A379697 For n >= 3, a(n) is the least k >= 0 such that (k + 1)*(2*n + k) / 2 is a triangular number (A000217).

0, 2, 5, 0, 1, 20, 4, 0, 2, 6, 8, 2, 0, 10, 119, 3, 4, 20, 0, 1, 5, 14, 2, 5, 1, 0, 32, 6, 7, 464, 20, 2, 8, 0, 24, 8, 2, 4, 65, 9, 10, 47, 0, 3, 11, 30, 17, 2, 3, 1, 59, 12, 0, 2, 21, 4, 14, 38, 40, 14, 4, 42, 101, 0, 16, 74, 2, 5, 17, 46, 48, 17, 5, 1, 11, 0, 19, 125, 10, 6, 20, 54, 1, 20, 6, 44, 272, 21, 0

Offset

3,2

Comments

Also for n >= 3, a(n) is the least k >= 0 such that the Sum_{i = 0..k} (n + i) is a triangular number (A000217). For n = 0, 1 the Sum is a triangular number for all n. For n = 2, there is no solution.

Links

Table of n, a(n) for n=3..91.

Formula

a(n) = 0 for n from A000217.

a(n) = 1 for n from A074377 AND n is not a triangular number.

Example

n = 4: the least k >= 0 such that (k + 1)*(8 + k)/2 is a triangular number is k = 2, thus a(4) = 2.
n = 6: the least k >= 0 such that (k + 1)*(12 + k)/2 is a triangular number is k = 0, thus a(6) = 0.

Mathematica

a[n_] := Module[{k = 0}, While[! IntegerQ[Sqrt[4*(k + 1)*(2*n + k) + 1]], k++]; k]; Array[a, 100, 3] (* Amiram Eldar, Dec 30 2024 *)

Prog

(PARI) a(n) = my(k=0); while (!ispolygonal((k + 1)*(2*n + k)/2, 3), k++); k; \\ Michel Marcus, Dec 30 2024

Crossrefs

Cf. A000217, A074377, A379676.

Sequence in context: A011183 A005671 A343605 * A296047 A307237 A127863

Adjacent sequences: A379692 A379693 A379694 * A379698 A379699 A379700

Keyword

nonn,new

Author

Ctibor O. Zizka, Dec 30 2024

Status

approved