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

Author

Zhi-Wei Sun, Oct 24 2013

Keywords

• nonn

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.

Examples

			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.
		

Links

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

Crossrefs

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

Programs

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}]