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A248234 a(n) = floor(1/(zeta(5) - Sum_{h=1..n} 1/h^5)).

%I A248234 #20 Jul 17 2018 02:42:29
%S A248234 27,176,639,1706,3759,7279,12842,21119,32879,48986,70399,98175,133466,
%T A248234 177519,231679,297386,376175,469679,579626,707839,856239,1026842,
%U A248234 1221759,1443199,1693466,1974959,2290175,2641706,3032239,3464559,3941546,4466175,5041519
%N A248234 a(n) = floor(1/(zeta(5) - Sum_{h=1..n} 1/h^5)).
%C A248234 This sequence provides insight into the manner of convergence of Sum_{h=1..n} 1/h^5.
%H A248234 Clark Kimberling, <a href="/A248234/b248234.txt">Table of n, a(n) for n = 1..2000</a>
%H A248234 Soumyadip Sahu, <a href="https://arxiv.org/abs/1807.05454">On Certain Reciprocal Sums</a>, arXiv:1807.05454 [math.NT], 2018.
%F A248234 Empirically, a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - 5*a(n-4) + 6*a(n-5) - 4*a(n-6) + a(n-7), for n >= 10.
%F A248234 Conjecture (for n >= 3): (12*n*(1+n)*(4+3*n+3*n^2) - 8 - cos(2*n*Pi/3) + sqrt(3)*sin(2*n*Pi/3))/9. - _Vaclav Kotesovec_, Oct 09 2014
%t A248234 z = 400; p[k_] := p[k] = Sum[1/h^5, {h, 1, k}]; N[Table[Zeta[5] - p[n], {n, 1, z/10}]]
%t A248234 f[n_] := f[n] = Select[Range[z], Zeta[5] - p[#] < 1/n^4 &, 1]
%t A248234 u = Flatten[Table[f[n], {n, 1, z}]]   (* A248231 *)
%t A248234 Flatten[Position[Differences[u], 0]]  (* A248232 *)
%t A248234 Flatten[Position[Differences[u], 1]]  (* A248233 *)
%t A248234 Table[Floor[1/(Zeta[5] - p[n])], {n, 1, z}]  (* A248234 *)
%Y A248234 Cf. A248231, A248232, A248233, A248227, A013663.
%K A248234 nonn,easy
%O A248234 1,1
%A A248234 _Clark Kimberling_, Oct 05 2014