cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A248231 Least k such that zeta(5) - Sum_{h = 1..k} 1/h^5 < 1/n^4.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 8, 9, 10, 11, 11, 12, 13, 13, 14, 15, 16, 16, 17, 18, 18, 19, 20, 20, 21, 22, 23, 23, 24, 25, 25, 26, 27, 28, 28, 29, 30, 30, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 40, 40, 41, 42, 42, 43, 44, 45, 45, 46, 47, 47, 48
Offset: 1

Views

Author

Clark Kimberling, Oct 05 2014

Keywords

Comments

This sequence and A248234 provide insight into the manner of convergence of Sum_{h=0..k} 1/h^5. Since a(n+1) - a(n) is in {0,1} for n >= 1, A248232 and A248233 are complementary.

Examples

			Let s(n) = Sum_{h = 1..n} 1/h^5.
Approximations are shown here:
n ... zeta(5) - s(n) ... 1/n^4
1 ... 0.0369278 .... 1
2 ... 0.0056777 .... 0.0625
3 ... 0.0015625 .... 0.0123
4 ... 0.0005859 .... 0.0039
5 ... 0.0002659 .... 0.0016
6 ... 0.0001373 .... 0.0007
a(6) = 4 because zeta(5) - s(4) < 1/6^4 < zeta(5) - s(3).

Links

Crossrefs

Cf. A013663, A248232, A248233, A248234, A248227.

Programs

  • Mathematica
    z = 400; p[k_] := p[k] = Sum[1/h^5, {h, 1, k}]; N[Table[Zeta[5] - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], Zeta[5] - p[#] < 1/n^4 &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]   (* A248231 *)
Flatten[Position[Differences[u], 0]]  (* A248232 *)
Flatten[Position[Differences[u], 1]]  (* A248233 *)
Table[Floor[1/(Zeta[5] - p[n])], {n, 1, z}]  (* A248234 *)

Formula

a(n) ~ n / sqrt(2). - Vaclav Kotesovec, Oct 09 2014
Conjecture: a(n) = floor(sqrt(n^2/2 - 1) + 1/2), for n>1. - Ridouane Oudra, Sep 06 2023

A248232 Numbers k such that A248231(k+1) = A248231(k).

Original entry on oeis.org

1, 4, 8, 11, 15, 18, 22, 25, 28, 32, 35, 39, 42, 46, 49, 52, 56, 59, 63, 66, 69, 73, 76, 80, 83, 87, 90, 93, 97, 100, 104, 107, 110, 114, 117, 121, 124, 128, 131, 134, 138, 141, 145, 148, 151, 155, 158, 162, 165, 168, 172, 175, 179, 182, 186, 189, 192, 196
Offset: 1

Views

Author

Clark Kimberling, Oct 05 2014

Keywords

Comments

Since A248231(k+1) - A248232(k) is in {0,1} for k >= 1, A248232 and A248233 are complementary.

Examples

			The difference sequence of A248231 is (0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, ...), so that A248232 = (1, 4, 8, 11, 15, 18, 22, 25, 28,...) and A248233 = (2, 3, 5, 6, 7, 9, 10, 12, 13, 14, 16, 17,...), the complement of A248232.

Links

Crossrefs

Cf. A248231, A248233, A248234, A248227.

Programs

  • Mathematica
    z = 400; p[k_] := p[k] = Sum[1/h^5, {h, 1, k}]; N[Table[Zeta[5] - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], Zeta[5] - p[#] < 1/n^4 &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]   (* A248231 *)
Flatten[Position[Differences[u], 0]]  (* A248232 *)
Flatten[Position[Differences[u], 1]]  (* A248233 *)
Table[Floor[1/(Zeta[5] - p[n])], {n, 1, z}]  (* A248234 *)

A248234 a(n) = floor(1/(zeta(5) - Sum_{h=1..n} 1/h^5)).

Original entry on oeis.org

27, 176, 639, 1706, 3759, 7279, 12842, 21119, 32879, 48986, 70399, 98175, 133466, 177519, 231679, 297386, 376175, 469679, 579626, 707839, 856239, 1026842, 1221759, 1443199, 1693466, 1974959, 2290175, 2641706, 3032239, 3464559, 3941546, 4466175, 5041519
Offset: 1

Views

Author

Clark Kimberling, Oct 05 2014

Keywords

Comments

This sequence provides insight into the manner of convergence of Sum_{h=1..n} 1/h^5.

Links

Crossrefs

Cf. A248231, A248232, A248233, A248227, A013663.

Programs

  • Mathematica
    z = 400; p[k_] := p[k] = Sum[1/h^5, {h, 1, k}]; N[Table[Zeta[5] - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], Zeta[5] - p[#] < 1/n^4 &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]   (* A248231 *)
Flatten[Position[Differences[u], 0]]  (* A248232 *)
Flatten[Position[Differences[u], 1]]  (* A248233 *)
Table[Floor[1/(Zeta[5] - p[n])], {n, 1, z}]  (* A248234 *)

Formula

Empirically, a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - 5*a(n-4) + 6*a(n-5) - 4*a(n-6) + a(n-7), for n >= 10.
Conjecture (for n >= 3): (12*n*(1+n)*(4+3*n+3*n^2) - 8 - cos(2*n*Pi/3) + sqrt(3)*sin(2*n*Pi/3))/9. - Vaclav Kotesovec, Oct 09 2014
Showing 1-3 of 3 results.

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