A101194 G.f. defined as the limit: A(x) = lim_{n->oo} F(n)^(1/5^(n-1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^5 + (5x)^((5^n-1)/4) for n >= 1.

1, 5, 0, 0, 0, 0, 3125, -62500, 781250, -7812500, 68359375, -546875000, 4082031250, -28417968750, 179443359375, -939941406250, 2685546875000, 23010253906250, -569122314453125, 7669982910156250, -84739685058593750, 836715698242187500, -7611751556396484375

Offset: 0

Author

Paul D. Hanna, Dec 07 2004

Keywords

• sign

Comments

The Euler transform of the power series A(x) at x=1/5 converges to the constant: c = Sum_{n>=0} (Sum_{k=0..n} C(n,k)*a(k)/5^k) / 2^(n+1) = 2.012346619142363112612326559... which is the limit of S(n)^(1/5^(n-1)) where S(0)=1, S(n+1) = S(n)^5 + 1.

Examples

			The iteration begins:
F(0) = 1,
F(1) = 1 + 5*x
F(2) = 1 + 25*x + 250*x^2 + 1250*x^3 + 3125*x^4 + 3125*x^5 + 15625*x^6
F(3) = 1 + 125*x + 7500*x^2 + 287500*x^3 + ... + 5^31*x^31.
The 5^(n-1)-th root of F(n) tend to the limit of A(x):
F(1)^(1/5^0) = 1 + 5*x
F(2)^(1/5^1) = 1 + 5*x + 3125*x^6 - 62500*x^7 + 781250*x^8 + ...
F(3)^(1/5^2) = 1 + 5*x + 3125*x^6 - 62500*x^7 + 781250*x^8 + ...
		

Crossrefs

Cf. A101189, A101192, A101193.

Programs

PARI

{a(n)=local(F=1,A,L);if(n==0,A=1,L=ceil(log(n+1)/log(5)); for(k=1,L,F=F^5+(5*x)^((5^k-1)/4)); A=polcoeff((F+x*O(x^n))^(1/5^(L-1)),n));A}

Formula

G.f. begins: A(x) = (1+m*x) + m^m*x^(m+1)/(1+m*x)^(m-1) + ... at m=5.