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%I #21 Oct 12 2023 10:26:49
%S 1,1,5,25,137,801,4893,30857,199377,1313089,8782389,59491257,
%T 407308377,2814044897,19594237133,137364464681,968743846561,
%U 6868059398273,48921561805413,349942779608153,2512722402972457,18104571857859233,130856263145140861,948520413875412681
%N Expansion of g.f. A(x) satisfying A(x) = 1 + x*A(x)/(1 - 2*x*A(x))^2.
%C Related identities which hold formally for all Maclaurin series F(x):
%C (1) F(x) = (1/x) * Sum{n>=1} n^(n-1) * x^n * F(x)^n / (1 + n*x*F(x))^(n+1),
%C (2) F(x) = (2/x) * Sum{n>=1} n*(n+1)^(n-2) * x^n * F(x)^n / (1 + (n+1)*x*F(x))^(n+1),
%C (3) F(x) = (3/x) * Sum{n>=1} n*(n+2)^(n-2) * x^n * F(x)^n / (1 + (n+2)*x*F(x))^(n+1),
%C (4) F(x) = (4/x) * Sum{n>=1} n*(n+3)^(n-2) * x^n * F(x)^n / (1 + (n+3)*x*F(x))^(n+1),
%C (5) F(x) = (k/x) * Sum{n>=1} n*(n+k-1)^(n-2) * x^n * F(x)^n / (1 + (n+k-1)*x*F(x))^(n+1) for all fixed nonzero k.
%H Paul D. Hanna, <a href="/A365772/b365772.txt">Table of n, a(n) for n = 0..400</a>
%F a(n) = Sum{k=0..n} binomial(n+1, n-k)/(n+1) * binomial(2*n-k-1, k) * 2^k.
%F Let A(x)^m = Sum_{n>=0} a(n,m) * x^n then a(n,m) = Sum_{k=0..n} binomial(n+m, n-k)*m/(n+m) * binomial(2*n-k-1, k) * 2^k.
%F G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
%F (1) A(x) = 1 + x*A(x)/(1 - 2*x*A(x))^2.
%F (2) A(x) = (1/x) * Series_Reversion( x/(1 + x/(1 - 2*x)^2) ).
%F (3) A( x/(1 + x/(1 - 2*x)^2) ) = 1 + x/(1 - 2*x)^2.
%F (4) A(x) = 1 + (m+1) * Sum{n>=1} n*(n+m)^(n-2) * x^n * A(x)^n / (1 + (n+m-2)*x*A(x))^(n+1) for all fixed nonnegative m.
%F (4.a) A(x) = 1 + Sum{n>=1} n^(n-1) * x^n * A(x)^n / (1 + (n-2)*x*A(x))^(n+1).
%F (4.b) A(x) = 1 + 2 * Sum{n>=1} n*(n+1)^(n-2) * x^n * A(x)^n / (1 + (n-1)*x*A(x))^(n+1).
%F (4.c) A(x) = 1 + 3 * Sum{n>=1} n*(n+2)^(n-2) * x^n * A(x)^n / (1 + n*x*A(x))^(n+1).
%F (4.d) A(x) = 1 + 4 * Sum{n>=1} n*(n+3)^(n-2) * x^n * A(x)^n / (1 + (n+1)*x*A(x))^(n+1).
%F a(n) ~ 7^(n + 3/2) * sqrt(3/((-1916 + (1833997600 - 95194848*sqrt(69))^(1/3) + 2^(5/3)*(57312425 + 2974839*sqrt(69))^(1/3))*Pi)) / (2 * n^(3/2) * (1 - 53*(2/(3*(-45 + 161*sqrt(69))))^(1/3) + ((-45 + 161*sqrt(69))/2)^(1/3)/3^(2/3))^n). - _Vaclav Kotesovec_, Oct 05 2023
%e G.f.: A(x) = 1 + x + 5*x^2 + 25*x^3 + 137*x^4 + 801*x^5 + 4893*x^6 + 30857*x^7 + 199377*x^8 + 1313089*x^9 + 8782389*x^10 + ...
%e where A(x) satisfies A(x) = 1 + x*A(x)/(1 - 2*x*A(x))^2
%e also
%e A(x) = 1 + x*A(x)/(1 + (-1)*x*A(x))^2 + 2*x^2*A(x)^2/(1 + 0*x*A(x))^3 + 3^2*x^3*A(x)^3/(1 + 1*x*A(x))^4 + 4^3*x^4*A(x)^4/(1 + 2*x*A(x))^5 + 5^4*x^5*A(x)^5/(1 + 3*x*A(x))^6 + 6^5*x^6*A(x)^6/(1 + 4*x*A(x))^7 + ...
%e and
%e A(x) = 1 + 3*1*3^(-1)*x*A(x)/(1 + 1*x*A(x))^2 + 3*2*4^0*x^2*A(x)^2/(1 + 2*x*A(x))^3 + 3*3*5^1*x^3*A(x)^3/(1 + 3*x*A(x))^4 + 3*4*6^2*x^4*A(x)^4/(1 + 4*x*A(x))^5 + 3*5*7^3*x^5*A(x)^5/(1 + 5*x*A(x))^6 + ...
%t nmax = 30; A[_] = 0; Do[A[x_] = 1 + x*A[x]/(1 - 2*x*A[x])^2 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] (* _Vaclav Kotesovec_, Oct 05 2023 *)
%o (PARI) {a(n) = sum(k=0, n, binomial(n+1, n-k)/(n+1) * binomial(2*n-k-1, k) * 2^k)}
%o for(n=0,30, print1(a(n),", "))
%o (PARI) {a(n) = my(A = (1/x) * serreverse( x/(1 + x/(1 - 2*x +O(x^(n+2)) )^2) ) ); polcoeff(A,n)}
%o for(n=0,30, print1(a(n),", "))
%Y Cf. A365770, A109081, A365773, A365774, A365775.
%Y Cf. A366232 (dual).
%K nonn
%O 0,3
%A _Paul D. Hanna_, Oct 04 2023
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