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%I #9 Apr 30 2023 04:17:12
%S 1,5,0,0,0,0,3125,-62500,781250,-7812500,68359375,-546875000,
%T 4082031250,-28417968750,179443359375,-939941406250,2685546875000,
%U 23010253906250,-569122314453125,7669982910156250,-84739685058593750,836715698242187500,-7611751556396484375
%N 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.
%C 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.
%F G.f. begins: A(x) = (1+m*x) + m^m*x^(m+1)/(1+m*x)^(m-1) + ... at m=5.
%e The iteration begins:
%e F(0) = 1,
%e F(1) = 1 + 5*x
%e F(2) = 1 + 25*x + 250*x^2 + 1250*x^3 + 3125*x^4 + 3125*x^5 + 15625*x^6
%e F(3) = 1 + 125*x + 7500*x^2 + 287500*x^3 + ... + 5^31*x^31.
%e The 5^(n-1)-th root of F(n) tend to the limit of A(x):
%e F(1)^(1/5^0) = 1 + 5*x
%e F(2)^(1/5^1) = 1 + 5*x + 3125*x^6 - 62500*x^7 + 781250*x^8 + ...
%e F(3)^(1/5^2) = 1 + 5*x + 3125*x^6 - 62500*x^7 + 781250*x^8 + ...
%o (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}
%Y Cf. A101189, A101192, A101193.
%K sign
%O 0,2
%A _Paul D. Hanna_, Dec 07 2004
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