OFFSET
1,1
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.
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
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 05 2014
STATUS
approved