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%I #22 Sep 07 2023 07:15:04
%S 1,1,2,3,3,4,5,6,6,7,8,8,9,10,11,11,12,13,13,14,15,16,16,17,18,18,19,
%T 20,20,21,22,23,23,24,25,25,26,27,28,28,29,30,30,31,32,33,33,34,35,35,
%U 36,37,37,38,39,40,40,41,42,42,43,44,45,45,46,47,47,48
%N Least k such that zeta(5) - Sum_{h = 1..k} 1/h^5 < 1/n^4.
%C This sequence and A248234 provide insight into the manner of convergence of Sum_{h=0..k} 1/h^5. Since a(n+1) - a(n) is in {0,1} for n >= 1, A248232 and A248233 are complementary.
%H Clark Kimberling, <a href="/A248231/b248231.txt">Table of n, a(n) for n = 1..2000</a>
%F a(n) ~ n / sqrt(2). - _Vaclav Kotesovec_, Oct 09 2014
%F Conjecture: a(n) = floor(sqrt(n^2/2 - 1) + 1/2), for n>1. - _Ridouane Oudra_, Sep 06 2023
%e Let s(n) = Sum_{h = 1..n} 1/h^5.
%e Approximations are shown here:
%e n ... zeta(5) - s(n) ... 1/n^4
%e 1 ... 0.0369278 .... 1
%e 2 ... 0.0056777 .... 0.0625
%e 3 ... 0.0015625 .... 0.0123
%e 4 ... 0.0005859 .... 0.0039
%e 5 ... 0.0002659 .... 0.0016
%e 6 ... 0.0001373 .... 0.0007
%e a(6) = 4 because zeta(5) - s(4) < 1/6^4 < zeta(5) - s(3).
%t z = 400; p[k_] := p[k] = Sum[1/h^5, {h, 1, k}]; N[Table[Zeta[5] - p[n], {n, 1, z/10}]]
%t f[n_] := f[n] = Select[Range[z], Zeta[5] - p[#] < 1/n^4 &, 1]
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248231 *)
%t Flatten[Position[Differences[u], 0]] (* A248232 *)
%t Flatten[Position[Differences[u], 1]] (* A248233 *)
%t Table[Floor[1/(Zeta[5] - p[n])], {n, 1, z}] (* A248234 *)
%Y Cf. A013663, A248232, A248233, A248234, A248227.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, Oct 05 2014