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 1-7 of 7 results.

A247912 Numbers k such that A247911(k+1) = A247911(k).

Original entry on oeis.org

2, 4, 7, 9, 12, 15, 17, 20, 22, 25, 27, 30, 32, 35, 37, 40, 42, 45, 47, 50, 52, 55, 57, 60, 62, 64, 67, 69, 72, 74, 77, 79, 81, 84, 86, 89, 91, 93, 96, 98, 101, 103, 105, 108, 110, 113, 115, 117, 120, 122, 125, 127, 129, 132, 134, 137, 139, 141, 144, 146
Offset: 1

Views

Author

Clark Kimberling, Sep 27 2014

Keywords

Comments

Complement of A247913.

Examples

			A247911(n+1) - A247911(n) = (1,0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,...), and a(n) is the position of the n-th 0.

Links

Crossrefs

Cf. A247911, A247913, A247908, A247914.

Programs

  • Mathematica
    $RecursionLimit = 1000; $MaxExtraPrecision = 1000;
z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);
f[n_] := f[n] = Select[Range[z], (2 # + 1)/u[2 # + 1] - E < n^-n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247911 *)
w = Differences[u]
Flatten[Position[w, 0]] (* A247912 *)
Flatten[Position[w, 1]] (* A247913 *)

A247913 Numbers k such that A247911(k+1) = A247911(k) + 1.

Original entry on oeis.org

1, 3, 5, 6, 8, 10, 11, 13, 14, 16, 18, 19, 21, 23, 24, 26, 28, 29, 31, 33, 34, 36, 38, 39, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 58, 59, 61, 63, 65, 66, 68, 70, 71, 73, 75, 76, 78, 80, 82, 83, 85, 87, 88, 90, 92, 94, 95, 97, 99, 100, 102, 104, 106, 107
Offset: 1

Views

Author

Clark Kimberling, Sep 27 2014

Keywords

Comments

Complement of A247912.

Examples

			A247911(n+1) - A247911(n) = (1,0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,...), and a(n) is the position of the n-th 1.

Links

Crossrefs

Cf. A247911, A247912, A247908, A247914.

Programs

  • Mathematica
    $RecursionLimit = 1000; $MaxExtraPrecision = 1000;
z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);
f[n_] := f[n] = Select[Range[z], (2 # + 1)/u[2 # + 1] - E < n^-n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247911 *)
w = Differences[u]
Flatten[Position[w, 0]] (* A247912 *)
Flatten[Position[w, 1]] (* A247913 *)

A247908 Least number k such that e - 2*k/u(2*k) < 1/n^n, where u is defined as in Comments.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Sep 27 2014

Keywords

Comments

The sequence u is define recursively by u(n) = u(n-1) + u(n-2)/(n-2), with u(1) = 0 and u(2) = 1. Let d(n) = a(n+1) - a(n). It appears that d(n) is in {0,1} for n >= 1, that d(n+1) - d(n) is in {2,3}, and that similar bounds hold for higher differences.

Examples

			Approximations for the first few terms of e - 2*n/u(2*n) and 1/n^n are shown here:
n ... e-2*n/u(2*n) .... 1/n^n
1 ... 0.71828 ........  1
2 ... 0.0516152 ....... 0.25
3 ... 0.0013007 ....... 0.037037
4 ... 0.0000184967 .... 0.00390625
a(2) = 2 because e - 4/u(4) < 1/2^2 < e - 2/u(2).

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 19.

Links

Crossrefs

Cf. A247909, A247910, A247911, A247914.

Programs

  • Mathematica
    $RecursionLimit = 1000; $MaxExtraPrecision = 1000;
z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);
f[n_] := f[n] = Select[Range[z], E - 2 #/u[2 #] < 1/n^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247908 *)
w = Differences[u]
Flatten[Position[w, 0]] (* A247909 *)
Flatten[Position[w, 1]] (* A247910 *)

A247914 Least number k such that |(k+1)/u(k+1) - e| < 1/n^n, where u is defined as in Comments.

Original entry on oeis.org

1, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26, 27, 28, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 80
Offset: 1

Views

Author

Clark Kimberling, Sep 27 2014

Keywords

Comments

The sequence u is define recursively by u(n) = u(n-1) + u(n-2)/(n-2), with u(1) = 0 and u(2) = 1. Let d(n) = a(n+1) - a(n). It appears that d(n) is in {1,2} for n >= 1, that d(n+1) - d(n) is in {-1,0,1}, and that similar bounds hold for higher differences.

Examples

			Approximations for the first few terms |(n+1)/u(n+1) - e| and 1/n^n are shown here:
n ... |(n+1)/u(n+1)-e| .. 1/n^n
1 ... 0.7182818285 ...... 1
2 ... 0.28171817 ........ 0.25
3 ... 0.051615161 ....... 0.037037
4 ... 0.0089908988 ...... 0.00390625
5 ... 0.0013006963 ...... 0.00032000
a(2) = 3 because |4/u(4) - e| < 1/2^2 < |3/u(3) - e|.

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 19.

Links

Crossrefs

Cf. A247908, A247911, A247915, A247916.

Programs

  • Mathematica
    $RecursionLimit = Infinity; $MaxExtraPrecision = Infinity;
z = 500; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);
f[n_] := f[n] = Select[Range[z], Abs[(# + 1)/u[# + 1] - E] < n^-n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247914 *)
w = Differences[u]
f1 = Flatten[Position[w, 1]] (* A247915 *)
f2 = Flatten[Position[w, 2]] (* A247916 *)

A247915 Numbers k such that A247914(k+1) = A247914(k) + 1.

Original entry on oeis.org

2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 24, 25, 26, 27, 29, 30, 31, 32, 34, 35, 36, 37, 39, 40, 41, 42, 44, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 72, 73, 74, 75, 77, 78, 79, 80, 81, 83
Offset: 1

Views

Author

Clark Kimberling, Sep 27 2014

Keywords

Comments

Complement of A247916.

Examples

			A247914(n+1) - A247914(n) = (2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1,...), and a(n) is the position of the n-th 1.

Links

Crossrefs

Cf. A247908, A247911, A247914, A247916.

Programs

  • Mathematica
    $RecursionLimit = Infinity; $MaxExtraPrecision = Infinity;
z = 500; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);
f[n_] := f[n] = Select[Range[z], Abs[(# + 1)/u[# + 1] - E] < n^-n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247914 *)
w = Differences[u]
f1 = Flatten[Position[w, 1]] (* A247915 *)
f2 = Flatten[Position[w, 2]] (* A247916 *)

A247916 Numbers k such that A247914(k+1) = A247914(k) + 1.

Original entry on oeis.org

1, 6, 10, 14, 19, 23, 28, 33, 38, 43, 49, 54, 59, 65, 71, 76, 82, 88, 94, 100, 106, 112, 118, 124, 130, 136, 143, 149, 155, 162, 168, 174, 181, 187, 194, 201, 207, 214, 221, 227, 234, 241, 248, 254, 261, 268, 275, 282, 289, 296, 303, 310, 317, 324, 331, 338
Offset: 1

Views

Author

Clark Kimberling, Sep 27 2014

Keywords

Comments

Complement of A247915.

Examples

			A247914(n+1) - A247914(n) = (2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1,...), and a(n) is the position of the n-th 2.

Links

Crossrefs

Cf. A247908, A247911, A247914, A247915.

Programs

  • Mathematica
    $RecursionLimit = Infinity; $MaxExtraPrecision = Infinity;
z = 500; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);
f[n_] := f[n] = Select[Range[z], Abs[(# + 1)/u[# + 1] - E] < n^-n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247914 *)
w = Differences[u]
f1 = Flatten[Position[w, 1]] (* A247915 *)
f2 = Flatten[Position[w, 2]] (* A247916 *)

A247988 Least number k such that e - k/(k!)^(1/k) < 1/n.

Original entry on oeis.org

4, 11, 19, 27, 36, 45, 54, 64, 74, 84, 94, 105, 115, 126, 136, 147, 158, 169, 180, 191, 203, 214, 225, 237, 248, 260, 272, 283, 295, 307, 319, 331, 343, 355, 367, 379, 391, 403, 416, 428, 440, 452, 465, 477, 490, 502, 515, 527, 540, 552, 565, 578, 590, 603
Offset: 1

Views

Author

Clark Kimberling, Sep 29 2014

Keywords

Examples

			Let w(n) = e - n/(n!)^(1/n).  Approximations are shown here:
n .... w(n)  ...... 1/n
1 .... 1.71828 .... 1
2 .... 1.30407 .... 0.5
3 .... 1.06732 .... 0.333333
4 .... 0.911078 ... 0.25
5 .... 0.799022 ... 0.2
10 ... 0.510157 ... 0.1
11 ... 0.477609 ... 0.090909
a(2) = 11 because w(11) < 1/2 < w(10).

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 14.

Crossrefs

Cf. A247778, A247908, A247911, A247914, A247985.

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity;
z = 1000; p[k_] := p[k] = k/(k!)^(1/k) (* Finch p. 14 *)
N[Table[E - p[n], {n, 1, z}]];
f[n_] := f[n] = Select[Range[z], E - p[#] < 1/n &, 1];
u = Flatten[Table[f[n], {n, 1, z/10}]]  (* A247988 *)
Showing 1-7 of 7 results.

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