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Least k such that zeta(4) - sum{1/h^4, h = 1..k} < 1/n^3.
8

%I #9 Oct 09 2014 07:13:15

%S 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,

%T 19,20,21,21,22,23,24,24,25,26,26,27,28,28,29,30,31,31,32,33,33,34,35,

%U 35,36,37,37,38,39,40,40,41,42,42,43,44,44,45,46,46,47

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

%C 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.

%H Clark Kimberling, <a href="/A248227/b248227.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) ~ 3^(-1/3) * n. - _Vaclav Kotesovec_, Oct 09 2014

%e Let s(n) = sum{1/h^4, h = 1..n}. Approximations are shown here:

%e n ... zeta(4) - s(n) ... 1/n^3

%e 1 ... 0.0823232 .... 1

%e 2 ... 0.0198232 .... 0.125

%e 3 ... 0.0074775 .... 0.037

%e 4 ... 0.0035713 .... 0.015

%e 5 ... 0.0019713 .... 0.008

%e 6 ... 0.0011997 .... 0.005

%e a(6) = 4 because zeta(4) - s(4) < 1/216 < zeta(4) - s(3).

%t $MaxExtraPrecision = Infinity; z = 400; p[k_] := p[k] = Sum[1/h^4, {h, 1, k}];

%t N[Table[Zeta[4] - p[n], {n, 1, z/10}]]

%t f[n_] := f[n] = Select[Range[z], Zeta[4] - p[#] < 1/n^3 &, 1];

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248227 *)

%t Flatten[Position[Differences[u], 0]] (* A248228 *)

%t Flatten[Position[Differences[u], 1]] (* A248229 *)

%t f = Table[Floor[1/(Zeta[4] - p[n])], {n, 1, z}] (* A248230 *)

%Y Cf. A248228, A248229, A248230, A013662.

%K nonn,easy

%O 1,3

%A _Clark Kimberling_, Oct 05 2014