OFFSET
0,2
COMMENTS
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.
FORMULA
G.f. begins: A(x) = (1+m*x) + m^m*x^(m+1)/(1+m*x)^(m-1) + ... at m=5.
EXAMPLE
The iteration begins:
F(0) = 1,
F(1) = 1 + 5*x
F(2) = 1 + 25*x + 250*x^2 + 1250*x^3 + 3125*x^4 + 3125*x^5 + 15625*x^6
F(3) = 1 + 125*x + 7500*x^2 + 287500*x^3 + ... + 5^31*x^31.
The 5^(n-1)-th root of F(n) tend to the limit of A(x):
F(1)^(1/5^0) = 1 + 5*x
F(2)^(1/5^1) = 1 + 5*x + 3125*x^6 - 62500*x^7 + 781250*x^8 + ...
F(3)^(1/5^2) = 1 + 5*x + 3125*x^6 - 62500*x^7 + 781250*x^8 + ...
PROG
(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}
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Dec 07 2004
STATUS
approved