A248227 Least k such that zeta(4) - sum{1/h^4, h = 1..k} < 1/n^3.

1, 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 40, 40, 41, 42, 42, 43, 44, 44, 45, 46, 46, 47

Offset: 1

Author

Clark Kimberling, Oct 05 2014

Keywords

• nonn
• easy

Comments

This sequence and A248230 provide insight into the manner of convergence of sum{1/h^4, h = 0..k}. Since a(n+1) - a(n) is in {0,1} for n >= 1, A248228 and A248229 are complementary.

Examples

			Let s(n) = sum{1/h^4, h = 1..n}.  Approximations are shown here:
n ... zeta(4) - s(n) ... 1/n^3
1 ... 0.0823232 .... 1
2 ... 0.0198232 .... 0.125
3 ... 0.0074775 .... 0.037
4 ... 0.0035713 .... 0.015
5 ... 0.0019713 .... 0.008
6 ... 0.0011997 .... 0.005
a(6) = 4 because zeta(4) - s(4) < 1/216 < zeta(4) - s(3).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A248228, A248229, A248230, 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

a(n) ~ 3^(-1/3) * n. - Vaclav Kotesovec, Oct 09 2014