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%I #31 Oct 30 2013 04:58:25
%S 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,
%T 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,
%U 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
%N 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.
%C Conjecture: a(n) > 0 for all n > 4.
%C 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.
%C 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.
%H Zhi-Wei Sun, <a href="/A227877/b227877.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;4704f25.1310">A new conjecture on triangular numbers</a>, a message to Number Theory List, Oct. 25, 2013.
%e 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.
%e 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.
%t TQ[n_]:=IntegerQ[Sqrt[8n+1]]
%t 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}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000040, A000217, A132399, A229166, A230121, A230451, A230596.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Oct 25 2013
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