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G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} T(n,k)^2 * x^k] / A(x)^n * x^n/n ), where T(n,k) is the coefficient of x^k in (1 + x + 2*x^2)^n.
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%I #5 Mar 03 2015 03:30:57

%S 1,1,1,5,8,8,16,28,48,80,128,208,320,512,768,1216,1792,2816,4096,6400,

%T 9216,14336,20480,31744,45056,69632,98304,151552,212992,327680,458752,

%U 704512,983040,1507328,2097152,3211264,4456448,6815744,9437184,14417920,19922944,30408704,41943040

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

%C More generally, if G(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} T(n,k)^2 * x^k] / G(x)^n * x^n/n ), where T(n,k) is the coefficient of x^k in (p + q*x + r*x^2)^n, then G(x) = (1 + p^2*x)*(1 + r^2*x^3)*(1 + (q^2-2*p*r)*x^2 + p^2*r^2*x^4) / (1-p*r*x^2)^2.

%F G.f.: (1 + x)*(1 + 4*x^3)*(1 - 3*x^2 + 4*x^4) / (1 - 2*x^2)^2.

%e G.f.: A(x) = 1 + x + x^2 + 5*x^3 + 8*x^4 + 8*x^5 + 16*x^6 + 28*x^7 +...

%e where

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

%e (1 + 2^2*x + 5^2*x^2 + 4^2*x^3 + 4^2*x^4)/A(x)^2 * x^2/2 +

%e (1 + 3^2*x + 9^2*x^2 + 13^2*x^3 + 18^2*x^4 + 12^2*x^5 + 8^2*x^6)/A(x)^3 * x^3/3 +

%e (1 + 4^2*x + 14^2*x^2 + 28^2*x^3 + 49^2*x^4 + 56^2*x^5 + 56^2*x^6 + 32^2*x^7 + 16^2*x^8)/A(x)^4 * x^4/4 +

%e (1 + 5^2*x + 20^2*x^2 + 50^2*x^3 + 105^2*x^4 + 161^2*x^5 + 210^2*x^6 + 200^2*x^7 + 160^2*x^8 + 80^2*x^9 + 32^2*x^10)/A(x)^5 * x^5/5 +...

%e which involves the squares of coefficients in (1 + x + 2*x^2)^n - see triangle A084600.

%o (PARI) /* By Definition: */

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

%o for(n=0, 50, print1(a(n), ", "))

%o (PARI) /* By G.F. Identity (faster): */

%o {a(n,p=1,q=1,r=2)=polcoeff( (1 + p^2*x)*(1 + r^2*x^3)*(1 + (q^2-2*p*r)*x^2 + p^2*r^2*x^4) / ((1-p*r*x^2)^2 +x*O(x^n)), n)}

%o for(n=0, 50, print1(a(n), ", "))

%Y Cf. A084600, A251688, A251689, A200537.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Mar 03 2015