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.

A248563 Numbers k such that A248562(k+1) = A248562(k) + 1.

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 12, 15, 17, 19, 22, 24, 27, 29, 32, 34, 37, 39, 42, 44, 47, 49, 52, 55, 57, 60, 62, 65, 68, 70, 73, 75, 78, 81, 83, 86, 89, 91, 94, 96, 99, 102, 104, 107, 110, 112, 115, 118, 120, 123, 126, 128, 131, 134, 136, 139, 142, 144, 147, 150, 152
Offset: 1

Views

Author

Clark Kimberling, Oct 09 2014

Keywords

Examples

			(A248562(k+1) - A248562(k)) = (1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2,...), so that A248563 = (1, 2, 4, 6, 8, 10, 12, 15, 17, ..) and A248564 = (3, 5, 7, 9, 11, 13, 14, 16, ...).

Links

Crossrefs

Cf. A248562, A248564, A248560, A248566.

Programs

  • Mathematica
    z = 300; p[k_] := p[k] = Sum[1/(h*3^h), {h, 1, k}];
N[Table[Log[3/2] - p[n], {n, 1, z/5}]]
f[n_] := f[n] = Select[Range[z], Log[3/2] - p[#] < 1/6^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]    (* A248562 *)
Flatten[Position[Differences[u], 1]]   (* A248563 *)
Flatten[Position[Differences[u], 2]]   (* A248564 *)

A248564 Numbers k such that A248562(k+1) = A248562(k) + 2.

Original entry on oeis.org

3, 5, 7, 9, 11, 13, 14, 16, 18, 20, 21, 23, 25, 26, 28, 30, 31, 33, 35, 36, 38, 40, 41, 43, 45, 46, 48, 50, 51, 53, 54, 56, 58, 59, 61, 63, 64, 66, 67, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 92, 93, 95, 97, 98, 100, 101, 103, 105, 106, 108
Offset: 1

Views

Author

Clark Kimberling, Oct 09 2014

Keywords

Examples

			(A248562(k+1) - A248562(k)) = (1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2,...), so that A248563 = (1, 2, 4, 6, 8, 10, 12, 15, 17, ..) and A248564 = (3, 5, 7, 9, 11, 13, 14, 16, ...).

Links

Crossrefs

Cf. A248562, A248563, A248561, A248567.

Programs

  • Mathematica
    z = 300; p[k_] := p[k] = Sum[1/(h*3^h), {h, 1, k}];
N[Table[Log[3/2] - p[n], {n, 1, z/5}]]
f[n_] := f[n] = Select[Range[z], Log[3/2] - p[#] < 1/6^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]    (* A248562 *)
Flatten[Position[Differences[u], 1]]   (* A248563 *)
Flatten[Position[Differences[u], 2]]   (* A248564 *)

A248560 Numbers k such that A248559(k+1) = A248559(k) + 1.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 10, 11, 13, 15, 17, 19, 21, 23, 25, 28, 30, 32, 34, 36, 38, 41, 43, 45, 47, 50, 52, 54, 56, 59, 61, 63, 65, 68, 70, 72, 75, 77, 79, 82, 84, 86, 88, 91, 93, 95, 98, 100, 102, 105, 107, 109, 112, 114, 116, 119, 121, 123, 126, 128, 130
Offset: 1

Views

Author

Clark Kimberling, Oct 09 2014

Keywords

Examples

			(A248559(k+1) - A248559(k)) = (1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2,...), so that A248561 = (1, 2, 3, 5, 6, 8, 10, 11, 13, ...) and A248562 = (4, 7, 9, 12, 14, 16, 18, 20, ...).

Links

Crossrefs

Cf. A248559, A248561.

Programs

  • Mathematica
    z = 200; p[k_] := p[k] = Sum[1/(h*2^h), {h, 1, k}]
N[Table[Log[2] - p[n], {n, 1, z/5}]]
f[n_] := f[n] = Select[Range[z], Log[2] - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]    (* A248559 *)
Flatten[Position[Differences[u], 1]]   (* A248560 *)
Flatten[Position[Differences[u], 2]]   (* A248561 *)

A248561 Numbers k such that A248559(k+1) = A248559(k) + 2.

Original entry on oeis.org

4, 7, 9, 12, 14, 16, 18, 20, 22, 24, 26, 27, 29, 31, 33, 35, 37, 39, 40, 42, 44, 46, 48, 49, 51, 53, 55, 57, 58, 60, 62, 64, 66, 67, 69, 71, 73, 74, 76, 78, 80, 81, 83, 85, 87, 89, 90, 92, 94, 96, 97, 99, 101, 103, 104, 106, 108, 110, 111, 113, 115, 117, 118
Offset: 1

Views

Author

Clark Kimberling, Oct 09 2014

Keywords

Examples

			(A248559(k+1) - A248559(k)) = (1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2,...), so that A248561 = (1, 2, 3, 5, 6, 8, 10, 11, 13, ...) and A248562 = (4, 7, 9, 12, 14, 16, 18, 20, ...).

Links

Crossrefs

Cf. A248559, A248560.

Programs

  • Mathematica
    z = 200; p[k_] := p[k] = Sum[1/(h*2^h), {h, 1, k}]
N[Table[Log[2] - p[n], {n, 1, z/5}]]
f[n_] := f[n] = Select[Range[z], Log[2] - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]    (* A248559 *)
Flatten[Position[Differences[u], 1]]   (* A248560 *)
Flatten[Position[Differences[u], 2]]   (* A248561 *)
Showing 1-4 of 4 results.

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