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 1-4 of 4 results.

A226184 Least positive integer k such that 1 + 1/2 + ... + 1/n < 1/(n+1) + ... + 1/(n+k).

Original entry on oeis.org

3, 9, 19, 32, 49, 69, 93, 121, 152, 187, 225, 266, 312, 361, 413, 469, 529, 592, 658, 729, 802, 880, 961, 1045, 1133, 1225, 1320, 1419, 1521, 1627, 1736, 1849, 1966, 2086, 2210, 2337, 2468, 2602, 2740, 2882, 3027, 3175, 3327, 3483, 3642, 3805, 3972, 4142
Offset: 1

Views

Author

Clark Kimberling, May 30 2013

Keywords

Examples

			a(3) = 19 because 1/4 + 1/5 + ... + 1/(3+18) < 1 + 1/2 + 1/3 < 1/4 + 1/5 + ... + 1/(3+19).

Links

Crossrefs

Cf. A226183, A289183.

Programs

  • Mathematica
    z = 55; f[n_] := 1/n; p[n_] := p[n] = Sum[f[k], {k, 1, n}]; Do[s = 0; a[n] = NestWhile[# + 1 &, 1, ! (s += f[#]) >= 2 p[n] &], {n, 1, z}]; m = Map[a, Range[z]]  (* A226183 *)
m1 = Table[m[[n]] - n, {n, 1, z}] (* A226184 *)

Formula

a(n) = A226183(n) - n. - Michel Marcus, Sep 09 2021

A226189 Least positive integer k such that 1 + 1/2 + ... + 1/k >= sqrt(n).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 11, 13, 15, 18, 21, 24, 27, 31, 35, 39, 44, 49, 55, 61, 68, 75, 83, 92, 101, 112, 122, 134, 147, 161, 175, 191, 208, 227, 246, 267, 289, 313, 339, 366, 396, 427, 460, 495, 533, 573, 616, 661, 709, 760, 815, 872, 934, 998, 1067, 1140
Offset: 1

Views

Author

Clark Kimberling, May 30 2013

Keywords

Examples

			a(12) = 18 because 1 + 1/2 + ... + 1/17 < sqrt(12) < 1 + 1/2 + ... + 1/18.

Links

Crossrefs

Cf. A226183, A226190.

Programs

  • Mathematica
    z = 80; f[n_] := 1/n; Do[s = 0; a[n] = NestWhile[# + 1 &, 1, ! (s += f[#]) >= Sqrt[n] &], {n, 1, z}]; m = Map[a, Range[z]]
Table[Ceiling[x /. FindInstance[HarmonicNumber[x] == Sqrt[n] && x > 0, x][[1]]], {n, 80}] (* Vladimir Reshetnikov, Aug 06 2019 *)

A226190 Least positive integer k such that 1 + 1/2 + ... + 1/k >= log(n).

Original entry on oeis.org

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

Views

Author

Clark Kimberling, May 30 2013

Keywords

Examples

			a(9) = 5 because 1 + 1/2 + 1/3 + 1/4 < log(9) < 1 + 1/2 + 1/3 + 1/4 + 1/5.

Links

Crossrefs

Cf. A226183, A226189, A004081.

Programs

  • Mathematica
    z = 80; f[n_] := 1/n; Do[s = 0; a[n] = NestWhile[# + 1 &, 1, ! (s += f[#]) > Log[n] &], {n, 1, z}]; m = Map[a, Range[z]]

A226185 Least positive integer k such that 1 + 1/2 + ... + 1/n < 1 + 1/3 + 1/5 + ... + 1/(2k-1).

Original entry on oeis.org

1, 3, 6, 10, 14, 19, 26, 33, 41, 50, 59, 70, 82, 94, 108, 122, 137, 153, 170, 188, 206, 226, 246, 268, 290, 313, 337, 362, 388, 415, 442, 471, 500, 531, 562, 594, 627, 661, 695, 731, 767, 805, 843, 882, 922, 963, 1005, 1048, 1092, 1136, 1181, 1228, 1275
Offset: 1

Views

Author

Clark Kimberling, May 30 2013

Keywords

Examples

			a(2) = 3 because 1 + 1/3 < 1 + 1/2 < 1 + 1/3 + 1/5.

Links

Crossrefs

Cf. A226183.

Programs

  • Mathematica
    z = 54; f[n_] := 1/n; p[n_] := p[n] = Sum[f[k], {k, 1, n}]; q[n_] := 1/(2 n - 1); Do[s = 0;   a[n] = NestWhile[# + 1 &, 1, ! (s += q[#]) >= p[n] &], {n, 1, z}]; m = Map[a, Range[z]]
Showing 1-4 of 4 results.

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