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A101193 G.f. defined as the limit: A(x) = lim_{n->oo} F(n)^(1/4^(n-1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^4 + (4x)^((4^n-1)/3) for n >= 1. 2
1, 4, 0, 0, 0, 256, -3072, 24576, -163840, 983040, -5603328, 32112640, -195035136, 1283457024, -8975810560, 64281903104, -458387095552, 3216662069248, -22225382014976, 152271623028736, -1043452104015872, 7199883459035136, -50175319780360192, 353054558068408320 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
The Euler transform of the power series A(x) at x=1/4 converges to the constant: c = Sum_{n>=0} (Sum_{k=0..n} C(n,k)*a(k)/4^k)/2^(n+1) = 2.030544704345910171947313128... which is the limit of S(n)^(1/4^(n-1)) where S(0)=1, S(n+1) = S(n)^4 +1.
LINKS
FORMULA
G.f. begins: A(x) = (1+m*x) + m^m*x^(m+1)/(1+m*x)^(m-1) + ... at m=4.
EXAMPLE
The iteration begins:
F(0) = 1,
F(1) = 1 + 4*x,
F(2) = 1 + 16*x + 96*x^2 + 256*x^3 + 256*x^4 + 1024*x^5,
F(3) = 1 + 64*x + 1920*x^2 + 35840*x^3 + ... + 4398046511104*x^21.
The 4^(n-1)-th roots of F(n) tend to the limit of A(x):
F(1)^(1/4^0) = 1 + 4*x
F(2)^(1/4^1) = 1 + 4*x + 256*x^5 - 3072*x^6 + 24576*x^7 - 163840*x^8 + ...
F(3)^(1/4^2) = 1 + 4*x + 256*x^5 - 3072*x^6 + 24576*x^7 - 163840*x^8 + ...
PROG
(PARI) {a(n)=local(F=1, A, L); if(n==0, A=1, L=ceil(log(n+1)/log(4)); for(k=1, L, F=F^4+(4*x)^((4^k-1)/3)); A=polcoeff((F+x*O(x^n))^(1/4^(L-1)), n)); A}
CROSSREFS
Sequence in context: A222931 A258645 A222953 * A258646 A258647 A013334
KEYWORD
sign
AUTHOR
Paul D. Hanna, Dec 07 2004
STATUS
approved

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Last modified March 29 05:48 EDT 2024. Contains 371265 sequences. (Running on oeis4.)