A304639 - OEIS

%I #23 Dec 27 2021 20:42:00

%S 1,1,2,11,117,1735,31853,689043,17079221,476238926,14742680162,

%T 501584454703,18605089712174,747393133162471,32332767332220442,

%U 1498961537925543920,74153115616699819304,3899494667155151052688,217246028175467702590241,12783023090792392539557926,792236994094236725330142276,51585659784100723438219893047,3520987513029712770759434038820

%N G.f. A(x) satisfies: 1 = Sum_{n>=0} ( 1/(1-x)^n - A(x) )^n.

%H Paul D. Hanna, <a href="/A304639/b304639.txt">Table of n, a(n) for n = 0..200</a>

%F G.f. A(x) satisfies:

%F (1) 1 = Sum_{n>=0} ( 1/(1-x)^n - A(x) )^n.

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

%F (3) 1 = Sum_{n>=0} (1-x)^n / ( (1-x)^n + A(x) )^(n+1).

%F a(n) ~ c * d^n * n! / sqrt(n), where d = A317855 = 3.16108865386542881383... and c = 0.16107844724485... - _Vaclav Kotesovec_, Oct 14 2020

%e G.f.: A(x) = 1 + x + 2*x^2 + 11*x^3 + 117*x^4 + 1735*x^5 + 31853*x^6 + 689043*x^7 + 17079221*x^8 + 476238926*x^9 + 14742680162*x^10 + 501584454703*x^11 + ...

%e is such that

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

%e Also,

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

%e PARTICULAR VALUES.

%e Although the power series A(x) diverges at x = -1, it may be evaluated formally.

%e Let t = A(-1) = 0.5452189736359494312349502450349441069576127988881794567242641...

%e then t satisfies

%e (1) 1 = Sum_{n>=0} ( 1/2^n - t )^n.

%e (2) 1 = Sum_{n>=0} ( 1 - 2^n*t )^n / 2^(n^2).

%e (3) 1 = Sum_{n>=0} 2^n / ( 2^n + t )^(n+1).

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

%o for(n=0, 30, print1(a(n), ", "))

%Y Cf. A326262, A326263, A326264, A326265.

%Y Cf. A303056.

%K nonn

%O 0,3

%A _Paul D. Hanna_, May 16 2018