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A227877
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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.
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3
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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
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OFFSET
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1,5
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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