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%I #42 Oct 27 2023 21:11:19
%S 1,4,12,27,57,119,245,497,1005,2023,4063,8149,16327,32692,65435,
%T 130938,261965,524050,1048259,2096730,4193742,8387859,16776218,
%U 33553102,67107091,134215364,268432305,536866711,1073736223,2147476180,4294957340,8589921317,17179851485
%N Second term in continued fraction for zeta(n).
%H Alois P. Heinz, <a href="/A013697/b013697.txt">Table of n, a(n) for n = 2..1000</a> (terms n = 2..100 from Vincenzo Librandi)
%H Tal Barnea, <a href="https://arxiv.org/abs/1808.06653">On the Riemann Zeta Function and the fractional part of rational powers</a>, arXiv:1808.06653 [math.NT], 2018.
%H Tal Barnea, <a href="https://www.emis.de/journals/JIS/VOL22/Barnea/barnea4.html">The Riemann Zeta Function and the Fractional Part of Rational Powers</a>, J. Int. Seq., Vol. 22 (2019), Article 19.3.6.
%F From _Franklin T. Adams-Watters_, Mar 23 2010: (Start)
%F a(n) = floor(1/(zeta(n)-1)).
%F a(n) = 2^n - (4/3)^n + O(1). It appears that a(n) = 2^n - floor((4/3)^n) - k, where k is usually 2 but is sometimes 1. Up to n=1000, the only values of n where k = 1 are 4, 5, 13, 14, and 17. (End)
%t a[n_] := ContinuedFraction[ Zeta[n], 2] // Last; Table[a[n], {n, 2, 31}] (* _Jean-François Alcover_, Feb 26 2013 *)
%o (Maxima) A013697(n):=floor(1/(zeta(n)-1))$
%o makelist(A013697(n),n,2,30); /* _Martin Ettl_, Nov 03 2012 */
%o (Python)
%o from sympy import zeta
%o print([1//(zeta(n) - 1) for n in range(2, 32)]) # _Karl V. Keller, Jr._, Jul 21 2020
%Y Bisections: A190297, A190584.
%K nonn,easy
%O 2,2
%A _N. J. A. Sloane_.
%E More terms from _Vladeta Jovovic_, Apr 22 2001
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