A303056 G.f. A(x) satisfies: 1 = Sum_{n>=0} ((1+x)^n - A(x))^n.

1, 1, 1, 8, 89, 1326, 24247, 521764, 12867985, 357229785, 11017306489, 373675921093, 13825260663882, 554216064798423, 23934356706763264, 1108017262467214486, 54747529760516714323, 2876096694574711401525, 160092696678371426933342, 9413031424290635395882462, 583000844360279565483710624

Offset: 0

Author

Paul D. Hanna, Apr 19 2018

Keywords

• nonn

Comments

More generally, the following sums are equal:

(1) Sum_{n>=0} binomial(n+k-1, n) * r^n * (p + q^n)^n,

(2) Sum_{n>=0} binomial(n+k-1, n) * r^n * q^(n^2) / (1 - r*p*q^n)^(n+k),

for any fixed integer k; here, k = 1 with r = 1, p = -A(x), q = (1+x). - Paul D. Hanna, Jun 22 2019

Examples

			G.f.: A(x) = 1 + x + x^2 + 8*x^3 + 89*x^4 + 1326*x^5 + 24247*x^6 + 521764*x^7 + 12867985*x^8 + 357229785*x^9 + 11017306489*x^10 + ...
such that
1 = 1  +  ((1+x) - A(x))  +  ((1+x)^2 - A(x))^2  +  ((1+x)^3 - A(x))^3  +  ((1+x)^4 - A(x))^4  +  ((1+x)^5 - A(x))^5  +  ((1+x)^6 - A(x))^6  +  ((1+x)^7 - A(x))^7 + ...
Also,
1 = 1/(1 + A(x))  +  (1+x)/(1 + (1+x)*A(x))^2  +  (1+x)^4/(1 + (1+x)^2*A(x))^3  +  (1+x)^9/(1 + (1+x)^3*A(x))^4  +  (1+x)^16/(1 + (1+x)^4*A(x))^5  +  (1+x)^25/(1 + (1+x)^5*A(x))^6  +  (1+x)^36/(1 + (1+x)^6*A(x))^7 + ...
RELATED SERIES.
log(A(x)) = x + x^2/2 + 22*x^3/3 + 325*x^4/4 + 6186*x^5/5 + 137380*x^6/6 + 3478651*x^7/7 + 98674253*x^8/8 + 3096911434*x^9/9 + ...
PARTICULAR VALUES.
Although the power series A(x) diverges at x = -1/2, it may be evaluated formally.
Let t = A(-1/2) = 0.545218973635949431234950245034944106957612798888179456724264...
then t satisfies
(1) 1 = Sum_{n>=0} ( 1/2^n - t )^n.
(2) 1 = Sum_{n>=0} 2^n / ( 2^n + t )^(n+1).
Also,
A(r) = 1/2 at r = -0.54683649902292991492196620520872286547799291909992048564578...
where
(1) 1 = Sum_{n>=0} ( (1+r)^n - 1/2 )^n.
(2) 1 = Sum_{n>=0} (1+r)^(-n) / ( 1/(1+r)^n + 1/2 )^(n+1).
		

Links

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

Crossrefs

Cf. A304642, A304639, A303926.

Cf. A321602, A321603, A321604, A321605.

Cf. A326282, A326283, A326284.

Cf. A337755, A337756, A337757.

Programs

PARI

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

Formula

G.f. A(x) satisfies:

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

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

a(n) ~ c * d^n * n! / sqrt(n), where d = A317855 = 3.1610886538654... and c = 0.11739505492506... - Vaclav Kotesovec, Sep 26 2020