A101192 G.f. defined as the limit: A(x) = lim_{n->oo} F(n)^(1/3^(n-1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^3 + (3x)^((3^n-1)/2) for n >= 1.

1, 3, 0, 0, 27, -162, 729, -2916, 10206, -28431, 39366, 216513, -2506302, 16395939, -87687765, 419838390, -1879883964, 8098629399, -33997343652, 136405492911, -478000355922, 987247848321, 4754553381171, -85842565710012, 782970953914944, -5641921802462517, 34830591205459716

Offset: 0

Author

Paul D. Hanna, Dec 07 2004

Keywords

• sign

Comments

The Euler transform of the power series A(x) at x=1/3 converges to the constant: c = Sum_{n>=0} (Sum_{k=0..n} C(n,k)*a(k)/3^k)/2^(n+1) = 2.080400667750319352117745232... which is the limit of S(n)^(1/3^(n-1)) where S(0)=1, S(n+1) = S(n)^3 + 1.

Examples

			The iteration begins:
F(0) = 1,
F(1) = 1 +  3*x,
F(2) = 1 +  9*x +  27*x^2 +   27*x^3 +    81*x^4,
F(3) = 1 + 27*x + 324*x^2 + 2268*x^3 + 10449*x^4 + ... + 1594323*x^13.
The 3^(n-1)-th roots of F(n) tend to the limit of A(x):
F(1)^(1/3^0) = 1 + 3*x
F(2)^(1/3^1) = 1 + 3*x + 27*x^4 - 162*x^5 + 729*x^6 - 2916*x^7 + ...
F(3)^(1/3^2) = 1 + 3*x + 27*x^4 - 162*x^5 + 729*x^6 - 2916*x^7 + ...
		

Crossrefs

Cf. A101189, A101193, A101194.

Programs

PARI

{a(n)=local(F=1,A,L);if(n==0,A=1,L=ceil(log(n+1)/log(3)); for(k=1,L,F=F^3+(3*x)^((3^k-1)/2)); A=polcoeff((F+x*O(x^n))^(1/3^(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=3.