OFFSET
1,5
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.
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.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 25 2013
STATUS
approved