,5304,50232,498360,5112756,53796820,577370508,

%T 6295961100,69557631936,776913430272,8758443555360,99527014659360,

%U 1138832618425272,13110313153525272,151738042878341400,1764609260161776600

%N Self-convolution cube of A120590, such that a(n) = 4*A120590(n) for n>=2.

%e A(x) = 1 + 3*x + 12*x^2 + 76*x^3 + 600*x^4 + 5304*x^5 + 50232*x^6 +...

%e A(x)^(1/3) = 1 + x + 3*x^2 + 19*x^3 + 150*x^4 + 1326*x^5 + 12558*x^6 +...

%o (PARI) {a(n)=local(A=1+x+3*x^2+x*O(x^n));for(i=0,n,A=A-4*A+3+x+A^3);polcoeff(A^3,n)}

%Y Cf. A120590 (A(x)^(1/3)); A120588, A120592 - A120607.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Jun 16 2006