A321602 G.f. A(x) satisfies: 1 = Sum_{n>=0} ((1+x)^(2*n) - A(x))^n.

1, 2, 5, 68, 1521, 45328, 1660032, 71548008, 3533826841, 196432984748, 12128132342482, 823366216285428, 60966207548525287, 4890600994792550264, 422601696583826709492, 39142599000082019249968, 3869325702147169825040193, 406650337650126697706078146, 45281361448272561712508294157, 5325916931170845646048163850556, 659842223101960470758187538118437

Offset: 0

Author

Paul D. Hanna, Nov 14 2018

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + 2*x + 5*x^2 + 68*x^3 + 1521*x^4 + 45328*x^5 + 1660032*x^6 + 71548008*x^7 + 3533826841*x^8 + 196432984748*x^9 + 12128132342482*x^10 + ...
such that
1 = 1  +  ((1+x)^2 - A(x))  +  ((1+x)^4 - A(x))^2  +  ((1+x)^6 - A(x))^3  +  ((1+x)^8 - A(x))^4  +  ((1+x)^10 - A(x))^5  +  ((1+x)^12 - A(x))^6  +  ((1+x)^14 - A(x))^7 + ...
Also,
1 = 1/(1 + A(x))  +  (1+x)^2/(1 + (1+x)^2*A(x))^2  +  (1+x)^8/(1 + (1+x)^4*A(x))^3  +  (1+x)^18/(1 + (1+x)^6*A(x))^4  +  (1+x)^32/(1 + (1+x)^8*A(x))^5  +  (1+x)^50/(1 + (1+x)^10*A(x))^6  + ...
RELATED SERIES.
The logarithmic derivative of the g.f. begins
A'(x)/A(x) = 2 + 6*x + 182*x^2 + 5554*x^3 + 211172*x^4 + 9397920*x^5 + 476737830*x^6 + 27086036234*x^7 + 1702330030676*x^8 + ...
the coefficients of which are all even:
(1/2) * A'(x)/A(x) = 1 + 3*x + 91*x^2 + 2777*x^3 + 105586*x^4 + 4698960*x^5 + 238368915*x^6 + 13543018117*x^7 + 851165015338*x^8 + ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A303056, A321603, A321604, A321605.

Cf. A326262.

Programs

PARI

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

Formula

G.f. A(x) satisfies:

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

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