%I #15 Jul 06 2020 17:32:10
%S 1,2,8,45,297,2144,16398,130622,1072509,9015741,77229624,671868010,
%T 5921241337,52764270015,474699837123,4306641596007,39363068782364,
%U 362191362113221,3352866324085927,31210685632641522,292025240058727496,2745513045893833352,25929804402647536603,245958435266263341412,2342884864036837008480,22409497495190975013498
%N G.f. satisfies: A(x) = Sum_{n>=0} x^n / (1 - (n+1)*x*A(x)^2).
%H Paul D. Hanna, <a href="/A316367/b316367.txt">Table of n, a(n) for n = 0..300</a>
%H Sierra Brown, Spencer Daugherty, Eugene Fiorini, Barbara Maldonado, Diego Manzano-Ruiz, Sean Rainville, Riley Waechter, and Tony W. H. Wong, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Wong/wong24.html">Nimber Sequences of Node-Kayles Games</a>, J. Int. Seq., Vol. 23 (2020), Article 20.3.5.
%F G.f. A(x) satisfies:
%F (1) A(x) = Sum_{n>=0} x^n / (1 - (n+1)*x*A(x)^2).
%F (2) A(x) = Sum_{n>=0} n! * x^n/(1-x)^(n+1) * A(x)^(2*n) / Product_{k=1..n} (1 + k*x*A(x)^2).
%e G.f.: A(x) = 1 + 2*x + 8*x^2 + 45*x^3 + 297*x^4 + 2144*x^5 + 16398*x^6 + 130622*x^7 + 1072509*x^8 + 9015741*x^9 + 77229624*x^10 + ...
%e where we have the following series identity:
%e A(x) = 1/(1-x*A(x)^2) + x/(1-2*x*A(x)^2) + x^2/(1-3*x*A(x)^2) + x^3/(1-4*x*A(x)^2) + x^4/(1-5*x*A(x)^2) + x^5/(1-6*x*A(x)^2) + x^6/(1-7*x*A(x)^2) + ...
%e is equal to
%e A(x) = 1/(1-x) + x/(1-x)^2*A(x)^2/(1+x*A(x)^2) + 2!*x^2/(1-x)^3*A(x)^4/((1+x*A(x)^2)*(1+2*x*A(x)^2)) + 3!*x^3/(1-x)^4*A(x)^6/((1+x*A(x)^2)*(1+2*x*A(x)^2)*(1+3*x*A(x)^2)) + 4!*x^4/(1-x)^5*A(x)^8/((1+x*A(x)^2)*(1+2*x*A(x)^2)*(1+3*x*A(x)^2)*(1+4*x*A(x)^2)) + 5!*x^5/(1-x)^6*A(x)^10/((1+x*A(x)^2)*(1+2*x*A(x)^2)*(1+3*x*A(x)^2)*(1+4*x*A(x)^2)*(1+5*x*A(x)^2)) + ...
%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, x^m/(1 - (m+1)*x*A^2 +x*O(x^n)))); polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, m!*x^m*A^(2*m)/(1-x +x*O(x^n))^(m+1)/prod(k=1, m, 1 + k*x*A^2 +x*O(x^n)))); polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A026898, A245389.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jul 21 2018