A227877 Number of ways to write n = x + y + z (x, y, z > 0) such that x*y and x*z are triangular numbers, and 6*y-1 and 6*z+1 are both prime.

0, 0, 1, 0, 3, 2, 2, 3, 3, 7, 3, 6, 3, 3, 2, 3, 7, 6, 7, 5, 4, 5, 10, 2, 10, 4, 5, 2, 2, 9, 5, 9, 2, 4, 3, 4, 5, 7, 5, 11, 12, 5, 8, 11, 12, 5, 11, 3, 7, 11, 4, 10, 6, 2, 9, 11, 8, 7, 9, 8, 9, 4, 3, 4, 10, 6, 9, 15, 9, 17, 3, 3, 8, 12, 10, 5, 1, 7, 9, 16, 8, 17, 6, 8, 16, 6, 8, 8, 10, 1, 6, 4, 8, 5, 23, 11, 2, 9, 6, 14

Offset: 1

Author

Zhi-Wei Sun, Oct 25 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 4.

For n = 4*k - 1, we have n = (2k-1) + k + k with (2k-1)*k = 2k*(2k-1)/2 a triangular number. For n = 4*k + 1, we have n = (2k+1) + k + k with (2k+1)*k = 2k*(2k+1)/2 a triangular number. For n = 4*k + 2, we have n = (2k+1) + k + (k+1), and (2k+1)*k = 2k*(2k+1)/2 and (2k+1)*(k+1) = (2k+1)(2k+2)/2 are both triangular numbers.

For n = 5*k, we have n = k + (2k-1) + (2k+1), and k*(2k-1) = 2k*(2k-1)/2 and k*(2k+1) = 2k*(2k+1)/2 are both triangular numbers. For n = 5*k - 2, we have n = k + (2k-1) + (2k-1) with k*(2k-1) = 2k*(2k-1)/2 a triangular number. For n = 5*k + 2, we have n = k + (2k+1) + (2k+1) with k*(2k+1) = 2k*(2k+1)/2 a triangular number.

Examples

			a(77) = 1 since 77 = 1 + 10 + 66, and 1*10 = 4*5/2 and 1*66 = 11*12/2 are triangular numbers, and 6*10 - 1 = 59 and 6*66 + 1 = 397 are both prime.
a(90) = 1 since 90 = 45 + 22 + 23, and 45*22 = 44*45/2 and 45*23 = 45*46/2 are triangular numbers, and 6*22 - 1 = 131 and 6*23 + 1 = 139 are both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, A new conjecture on triangular numbers, a message to Number Theory List, Oct. 25, 2013.

Crossrefs

Cf. A000040, A000217, A132399, A229166, A230121, A230451, A230596.

Programs

Mathematica

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
a[n_]:=Sum[If[PrimeQ[6j-1]&&PrimeQ[6(n-i-j)+1]&&TQ[i*j]&&TQ[i(n-i-j)],1,0],{i,1,n-2},{j,1,n-1-i}]
Table[a[n],{n,1,100}]