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A248230 Floor(1/(zeta(4) - sum{1/(h^4, h = 1..n})). 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{1/(h^4, h = 1..n} .

LINKS

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

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 A145886

Adjacent sequences:  A248227 A248228 A248229 * A248231 A248232 A248233

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Oct 05 2014

STATUS

approved

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Last modified February 19 17:08 EST 2018. Contains 299356 sequences. (Running on oeis4.)