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A200475 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} A027907(n,k)^2 * x^k * A(x)^(2*k)] * x^n*A(x)^n/n ), where A027907 is the triangle of trinomial coefficients. 2

%I #5 Mar 30 2012 18:37:32

%S 1,1,3,13,65,350,1981,11627,70132,432090,2707595,17202779,110563543,

%T 717547090,4695774335,30952628861,205318395288,1369539030021,

%U 9180527051187,61813112864984,417850301293691,2834802846097200,19294989810689802,131723105933867817,901709774424393614

%N G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} A027907(n,k)^2 * x^k * A(x)^(2*k)] * x^n*A(x)^n/n ), where A027907 is the triangle of trinomial coefficients.

%C Trinomial coefficients satisfy: Sum_{k=0..2*n} A027907(n,k)*x^k = (1+x+x^2)^n.

%F G.f. satisfies: A(x) = (1 + x^3*A(x)^6)*(1 + x^6*A(x)^12)/((1 - x*A(x)^2)*(1 - x^4*A(x)^8)).

%e G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 65*x^4 + 350*x^5 + 1981*x^6 +...

%e Let A = g.f. A(x), then the logarithm of the g.f. equals the series:

%e log(A(x)) = (1 + x*A^2 + x^2*A^4)*x*A +

%e (1 + 2^2*x*A^2 + 3^2*x^2*A^4 + 2^2*x^3*A^6 + x^4*A^8)*x^2*A^2/2 +

%e (1 + 3^2*x*A^2 + 6^2*x^2*A^4 + 7^2*x^3*A^6 + 6^2*x^4*A^8 + 3^2*x^5*A^10 + x^6*A^12)*x^3*A^3/3 +

%e (1 + 4^2*x*A^2 + 10^2*x^2*A^4 + 16^2*x^3*A^6 + 19^2*x^4*A^8 + 16^2*x^5*A^10 + 10^2*x^6*A^12 + 4^2*x^7*A^14 + x^8*A^16)*x^4*A^4/4 +

%e (1 + 5^2*x*A^2 + 15^2*x^2*A^4 + 30^2*x^3*A^6 + 45^2*x^4*A^8 + 51^2*x^5*A^10 + 45^2*x^6*A^12 + 30^2*x^7*A^14 + 15^2*x^8*A^16 + 5^2*x^9*A^18 + x^10*A^20)*x^5*A^5/5 +...

%e which involves the squares of the trinomial coefficients A027907(n,k).

%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=(1-x*A^2+x^3*A^6-x^5*A^10+x^6*A^12)/(1-x*A^2+x*O(x^n))^2);polcoeff(A,n)}

%o (PARI) /* G.f. A(x) using the squares of the trinomial coefficients */

%o {A027907(n, k)=polcoeff((1+x+x^2)^n, k)}

%o {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=0, 2*m, A027907(m, k)^2 *x^k*(A+x*O(x^n))^(2*k))*x^m*A^m/m))); polcoeff(A, n)}

%Y Cf. A200377, A199248, A186236, A199257, A168592, A027907.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Nov 18 2011

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