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.

A248630 Numbers k such that A248630(k+1) = A248360(k) + 1.

Original entry on oeis.org

7, 11, 15, 18, 22, 25, 28, 31, 34, 37, 39, 42, 45, 48, 51, 53, 56, 59, 61, 64, 67, 69, 72, 75, 77, 80, 82, 85, 88, 90, 93, 95, 98, 101, 103, 106, 108, 111, 113, 116, 118, 121, 124, 126, 129, 131, 134, 136, 139, 141, 144, 146, 149, 151, 154, 156, 159, 161
Offset: 1

Views

Author

Clark Kimberling, Oct 10 2014

Keywords

Examples

			(A248630(k+1) = A248360(k)) = (2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, ...), so that A248630 = (7, 11, 15, 18, ...).
		

Crossrefs

Programs

  • Mathematica
    z = 300; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}]
    d = N[Table[6 - p[k], {k, 1, z/5}], 12]
    f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1]
    u = Flatten[Table[f[n], {n, 1, z}]]  (* A248629 *)
    d = Differences[u]
    v = Flatten[Position[d, 1]]  (* A248630 *)

A248631 Least k such that 3/2 - sum{(h^2)/3^h, h = 1..k} < 1/2^n.

Original entry on oeis.org

3, 4, 5, 6, 7, 8, 8, 9, 10, 11, 11, 12, 13, 14, 14, 15, 16, 16, 17, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 43, 43, 44, 45, 45, 46, 47, 47, 48, 49, 49
Offset: 1

Views

Author

Clark Kimberling, Oct 11 2014

Keywords

Comments

This sequence provides insight into the manner of convergence of sum{(h^2)/3^h, h = 1..k} to 3/2.

Examples

			Let s(n) = 3/2 - sum{(h^2)/3^h, h = 1..n}.  Approximations follow:
n ... s(n) ...... 1/2^n
1 ... 1.16666 ... 0.500000
2 ... 0.72222 ... 0.250000
3 ... 0.38888 ... 0.125000
4 ... 0.03909 ... 0.062500
5 ... 0.08847 ... 0.031250
6 ... 0.03909 ... 0.015625
7 ... 0.01668 ... 0.007812
a(5) = 7 because s(7) < 1/32 < s(6).
		

Crossrefs

Programs

  • Mathematica
    z = 200; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}];
    d = N[Table[6 - p[k], {k, 1, z/5}], 12];
    f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1];
    u = Flatten[Table[f[n], {n, 1, z}]]; (* A248629 *)
    d = Differences[u];
    Flatten[Position[d, 1]];  (* A248630 *)

A248633 Least k such that 20/27- sum{(h^2)/4^h, h = 1..k} < 1/8^n.

Original entry on oeis.org

3, 5, 7, 9, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 27, 29, 30, 32, 33, 35, 36, 38, 40, 41, 43, 44, 46, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 64, 66, 67, 69, 70, 72, 73, 75, 76, 78, 80, 81, 83, 84, 86, 87, 89, 90, 92, 93, 95, 96, 98, 99, 101, 102, 104
Offset: 1

Views

Author

Clark Kimberling, Oct 11 2014

Keywords

Comments

This sequence provides insight into the manner of convergence of sum{(h^2)/4^h, h = 1..k} to 20/27. The difference sequence of A248633 entirely of 1s and 2s, so that A248634 and A248635 partition the positive integers.

Examples

			Let s(n) = 20/27 - sum{(h^2)/4^h, h = 1..n}.  Approximations follow:
n ... s(n) ........ 1/8^n
1 ... 0.49074 ... 0.125000
2 ... 0.24074 ... 0.015625
3 ... 0.10011 ... 0.001953
4 ... 0.03761 ... 0.000244
5 ... 0.01320 ... 0.000030
a(2) = 5 because s(5) < 1/8^2 < s(2).
		

Crossrefs

Programs

  • Mathematica
    z= 300; p[k_] := p[k] = Sum[(h^2/4^h), {h, 1, k}];
    d = N[Table[20/27 - p[k], {k, 1, z/5}], 12];
    f[n_] := f[n] = Select[Range[z], 20/27 - p[#] < 1/8^n &, 1];
    u = Flatten[Table[f[n], {n, 1, z}]]  (* A248633 *)
    d = Differences[u]
    Flatten[Position[d, 1]]  (* A248634 *)
    Flatten[Position[d, 2]]  (* A248635 *)

A248636 Least k such that 33/8- sum{(h^3)/3^h, h = 1..k} < 1/4^n.

Original entry on oeis.org

7, 9, 10, 12, 13, 15, 17, 18, 19, 21, 22, 24, 25, 27, 28, 29, 31, 32, 34, 35, 36, 38, 39, 40, 42, 43, 44, 46, 47, 48, 50, 51, 52, 54, 55, 56, 58, 59, 60, 62, 63, 64, 66, 67, 68, 70, 71, 72, 73, 75, 76, 77, 79, 80, 81, 83, 84, 85, 87, 88, 89, 90, 92, 93, 94
Offset: 1

Views

Author

Clark Kimberling, Oct 11 2014

Keywords

Comments

This sequence provides insight into the manner of convergence of sum{(h^3)/3^h, h = 1..k} to 33/8. The difference sequence of A248636 entirely of 1s and 2s, so that A248637 and A248638 partition the positive integers.

Examples

			Let s(n) = 33/8- sum{(h^3)/3^h, h = 1..n}.  Approximations follow:
n ... s(n) ........ 1/4^n
1 ... 3.79167 ... 0.250000
2 ... 2.90278 ... 0.062500
3 ... 1.90278 ... 0.015625
4 ... 1.11265 ... 0.003906
5 ... 0.59825 ... 0.000976
6 ... 0.30195 ... 0.000244
7 ... 0.14511 ... 0.000061
8 ... 0.06798 ... 0.000015
9 ... 0.03004 ... 0.000004
a(2) = 9 because s(9) < 1/16 < s(8).
		

Crossrefs

Programs

  • Mathematica
    z = 300; p[k_] := p[k] = Sum[(h^3/3^h), {h, 1, k}];
    d = N[Table[33/8 - p[k], {k, 1, z/5}], 12]
    f[n_] := f[n] = Select[Range[z], 33/8 - p[#] < 1/4^n &, 1];
    u = Flatten[Table[f[n], {n, 1, z}]]  (* A248636 *)
    d = Differences[u]
    v = Flatten[Position[d, 1]] (* A248637 *)
    w = Flatten[Position[d, 2]] (* A248638 *)
Showing 1-4 of 4 results.