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%I #27 Oct 24 2020 17:33:45
%S 2,5,12,27,58,120,245,498,1006,2024,4064,8149,16327,32692,65435,
%T 130938,261966,524051,1048260,2096731,4193743,8387860,16776219,
%U 33553102,67107091,134215365,268432305,536866711,1073736223,2147476181
%N Integer part of zeta(zeta(n)).
%C Notice the previous term is almost 1/2 the next term. Conjecture: lim_{n -> infinity} zeta(zeta(n))/zeta(zeta(n+1)) = 1/2.
%C If this sequence were defined instead as a(n) = lim_{k->n+} floor(zeta(zeta(k))), then it would be defined at n=1, with a(1) = 1. - _Iain Fox_, Sep 16 2020
%H Iain Fox, <a href="/A111000/b111000.txt">Table of n, a(n) for n = 2..3321</a>
%F Zeta(s) = Sum_{n >= 1} 1/n^s.
%F For n>=2, a(n) = floor(2^n-(4/3)^n-1+gamma+(8/9)^n-(4/5)^n+(2/3)^n). - _Benoit Cloitre_, Oct 04 2005
%F 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 Euler-Mascheroni constant. - _Benoit Cloitre_, Oct 04 2005
%F 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)
%F From _Iain Fox_, Nov 22 2017: (Start)
%F For reference, b(n) = round(1/(zeta(n)-1)) and c(n) = ceiling(gamma(zeta(n)-1)).
%F 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)
%F a(n) - c(n) <= 1. (Proved by _Iain Fox_, Sep 16 2020)
%F (End)
%F Proof of above comments from _Iain Fox_, Sep 16 2020: (Start)
%F Consider a function f(x) = zeta(x) - 1/(x-1). Lim_{x->infinity} f(x) = 1 and lim_{x->1} f(x) = 0.577... is the Euler-Mascheroni 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).
%F Consider a function g(x) = 1/(x-1) - gamma(x-1). g(x) is greater than 0 strictly decreasing on the interval (1, 2). Lim_{x->1} g(x) = .577... is the Euler-Mascheroni 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).
%F Consider a function h(x) = zeta(x) - gamma(x-1). Lim_{x->1} h(x) = 1.154... is twice the Euler-Mascheroni constant. On the interval (1, 2], h(x) is strictly decreasing and has values on the interval [Pi^2/6-1 = 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.
%F (End)
%e a(100) ~ 1267650600228229398378720795167.
%e a(101) ~ 2535301200456458798836096530474.
%e a(100)/a(101) ~ 0.49999999999999999959005759..
%t IntegerPart[Zeta[Zeta[Range[2,40]]]] (* _Harvey P. Dale_, Sep 17 2019 *)
%o (PARI) zz(n) = for(x=2,n,print1(floor(zeta(zeta(x)))","))
%K easy,nonn
%O 2,1
%A _Cino Hilliard_, Sep 30 2005
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