OFFSET
0,1
COMMENTS
Let f(s) = zeta(zeta(s+1)) - 1, where zeta(s) is the Riemann zeta function. Then f(s) is a strictly increasing function from (0, +oo) to (0, +oo), lim_{s->0+} f(s) = 0, lim_{s->+oo} f(s) = +oo.
Conjecture:
(i) f(s) has a unique fixed point s = A069995 - 1 in (0, +oo);
(ii) Lim_{s->+oo} f(s)/2^s = 1, lim_{s->0+} f(s)/2^(-1/s) = exp(-2/5) = A344428.
If these are true, let s_0 be any real number > alpha, s_n = zeta(s_{n-1}) for n >= 1, where alpha = A069995 is the fixed point of zeta(s) in (1, +oo), then {s_{2n}} diverges quickly to +oo, {s_{2n+1}} converges quickly to 1.
This is because the derivative of zeta(zeta(s)) - s at s = alpha is (zeta'(alpha))^2 - 1 = A344427^2 - 1 > 0, so (i) implies that zeta(zeta(s)) > s for s > alpha and zeta(zeta(s)) < s for 1 < s < alpha, hence ... > s_{2n} > s_{2n-2} > ... > s_2 > s_0 > alpha > s_1 > s_3 > ... > s_{2n+1} > ..., and it follows from (i) that lim_{n->+oo} s_{2n} = +oo, lim_{n->+oo} s_{2n+1} = 1. By definition s_n - 1 = f(s_{n-2} - 1), n >= 2. For large n, s_{2n} - 1 is approximately equal to 2^(s_{2(n-1)} - 1), and 1/(s_{2n+1} - 1) is approximately equal to exp(2/5) * 2^(1/(s_{2(n-1)+1} - 1)).
EXAMPLE
exp(-2/5) = 0.67032004603563930074... In comparison, (zeta(zeta(0.001+1)) - 1) / 2^(-1/0.001) = 0.67022226725425164463...
MATHEMATICA
RealDigits[Exp[-2/5], 10, 100][[1]] (* Amiram Eldar, May 19 2021 *)
PROG
(PARI) default(realprecision, 100); exp(-2/5)
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Jianing Song, May 19 2021
STATUS
approved