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

27, 176, 639, 1706, 3759, 7279, 12842, 21119, 32879, 48986, 70399, 98175, 133466, 177519, 231679, 297386, 376175, 469679, 579626, 707839, 856239, 1026842, 1221759, 1443199, 1693466, 1974959, 2290175, 2641706, 3032239, 3464559, 3941546, 4466175, 5041519

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^5.

Links

• Clark Kimberling, Table of n, a(n) for n = 1..2000
• Soumyadip Sahu, On Certain Reciprocal Sums, arXiv:1807.05454 [math.NT], 2018.

Crossrefs

Cf. A248231, A248232, A248233, A248227, A013663.

Programs

Mathematica

z = 400; p[k_] := p[k] = Sum[1/h^5, {h, 1, k}]; N[Table[Zeta[5] - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], Zeta[5] - p[#] < 1/n^4 &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]   (* A248231 *)
Flatten[Position[Differences[u], 0]]  (* A248232 *)
Flatten[Position[Differences[u], 1]]  (* A248233 *)
Table[Floor[1/(Zeta[5] - p[n])], {n, 1, z}]  (* A248234 *)

Formula

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.

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