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%I #7 Apr 30 2023 16:27:49
%S 1,4,0,0,0,256,-3072,24576,-163840,983040,-5603328,32112640,
%T -195035136,1283457024,-8975810560,64281903104,-458387095552,
%U 3216662069248,-22225382014976,152271623028736,-1043452104015872,7199883459035136,-50175319780360192,353054558068408320
%N 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.
%C 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.
%F G.f. begins: A(x) = (1+m*x) + m^m*x^(m+1)/(1+m*x)^(m-1) + ... at m=4.
%e The iteration begins:
%e F(0) = 1,
%e F(1) = 1 + 4*x,
%e F(2) = 1 + 16*x + 96*x^2 + 256*x^3 + 256*x^4 + 1024*x^5,
%e F(3) = 1 + 64*x + 1920*x^2 + 35840*x^3 + ... + 4398046511104*x^21.
%e The 4^(n-1)-th roots of F(n) tend to the limit of A(x):
%e F(1)^(1/4^0) = 1 + 4*x
%e F(2)^(1/4^1) = 1 + 4*x + 256*x^5 - 3072*x^6 + 24576*x^7 - 163840*x^8 + ...
%e F(3)^(1/4^2) = 1 + 4*x + 256*x^5 - 3072*x^6 + 24576*x^7 - 163840*x^8 + ...
%o (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}
%Y Cf. A101189, A101192, A101194.
%K sign
%O 0,2
%A _Paul D. Hanna_, Dec 07 2004
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