A387517 For n >= 0, a(n) is the least s >= 0 such that (s + 1)*(s + n) is a triangular number (A000217).

0, 0, 1, 0, 1, 2, 0, 4, 5, 6, 0, 8, 9, 1, 11, 0, 13, 1, 15, 16, 2, 0, 19, 20, 2, 22, 23, 3, 0, 26, 5, 3, 1, 2, 4, 32, 0, 34, 1, 36, 37, 5, 39, 40, 41, 0, 43, 44, 6, 2, 47, 48, 6, 50, 9, 0, 4, 54, 5, 1, 57, 4, 8, 60, 61, 62, 0, 1, 2, 9, 67, 13, 3, 9, 71

Offset

0,6

Comments

For n >= 3, 0 <= a(n) <= (n - 3).

a(n) are the least solutions s (together with some k) of the Diophantine equation 1 + ... + k = (k + n) + ... + (k + n + s) for k >= 1 and s >= 0, since that becomes k = ((2*s+1) + sqrt(8*(s + 1)*(s + n) + 1)) / 2 which is an integer iff (s + 1)*(s + n) is a triangular number.

Links

James C. McMahon, Table of n, a(n) for n = 0..1000

Ctibor O. Zizka, Plot of n vs. a(n)

Formula

For n >= 0, a(A000217(n)) = 0.

For n >= 1, a(A074378(n) - 1) = 1.

Example

For n = 7: (s + 1)*(s + 7) is a triangular number for the least s = 4, thus a(7) = 4.

Mathematica

a[n_]:=Module[{s=0}, While[!IntegerQ[(Sqrt[8*(s+1)*(s+n)+1]-1)/2], s++]; s]; Array[a, 75, 0] (* James C. McMahon, Sep 07 2025 *)

Prog

(PARI) a(n) = my(s=0); while (!ispolygonal((s + 1)*(s + n), 3), s++); s; \\ Michel Marcus, Sep 01 2025
(Python)
from sympy.abc import x, y, t
from sympy.solvers.diophantine.diophantine import diop_quadratic
def A387517(n): return min(d for d in (a[0].subs(t, 0) for a in diop_quadratic((x+1)*(x+n)*2-y*(y+1), t)) if d>=0) # Chai Wah Wu, Sep 07 2025

Crossrefs

Cf. A000217, A074378.

Sequence in context: A111677 A380703 A326186 * A326055 A276331 A049271

Adjacent sequences: A387514 A387515 A387516 * A387518 A387519 A387520

Keyword

nonn

Author

Ctibor O. Zizka, Sep 01 2025

Status

approved