Zeta(s) = Sum_{n >= 1} 1/n^s.
For n>=2, a(n) = floor(2^n(4/3)^n1+gamma+(8/9)^n(4/5)^n+(2/3)^n).  Benoit Cloitre, Oct 04 2005
a(n) = 2^n(4/3)^n+O(1) and more precisely lim_{n>infinity} zeta(zeta(n))2^n+(4/3)^n+1 = gamma where gamma is the EulerMascheroni constant.  Benoit Cloitre, Oct 04 2005
It appears that a(n) is always equal to or 1 greater than round(1/(zeta(n)1)).  Iain Fox, Oct 27 2017 (edited Nov 22 2017)
From Iain Fox, Nov 22 2017: (Start)
For reference, b(n) = round(1/(zeta(n)1)) and c(n) = ceiling(gamma(zeta(n)1)).
a(n) >= b(n) >= c(n). By The Squeeze Theorem, if a(n) = c(n) for some integer n, then a(n) = b(n). (Proved by Iain Fox, Sep 16 2020)
a(n)  c(n) <= 1. (Proved by Iain Fox, Sep 16 2020)
(End)
Proof of above comments from Iain Fox, Sep 16 2020: (Start)
Consider a function f(x) = zeta(x)  1/(x1). Lim_{x>infinity} f(x) = 1 and lim_{x>1} f(x) = 0.577... is the EulerMascheroni constant. f(x) is strictly increasing, so the range of f(x) on the interval (1, infinity) is (0.577..., 1). a(n) is necessarily greater than or equal to b(n) when f(zeta(n)) >= 1/2. This is the case for n>1. Therefore, a(n) >= b(n).
Consider a function g(x) = 1/(x1)  gamma(x1). g(x) is greater than 0 strictly decreasing on the interval (1, 2). Lim_{x>1} g(x) = .577... is the EulerMascheroni constant. b(n) is necessarily greater than or equal to c(n) when g(zeta(n)) > 1/2. This is the case for n>3. Since b(2) = c(2) and b(3) = c(3), this proves b(n) >= c(n).
Consider a function h(x) = zeta(x)  gamma(x1). Lim_{x>1} h(x) = 1.154... is twice the EulerMascheroni constant. On the interval (1, 2], h(x) is strictly decreasing and has values on the interval [Pi^2/61 = 0.644..., 1.154...). a(n)  c(n) is necessarily less than or equal to 1 when h(zeta(n)) < 2. This is the case for n>1. Therefore a(n)  c(n) <= 1.
(End)
