%I #37 Apr 04 2024 09:47:34
%S 1,1,3,18,121,896,7028,57406,483080,4159169,36462855,324391132,
%T 2921210383,26576350332,243901358678,2255283941595,20991223674553,
%U 196508265126327,1849038158249933,17478100523106657,165891345107764059,1580380321767062796,15106335141526197636,144839560162346664092,1392621873057558622860,13424503737125805253734
%N G.f. A(x) satisfies A(x)^2 = 1 + x + x*A(x)^7.
%C Terms appear to equal A011792, apart from offset and an initial 1.
%C Note that the function G(x) = 1 + x*G(x)^3 (g.f. of A001764) also satisfies the condition: G(x) = 1/G(-x*G(x)^5).
%H Paul D. Hanna, <a href="/A295537/b295537.txt">Table of n, a(n) for n = 0..500</a>
%F G.f. A(x) satisfies:
%F (1) 1 + Series_Reversion( x/(1 + 3*x + 9*x^2 + 13*x^3 + 11*x^4 + 5*x^5 + x^6) ).
%F (2) F(A(x)) = x such that F(x) = -(1 - x^2)/(1 + x^7).
%F (3) A(x) = 1 / A(-x*A(x)^5).
%F a(n) ~ sqrt((1 + s^7)/(5*Pi)) / (2*n^(3/2)*r^(n - 1/2)), where r = 0.09760952485001949704447959455483068408354114893087... and s = 1.239618221385127629837391742046923301758368685253... are real roots of the system of equations 1 + r + r*s^7 = s^2, 7*r*s^5 = 2. - _Vaclav Kotesovec_, Nov 28 2017
%F From _Seiichi Manyama_, Apr 04 2024: (Start)
%F G.f. A(x) satisfies A(x) = 1 + x * (1 - A(x) + A(x)^2 - A(x)^3 + A(x)^4 - A(x)^5 + A(x)^6).
%F a(n) = Sum_{k=0..n} binomial(n,k) * binomial(7*k/2+1/2,n)/(7*k+1). (End)
%e G.f.: A(x) = 1 + x + 3*x^2 + 18*x^3 + 121*x^4 + 896*x^5 + 7028*x^6 + 57406*x^7 + 483080*x^8 + 4159169*x^9 + 36462855*x^10 + 324391132*x^11 + 2921210383*x^12 + 26576350332*x^13 + 243901358678*x^14 + 2255283941595*x^15 + ...
%e such that A(x)^2 = 1+x + x*A(x)^7.
%e RELATED SERIES.
%e A(x)^2 = 1 + 2*x + 7*x^2 + 42*x^3 + 287*x^4 + 2142*x^5 + 16898*x^6 + 138600*x^7 + 1170037*x^8 + 10098774*x^9 + 88712736*x^10 + ...
%e A(x)^7 = 1 + 7*x + 42*x^2 + 287*x^3 + 2142*x^4 + 16898*x^5 + 138600*x^6 + 1170037*x^7 + 10098774*x^8 + 88712736*x^9 + ...
%e A(-x*A(x)^5) = 1 - x - 2*x^2 - 13*x^3 - 84*x^4 - 616*x^5 - 4788*x^6 - 38865*x^7 - 325489*x^8 - 2791845*x^9 - 24401730*x^10 + ...
%e which equals 1/A(x).
%p a:= n-> coeff(series(RootOf(x*_Z^6-x*_Z^5+x*_Z^4-x*_Z^3
%p +x*_Z^2-(1+x)*_Z+1+x), x, n+1), x, n):
%p seq(a(n), n=0..25); # _Alois P. Heinz_, Dec 06 2017
%t m = 26; A[_] = 0;
%t Do[A[x_] = Sqrt[1 + x + x A[x]^7] + O[x]^m, {m}];
%t CoefficientList[A[x], x] (* _Jean-François Alcover_, Oct 02 2019 *)
%o (PARI) {a(n) = my(A=1+x); for(i=1, n, A = sqrt(1+x + x*A^7 +x*O(x^n)) ); polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) A295537(N=20)=Vec(serreverse('x/Ser(Polrev([1,3,9,13,11,5,1]),,N))+1) \\ Yields a vector with N terms. To compute only a(n) use polcoeff(...,n) instead of Vec(), and N = n+1. - _M. F. Hasler_, Mar 16 2018
%Y Cf. A011792, A259757, A295538.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Nov 27 2017