OFFSET
0,2
COMMENTS
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=5)^2.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..50
FORMULA
Let q = 25/sqrt(8), then g.f. A(x) satisfies:
(1) 1 + 5*x = Sum_{n>=0} (5^n + q * sqrt(A(x)))^n * x^n / (1 + 5^n * q * x*sqrt(A(x)))^(n+1).
(2) 1 + 5*x = Sum_{n>=0} (5^n - q * sqrt(A(x)))^n * x^n / (1 - 5^n * q * x*sqrt(A(x)))^(n+1).
EXAMPLE
G.f.: A(x) = 1 + 2750*x + 240792500*x^2 + 476435807656250*x^3 + 23281990342550349218750*x^4 + 28421639578439751858524316406250*x^5 + ...
Let q = 25/sqrt(8) and B = A(x)^(1/2), then
1 + 5*x = 1/(1 + x*q*B) + (5 + q*B)*x/(1 + 5*x*q*B)^2 + (5^2 + q*B)^2*x^2/(1 + 5^2*x*q*B)^3 + (5^3 + q*B)^3*x^3/(1 + 5^3*x*q*B)^4 + (5^4 + q*B)^4*x^4/(1 + 5^4*x*q*B)^5 + (5^5 + q*B)^5*x^5/(1 + 5^5*x*q*B)^6 + (5^6 + q*B)^6*x^6/(1 + 5^6*x*q*B)^7 + ...
and also
1 + 5*x = 1/(1 - x*q*B) + (5 - q*B)*x/(1 - 5*x*q*B)^2 + (5^2 - q*B)^2*x^2/(1 - 5^2*x*q*B)^3 + (5^3 - q*B)^3*x^3/(1 - 5^3*x*q*B)^4 + (5^4 - q*B)^4*x^4/(1 - 5^4*x*q*B)^5 + (5^5 - q*B)^5*x^5/(1 - 5^5*x*q*B)^6 + (5^6 - q*B)^6*x^6/(1 - 5^6*x*q*B)^7 + ...
PROG
(PARI) /* Set k = 5 to generate this sequence (requires high precision) */
{a(n, k) = my(q = k^2/sqrt(2*k-2), A=[1, k^3*(k^2-3), 0]); for(i=0, n,
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]}
for(n=0, 20, print1(a(n, k=5), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 26 2019
STATUS
approved