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 1-2 of 2 results.

A076117 Cubes (or 0) from A076116.

Original entry on oeis.org

1, 27, 27, 0, 125, 27, 343, 0, 216, 125, 1331, 0, 2197, 343, 3375, 216, 4913, 729, 6859, 0, 9261, 1331, 12167, 0, 1000, 2197, 729, 0, 24389, 3375, 29791, 0, 35937, 4913, 42875, 0, 50653, 6859, 59319, 0, 68921, 9261, 79507, 0, 3375, 12167, 103823, 5832
Offset: 1

Views

Author

Amarnath Murthy, Oct 09 2002

Keywords

Crossrefs

Cf. A076116.
Cf. A076115.

Formula

Let c = A076116(n); if c > 0, then a(n) = n*(2c + n - 1)/2. If c = 0 then a(n) = 0. - David Wasserman, Apr 02 2005

Extensions

More terms from David Wasserman, Apr 02 2005

A379676 For n >= 0, a(n) is the least k >= 2 such that (n + 1)*(2*k + n) / 2 is a triangular number (A000217).

Original entry on oeis.org

3, 7, 4, 15, 7, 5, 10, 31, 13, 6, 16, 12, 19, 10, 7, 63, 25, 11, 28, 22, 8, 15, 34, 21, 37, 16, 40, 9, 43, 20, 46, 127, 14, 21, 18, 10, 55, 25, 15, 19, 61, 26, 64, 45, 11, 30, 70, 44, 73, 31, 21, 55, 79, 35, 12, 70, 22, 36, 88, 18, 91, 40, 34, 255, 31, 13, 100, 19, 28, 24, 106, 92, 109, 46, 29, 78, 25, 14, 118, 91, 121, 51, 124, 63, 42, 55, 35, 39, 133, 43, 15
Offset: 0

Views

Author

Ctibor O. Zizka, Dec 29 2024

Keywords

Comments

Also for n >= 0, a(n) is the least k >= 2 such that the Sum_{i = 0..n} (k + i) is a triangular number (A000217). For k = 0, 1 the Sum is a triangular number for all n. The sequences A076114 and A076116 are for square sum and cube sum.

Examples

			n = 4: the least k >= 2 such that (4 + 1)*(2*k + 4)/2 = 5*k + 10 is a triangular number is k = 7, thus a(4) = 7.
n = 5: the least k >= 2 such that (5 + 1)*(2*k + 5)/2 = 6*k + 15 is a triangular number is k = 5, thus a(5) = 5.

Crossrefs

Cf. A000217, A076114, A076116, A083390.

Programs

  • Mathematica
    a[n_] := Module[{k = 2}, While[! IntegerQ[Sqrt[4*(n + 1)*(2*k + n) + 1]], k++]; k]; Array[a, 100, 0] (* Amiram Eldar, Dec 30 2024 *)
  • PARI
    a(n) = my(k=2); while (!ispolygonal((n + 1)*(2*k + n)/2, 3), k++); k; \\ Michel Marcus, Dec 30 2024

Formula

For i >= 0, a(2^i - 1) = 2^(i + 2) - 1, max. values of a(n).
For i >= 0, a(i*(i + 3)/2) = i + 3, min. values of a(n).
For i >= 1, i is not from A083390, a(2*i) = (3*i + 1).
Showing 1-2 of 2 results.

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