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
Keywords
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 + ...
Programs
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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.
Comments