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 1-5 of 5 results.

A206807 Position of 3^n when {2^j} and {3^k} are jointly ranked; complement of A206805.

Original entry on oeis.org

2, 5, 7, 10, 12, 15, 18, 20, 23, 25, 28, 31, 33, 36, 38, 41, 43, 46, 49, 51, 54, 56, 59, 62, 64, 67, 69, 72, 74, 77, 80, 82, 85, 87, 90, 93, 95, 98, 100, 103, 105, 108, 111, 113, 116, 118, 121, 124, 126, 129, 131, 134, 137, 139, 142, 144, 147, 149, 152, 155
Offset: 1

Views

Author

Clark Kimberling, Feb 16 2012

Keywords

Comments

The joint ranking is for j >= 1 and k >= 1, so that the sets {2^j} and {3^k} are disjoint.

Examples

			The joint ranking begins with 2,3,4,8,9,16,27,32,64,81,128,243,256, so that
A206805 = (1,3,4,6,8,9,11,13,...)
A206807 = (2,5,7,10,12,...)

Crossrefs

Cf. A006899, A056576, A098294, A206805, A122437.

Programs

  • Mathematica
    f[n_] := 2^n; g[n_] := 3^n; z = 200;
c = Table[f[n], {n, 1, z}]; s = Table[g[n], {n, 1, z}];
j = Sort[Union[c, s]];
p[n_] := Position[j, f[n]]; q[n_] := Position[j, g[n]];
Flatten[Table[p[n], {n, 1, z}]]           (* A206805 *)
Table[n + Floor[n*Log[3, 2]], {n, 1, 50}] (* A206805 *)
Flatten[Table[q[n], {n, 1, z}]]           (* this sequence *)
Table[n + Floor[n*Log[2, 3]], {n, 1, 50}] (* this sequence as a table *)
  • PARI
    a(n) = logint(3^n, 2) + n; \\ Ruud H.G. van Tol, Dec 10 2023

Formula

a(n) = n + floor(n*log_2(3)).
A206805(n) = n + floor(n*log_3(2)).
a(n) = n + A056576(n). - Michel Marcus, Dec 12 2023
a(n) = A098294(n) + 2*n - 1. - Ruud H.G. van Tol, Jan 22 2024

A206911 Position of n-th partial sum of the harmonic series when all the partial sums are jointly ranked with the set {log(k+1)}; complement of A206912.

Original entry on oeis.org

2, 5, 8, 11, 13, 16, 19, 22, 24, 27, 30, 33, 36, 38, 41, 44, 47, 49, 52, 55, 58, 61, 63, 66, 69, 72, 74, 77, 80, 83, 86, 88, 91, 94, 97, 100, 102, 105, 108, 111, 113, 116, 119, 122, 125, 127, 130, 133, 136, 138, 141, 142, 143, 144, 145, 146, 147, 148, 149
Offset: 1

Views

Author

Clark Kimberling, Feb 13 2012

Keywords

Comments

Conjecture: the difference sequence of A206911 consists of 2s and 3s, and the ratio (number of 3s)/(number of 2s) tends to a number between 3.5 and 3.6.
Similar conjectures can be stated for difference sequences based on jointly ranked sets, such as A206903, A206906, A206928, A206805, A206812, and A206815.

Examples

			Let S(n)=1+1/2+1/3+...+1/n and L(n)=log(n+1).  Then
L(1)<S(1)<L(2)<L(3)<S(2)<L(4)<L(5)<S(3)<L(6)<..., so that
A206911=(2,5,8,...).

Crossrefs

Cf. A206912, A206815.

Programs

  • Mathematica
    f[n_] := Sum[1/k, {k, 1, n}];  z = 300;
g[n_] := N[Log[n + 1]];
c = Table[f[n], {n, 1, z}];
s = Table[g[n], {n, 1, z}];
j = Sort[Union[c, s]];
p[n_] := Position[j, f[n]]; q[n_] := Position[j, g[n]];
Flatten[Table[p[n], {n, 1, z}]]    (* A206911 *)
Flatten[Table[q[n], {n, 1, z}]]    (* A206912 *)

A206812 Position of 2^n in joint ranking of {2^i}, {3^j}, {5^k}.

Original entry on oeis.org

1, 3, 5, 7, 10, 11, 14, 16, 17, 20, 21, 24, 26, 28, 30, 32, 34, 36, 38, 40, 43, 44, 46, 49, 50, 53, 55, 57, 59, 60, 63, 65, 67, 69, 72, 73, 75, 77, 79, 82, 83, 86, 88, 89, 92, 94, 96, 98, 100, 102, 104, 106, 108, 111, 112, 115, 116, 118, 121, 122, 125, 127, 129
Offset: 1

Views

Author

Clark Kimberling, Feb 17 2012

Keywords

Comments

The exponents i,j,k range through the set N of positive integers, so that the position sequences (this sequence for 2^n, A206813 for 3^n, A206814 for 5^n) partition N.

Examples

			The joint ranking begins with 2,3,4,5,8,9,16,25,27,32,64,81,125,128,243 so that
this sequence = (1,3,5,7,10,11,...)
A206813 = (2,6,9,12,15,...)
A206814 = (4,8,13,18,23,...)

Crossrefs

Cf. A206805, A206813, A206814.

Programs

  • Mathematica
    f[1, n_] := 2^n; f[2, n_] := 3^n;
f[3, n_] := 5^n; z = 1000;
d[n_, b_, c_] := Floor[n*Log[b, c]];
t[k_] := Table[f[k, n], {n, 1, z}];
t = Sort[Union[t[1], t[2], t[3]]];
p[k_, n_] := Position[t, f[k, n]];
Flatten[Table[p[1, n], {n, 1, z/8}]] (* A206812 *)
Table[n + d[n, 3, 2] + d[n, 5, 2],
  {n, 1, 50}]                        (* A206812 *)
Flatten[Table[p[2, n], {n, 1, z/8}]] (* A206813 *)
Table[n + d[n, 2, 3] + d[n, 5, 3],
  {n, 1, 50}]                        (* A206813 *)
Flatten[Table[p[3, n], {n, 1, z/8}]] (* A206814 *)
Table[n + d[n, 2, 5] + d[n, 3, 5],
  {n, 1, 50}]                        (* A206814 *)
  • PARI
    a(n) = n + logint(2^n,3) + logint(2^n,5) \\ Ruud H.G. van Tol, Nov 16 2022

Formula

a(n) = n + [n*log_3(2)] + [n*log_5(2)],
A206813(n) = n + [n*log_2(3)] + [n*log_5(3)],
A206814(n) = n + [n*log_2(5)] + [n*log_3(5)],
where []=floor.

A206813 Position of 3^n in joint ranking of {2^i}, {3^j}, {5^k}.

Original entry on oeis.org

2, 6, 9, 12, 15, 19, 22, 25, 29, 31, 35, 39, 41, 45, 48, 51, 54, 58, 61, 64, 68, 71, 74, 78, 81, 84, 87, 91, 93, 97, 101, 103, 107, 110, 113, 117, 120, 123, 126, 130, 132, 136, 140, 143, 146, 149, 153, 156, 159, 163, 165, 169, 173, 175, 179, 182, 185, 188
Offset: 1

Views

Author

Clark Kimberling, Feb 17 2012

Keywords

Comments

The exponents i,j,k range through the set N of positive integers, so that the position sequences (A206812 for 2^n, A206813 for 3^n, A206814 for 5^n) partition N.

Examples

			The joint ranking begins with 2,3,4,5,8,9,16,25,27,32,64,81,125,128,243,256, so that
A205812=(1,3,5,7,10,11,14,...)
A205813=(2,6,9,12,15,...)
A205814=(4,8,13,18,23,...)

Crossrefs

Cf. A206805, A206812, A206814.

Programs

  • Mathematica
    f[1, n_] := 2^n; f[2, n_] := 3^n;
f[3, n_] := 5^n; z = 1000;
d[n_, b_, c_] := Floor[n*Log[b, c]];
t[k_] := Table[f[k, n], {n, 1, z}];
t = Sort[Union[t[1], t[2], t[3]]];
p[k_, n_] := Position[t, f[k, n]];
Flatten[Table[p[1, n], {n, 1, z/8}]] (* A206812 *)
Table[n + d[n, 3, 2] + d[n, 5, 2],
  {n, 1, 50}]                        (* A206812 *)
Flatten[Table[p[2, n], {n, 1, z/8}]] (* A206813 *)
Table[n + d[n, 2, 3] + d[n, 5, 3],
  {n, 1, 50}]                        (* A206813 *)
Flatten[Table[p[3, n], {n, 1, z/8}]] (* A206814 *)
Table[n + d[n, 2, 5] + d[n, 3, 5],
  {n, 1, 50}]                        (* A206814 *)

Formula

A205812(n) = n + [n*log(base 3)(2)] + [n*log(base 5)(2)],
A205813(n) = n + [n*log(base 2)(3)] + [n*log(base 5)(3)],
A205814(n) = n + [n*log(base 2)(5)] + [n*log(base 3)(5)],
where []=floor.

A206814 Position of 5^n in joint ranking of {2^i}, {3^j}, {5^k}.

Original entry on oeis.org

4, 8, 13, 18, 23, 27, 33, 37, 42, 47, 52, 56, 62, 66, 70, 76, 80, 85, 90, 95, 99, 105, 109, 114, 119, 124, 128, 134, 138, 142, 147, 152, 157, 161, 167, 171, 176, 181, 186, 190, 196, 200, 204, 210, 214, 219, 224, 229, 233, 239, 243, 248, 253, 258, 262
Offset: 1

Views

Author

Clark Kimberling, Feb 17 2012

Keywords

Comments

The exponents i,j,k range through the set N of positive integers, so that the position sequences (A206812 for 2^n, A206813 for 3^n, A206814 for 5^n) partition N.

Examples

			The joint ranking begins with 2,3,4,5,8,9,16,25,27,32,64,81,125,128,243,256, so that
A205812=(1,3,5,7,10,11,14,...)
A205813=(2,6,9,12,15,...)
A205814=(4,8,13,18,23,...)

Crossrefs

Cf. A206805, A206812, A206813.

Programs

  • Mathematica
    f[1, n_] := 2^n; f[2, n_] := 3^n;
f[3, n_] := 5^n; z = 1000;
d[n_, b_, c_] := Floor[n*Log[b, c]];
t[k_] := Table[f[k, n], {n, 1, z}];
t = Sort[Union[t[1], t[2], t[3]]];
p[k_, n_] := Position[t, f[k, n]];
Flatten[Table[p[1, n], {n, 1, z/8}]] (* A206812 *)
Table[n + d[n, 3, 2] + d[n, 5, 2],
  {n, 1, 50}]                        (* A206812 *)
Flatten[Table[p[2, n], {n, 1, z/8}]] (* A206813 *)
Table[n + d[n, 2, 3] + d[n, 5, 3],
  {n, 1, 50}]                        (* A206813 *)
Flatten[Table[p[3, n], {n, 1, z/8}]] (* A206814 *)
Table[n + d[n, 2, 5] + d[n, 3, 5],
  {n, 1, 50}]                        (* A206814 *)

Formula

A205812(n) = n + [n*log(base 3)(2)] + [n*log(base 5)(2)],
A205813(n) = n + [n*log(base 2)(3)] + [n*log(base 5)(3)],
A205814(n) = n + [n*log(base 2)(5)] + [n*log(base 3)(5)],
where []=floor.
Showing 1-5 of 5 results.

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