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G.f. A(x) satisfies: 1 + 2*x = Sum_{n>=0} (2^n + q*sqrt(A(x)))^n * x^n / (1 + 2^n*q*x*sqrt(A(x)))^(n+1), where q = sqrt(8).
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%I #17 Mar 29 2019 21:41:27

%S 1,8,2504,1721024,4004836544,34371837694976,1144244247852783104,

%T 150756716920877640704000,79170171472489778115703918592,

%U 166116745151581288207313820694446080,1393704864012762439255232236469539465232384,46767149231853952372202508147249496634017589755904,6277066566013980877709014174399846162076444100155305361408

%N G.f. A(x) satisfies: 1 + 2*x = Sum_{n>=0} (2^n + q*sqrt(A(x)))^n * x^n / (1 + 2^n*q*x*sqrt(A(x)))^(n+1), where q = sqrt(8).

%C For a given integer k > 1, there exists an integer series G(x,k)^2 that satisfies: 1 + k*x = Sum_{n>=0} (k^n + q*G(x,k))^n * x^n / (1 + k^n*q*x*G(x,k))^(n+1) iff q^2 = k^4/(2*k-2). In that case, G(x,k)^2 = 1 + k^3*(k^2-3)*x + k^4*(2*k^8 - 18*k^4 + 21*k^2 + 2*k + 1)*x^2/2 + ...; the g.f. for this sequence is A(x) = G(x,k=2)^2.

%H Paul D. Hanna, <a href="/A325062/b325062.txt">Table of n, a(n) for n = 0..50</a>

%F Let q = sqrt(8), then g.f. A(x) satisfies:

%F (1) 1 + 2*x = Sum_{n>=0} (2^n + q * sqrt(A(x)))^n * x^n / (1 + 2^n * q * x*sqrt(A(x)))^(n+1).

%F (2) 1 + 2*x = Sum_{n>=0} (2^n - q * sqrt(A(x)))^n * x^n / (1 - 2^n * q * x*sqrt(A(x)))^(n+1).

%e G.f.: A(x) = 1 + 8*x + 2504*x^2 + 1721024*x^3 + 4004836544*x^4 + 34371837694976*x^5 + 1144244247852783104*x^6 + 150756716920877640704000*x^7 + ...

%e Let q = sqrt(8) and B = A(x)^(1/2), then

%e 1 + 2*x = 1/(1 + x*q*B) + (2 + q*B)*x/(1 + 2*x*q*B)^2 + (2^2 + q*B)^2*x^2/(1 + 2^2*x*q*B)^3 + (2^3 + q*B)^3*x^3/(1 + 2^3*x*q*B)^4 + (2^4 + q*B)^4*x^4/(1 + 2^4*x*q*B)^5 + (2^5 + q*B)^5*x^5/(1 + 2^5*x*q*B)^6 + (2^6 + q*B)^6*x^6/(1 + 2^6*x*q*B)^7 + ...

%e and also

%e 1 + 2*x = 1/(1 - x*q*B) + (2 - q*B)*x/(1 - 2*x*q*B)^2 + (2^2 - q*B)^2*x^2/(1 - 2^2*x*q*B)^3 + (2^3 - q*B)^3*x^3/(1 - 2^3*x*q*B)^4 + (2^4 - q*B)^4*x^4/(1 - 2^4*x*q*B)^5 + (2^5 - q*B)^5*x^5/(1 - 2^5*x*q*B)^6 + (2^6 - q*B)^6*x^6/(1 - 2^6*x*q*B)^7 + ...

%o (PARI) /* Set k = 2 to generate this sequence (requires high precision) */

%o {a(n,k) = my(q = k^2/sqrt(2*k-2), A=[1, k^3*(k^2-3), 0]); for(i=0, n,

%o A=concat(A, 0); A[#A-1] = round( polcoeff( sum(m=0, #A, (k^m + q * Ser(A)^(1/2))^m * x^m / (1 + k^m * q * x*Ser(A)^(1/2))^(m+1) ), #A)/k^4)); A[n+1]}

%o for(n=0, 20, print1(a(n,k=2), ", "))

%Y Cf. A325063, A325064, A325065, A325066, A324299.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Mar 26 2019