A248231 Least k such that zeta(5) - Sum_{h = 1..k} 1/h^5 < 1/n^4.

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, 20, 20, 21, 22, 23, 23, 24, 25, 25, 26, 27, 28, 28, 29, 30, 30, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 40, 40, 41, 42, 42, 43, 44, 45, 45, 46, 47, 47, 48

Offset: 1

Author

Clark Kimberling, Oct 05 2014

Keywords

• nonn
• easy

Comments

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.

Examples

			Let s(n) = Sum_{h = 1..n} 1/h^5.
Approximations are shown here:
n ... zeta(5) - s(n) ... 1/n^4
1 ... 0.0369278 .... 1
2 ... 0.0056777 .... 0.0625
3 ... 0.0015625 .... 0.0123
4 ... 0.0005859 .... 0.0039
5 ... 0.0002659 .... 0.0016
6 ... 0.0001373 .... 0.0007
a(6) = 4 because zeta(5) - s(4) < 1/6^4 < zeta(5) - s(3).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..2000

Crossrefs

Cf. A013663, A248232, A248233, A248234, A248227.

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

a(n) ~ n / sqrt(2). - Vaclav Kotesovec, Oct 09 2014

Conjecture: a(n) = floor(sqrt(n^2/2 - 1) + 1/2), for n>1. - Ridouane Oudra, Sep 06 2023