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

12, 50, 133, 280, 507, 833, 1276, 1855, 2586, 3488, 4579, 5878, 7401, 9167, 11194, 13501, 16104, 19022, 22273, 25876, 29847, 34205, 38968, 44155, 49782, 55868, 62431, 69490, 77061, 85163, 93814, 103033, 112836, 123242, 134269, 145936, 158259, 171257, 184948

Offset: 1

Author

Clark Kimberling, Oct 05 2014

Keywords

• nonn
• easy

Comments

This sequence provides insight into the manner of convergence of Sum_{h=1..n} 1/h^4.

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000
• Soumyadip Sahu, On Certain Reciprocal Sums, arXiv:1807.05454 [math.NT], 2018.

Crossrefs

Cf. A248227, A248228, A248229, A013662.

Programs

Mathematica

$MaxExtraPrecision = Infinity; z = 400; p[k_] := p[k] = Sum[1/h^4, {h, 1, k}];
N[Table[Zeta[4] - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], Zeta[4] - p[#] < 1/n^3 &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]   (* A248227 *)
Flatten[Position[Differences[u], 0]]  (* A248228 *)
Flatten[Position[Differences[u], 1]]  (* A248229 *)
f = Table[Floor[1/(Zeta[4] - p[n])], {n, 1, z}]  (* A248230 *)

Formula

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).

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