A386665 G.f. satisfies: A(x) = Sum_{n>=0} ( (1+x)^n - A(x)^(-1/2) )^n / ( 2 - (1+x)^n * A(x)^(-1/2) )^(n+1).

1, 1, 8, 90, 1336, 24406, 530234, 13410942, 388841734, 12762735148, 469004980720, 19105730068460, 855146084504046, 41724450644602328, 2204075802189470532, 125300401263988607716, 7626356269363721248332, 494723229572772238087966, 34070289390944902842701094, 2482276670026891882801017812

Offset: 0

Author

Paul D. Hanna, Aug 29 2025

Keywords

• nonn
• new

Comments

It appears that lim_{n->oo} ( a(n+1)/a(n) )/(n+1) exists and is near 4.

Examples

			G.f.: A(x) = 1 + x + 8*x^2 + 90*x^3 + 1336*x^4 + 24406*x^5 + 530234*x^6 + 13410942*x^7 + 388841734*x^8 + 12762735148*x^9 + 469004980720*x^10 + ...
RELATED SERIES.
A(x)^(1/2) = 1 + 2*(x/4) + 62*(x/4)^2 + 2756*(x/4)^3 + 163574*(x/4)^4 + 11997852*(x/4)^5 + 1047984172*(x/4)^6 + 106571791752*(x/4)^7 + 12417003030694*(x/4)^8 + ...
A(x)^(-1/2) = 1 - 2*(x/4) - 58*(x/4)^2 - 2516*(x/4)^3 - 149434*(x/4)^4 - 11055996*(x/4)^5 - 976190180*(x/4)^6 - 100318703592*(x/4)^7 - 11796814729146*(x/4)^8 - ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..250

Crossrefs

Cf. A322735, A317350, A322737.

Programs

PARI

{a(n) = my(A=[1, 1]); for(i=1, n, A=concat(A, 0); A = Vec( sum(m=0, #A, ( (1+x)^m - Ser(A)^(-1/2) )^m / (2 - (1+x)^m*Ser(A)^(-1/2))^(m+1) ) ) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n along with B(x) = A(x)^(1/2) satisfies the following formulas.

(1) A(x) = Sum_{n>=0} ( (1+x)^n - 1/B(x) )^n / ( 2 - (1+x)^n/B(x) )^(n+1).

(2) A(x) = Sum_{n>=0} ( (1+x)^n + 1/B(x) )^n / ( 2 + (1+x)^n/B(x) )^(n+1).

(3) B(x) = Sum_{n>=0} ( (1+x)^n*B(x) - 1 )^n / ( 2*B(x) - (1+x)^n )^(n+1).

(4) B(x) = Sum_{n>=0} ( (1+x)^n*B(x) + 1 )^n / ( 2*B(x) + (1+x)^n )^(n+1).