The OEIS is supported by the many generous donors to the OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A248230 a(n) = floor(1/(zeta(4) - Sum_{h=1..n} 1/h^4)). 4
 12, 50, 133, 280, 507, 833, 1276, 1855, 2586, 3488, 4579, 5878, 7401, 9167, 11194, 13501, 16104, 19022, 22273, 25876, 29847, 34205, 38968, 44155, 49782, 55868, 62431, 69490, 77061, 85163, 93814, 103033, 112836, 123242, 134269, 145936, 158259, 171257, 184948 (list; graph; refs; listen; history; text; internal format)
 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 Cf. A248227, A248228, A248229, A013662. Sequence in context: A009927 A009938 A063491 * A083559 A051797 A342482 Adjacent sequences: A248227 A248228 A248229 * A248231 A248232 A248233 KEYWORD nonn,easy AUTHOR Clark Kimberling, Oct 05 2014 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified August 4 01:40 EDT 2024. Contains 374905 sequences. (Running on oeis4.)