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%I #11 Oct 24 2013 23:59:17
%S 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,
%T 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,
%U 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
%N 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.
%C Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 7.
%C (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.
%C 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.
%H Zhi-Wei Sun, <a href="/A230596/b230596.txt">Table of n, a(n) for n = 1..10000</a>
%e 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.
%e 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.
%t TQ[n_]:=IntegerQ[Sqrt[8n+1]]
%t 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}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000217, A000040, A132399, A229166, A230121, A230451.
%K nonn
%O 1,8
%A _Zhi-Wei Sun_, Oct 24 2013
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