A230596 Number of ways to write n = x + y + z with 0 < x <= y <= z such that x*y*z is a triangular number, and that x is a triangular number of the form (p^2 - 1)/8 with p an odd prime.

0, 0, 1, 0, 1, 1, 0, 2, 1, 1, 2, 2, 1, 1, 2, 2, 4, 1, 1, 2, 3, 2, 5, 1, 3, 3, 3, 3, 2, 8, 1, 4, 2, 2, 3, 5, 1, 3, 6, 3, 5, 3, 1, 6, 4, 5, 3, 3, 1, 6, 6, 3, 4, 2, 4, 3, 8, 3, 3, 8, 5, 2, 4, 4, 6, 6, 3, 6, 2, 3, 12, 7, 1, 10, 7, 3, 4, 5, 3, 7, 8, 2, 5, 4, 6, 4, 2, 5, 6, 6, 4, 4, 13, 6, 9, 6, 4, 10, 7, 4

Offset

1,8

Comments

Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 7.

(ii) For any integer n > 7, there are positive integers x, y, z with x + y + z = n such that x*y*z is a triangular number and x is among 1, 2, 3, 4, 5, 6.

Note that a(3k) and a(3k+2) are positive for every k = 1, 2, 3, .... In fact, 3k = 1 + k + (2k-1) with 1*k*(2k-1) = 2k*(2k-1)/2 a triangular number, and 3k+2 = 1 + k + (2k+1) with 1*k*(2k+1) = 2k(2k+1)/2 a triangular number.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Example

a(10) = 1 since 10 = 3 + 3 + 4, and 3 = (5^2-1)/8 with 5 an odd prime, and 3*3*4 = 8*9/2 is a triangular number.
a(31) = 1 since 31 = 3 + 11 + 17, and 3 = (5^2-1)/8 with 5 an odd prime, and 3*11*17 = 33*34/2 is a triangular number.

Mathematica

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
a[n_]:=Sum[If[TQ[(Prime[i]^2-1)/8*y*(n-(Prime[i]^2-1)/8-y)], 1, 0], {i, 2, PrimePi[Sqrt[8n/3+1]]}, {y, (Prime[i]^2-1)/8, (n-(Prime[i]^2-1)/8)/2}]
Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A000217, A000040, A132399, A229166, A230121, A230451.

Sequence in context: A054526 A113453 A245851 * A307079 A330190 A356300

Adjacent sequences: A230593 A230594 A230595 * A230597 A230598 A230599

Keyword

nonn

Author

Zhi-Wei Sun, Oct 24 2013

Status

approved