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A200535 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} C(2*n,k)^2 * x^k] / A(x)^n * x^n/n ).

%I A200535 #9 Jun 30 2023 18:54:40
%S A200535 1,1,4,5,9,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,
%T A200535 88,92,96,100,104,108,112,116,120,124,128,132,136,140,144,148,152,156,
%U A200535 160,164,168,172,176,180,184,188,192,196,200,204,208,212,216,220,224,228,232,236,240,244,248,252,256,260
%N A200535 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} C(2*n,k)^2 * x^k] / A(x)^n * x^n/n ).
%H A200535 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1).
%F A200535 G.f.: (1+x^2)^2*(1+x^3)/((1-x)*(1-x^2)).
%e A200535 G.f.: A(x) = 1 + x + 4*x^2 + 5*x^3 + 9*x^4 + 12*x^5 + 16*x^6 + 20*x^7 +...
%e A200535 where A(x) = G(x/A(x)) and G(x) = A(x*G(x)) is the g.f. of A199257:
%e A200535 G(x) = 1 + x + 5*x^2 + 18*x^3 + 86*x^4 + 408*x^5 + 2075*x^6 +...
%e A200535 ...
%e A200535 The logarithm of the g.f. A(x) equals the series:
%e A200535 log(A(x)) = (1 + 2^2*x + x^2)/A(x) * x +
%e A200535 (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4)/A(x)^2 * x^2/2 +
%e A200535 (1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)/A(x)^3 * x^3/3 +
%e A200535 (1 + 8^2*x + 28^2*x^2 + 56^2*x^3 + 70^2*x^4 + 56^2*x^5 + 28^2*x^6 + 8^2*x^7 + x^8)/A(x)^4 * x^4/4 +
%e A200535 (1 + 10^2*x + 45^2*x^2 + 120^2*x^3 + 210^2*x^4 + 252^2*x^5 + 210^2*x^6 + 120^2*x^7 + 45^2*x^8 + 10^2*x^9 + x^10)/A(x)^5 * x^5/5 +...
%e A200535 which involves the squares of binomial coefficients C(2*n,k).
%o A200535 (PARI) {a(n)=polcoeff((1+x^2)^2*(1+x^3)/((1-x)*(1-x^2) +x*O(x^n)),n)}
%o A200535 (PARI) {a(n)=local(A=1+x);for(i=1,n,A=exp(sum(m=1, n, sum(k=0, n, binomial(2*m, k)^2 *x^k)/(A+x*O(x^n))^m *x^m/m)));polcoeff(A, n)}
%Y A200535 Cf. A199257.
%K A200535 nonn
%O A200535 0,3
%A A200535 _Paul D. Hanna_, Nov 18 2011