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A248230 a(n) = floor(1/(zeta(4) - Sum_{h=1..n} 1/h^4)).

%I A248230 #17 Jul 17 2018 02:42:15
%S A248230 12,50,133,280,507,833,1276,1855,2586,3488,4579,5878,7401,9167,11194,
%T A248230 13501,16104,19022,22273,25876,29847,34205,38968,44155,49782,55868,
%U A248230 62431,69490,77061,85163,93814,103033,112836,123242,134269,145936,158259,171257,184948
%N A248230 a(n) = floor(1/(zeta(4) - Sum_{h=1..n} 1/h^4)).
%C A248230 This sequence provides insight into the manner of convergence of Sum_{h=1..n} 1/h^4.
%H A248230 Clark Kimberling, <a href="/A248230/b248230.txt">Table of n, a(n) for n = 1..1000</a>
%H A248230 Soumyadip Sahu, <a href="https://arxiv.org/abs/1807.05454">On Certain Reciprocal Sums</a>, arXiv:1807.05454 [math.NT], 2018.
%F A248230 Empirically, a(n) = 3*a(n-1) - a(n-2) + a(n-3) + a(n-4) - 3*a(n-5) + 3*a(n-6) - a(n-7).
%F A248230 Conjecture: a(n) = 1 + 7*n/2 + 9*n^2/2 + 3*n^3 + floor(n/4), holds for all n <= 10000. - _Vaclav Kotesovec_, Oct 09 2014
%t A248230 $MaxExtraPrecision = Infinity; z = 400; p[k_] := p[k] = Sum[1/h^4, {h, 1, k}];
%t A248230 N[Table[Zeta[4] - p[n], {n, 1, z/10}]]
%t A248230 f[n_] := f[n] = Select[Range[z], Zeta[4] - p[#] < 1/n^3 &, 1];
%t A248230 u = Flatten[Table[f[n], {n, 1, z}]]   (* A248227 *)
%t A248230 Flatten[Position[Differences[u], 0]]  (* A248228 *)
%t A248230 Flatten[Position[Differences[u], 1]]  (* A248229 *)
%t A248230 f = Table[Floor[1/(Zeta[4] - p[n])], {n, 1, z}]  (* A248230 *)
%Y A248230 Cf. A248227, A248228, A248229, A013662.
%K A248230 nonn,easy
%O A248230 1,1
%A A248230 _Clark Kimberling_, Oct 05 2014