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 A230596 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. 4
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,8 COMMENTS Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 7. (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. 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. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 EXAMPLE 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. 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. MATHEMATICA TQ[n_]:=IntegerQ[Sqrt[8n+1]] 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}] Table[a[n], {n, 1, 100}] CROSSREFS Cf. A000217, A000040, A132399, A229166, A230121, A230451. Sequence in context: A054526 A113453 A245851 * A307079 A330190 A356300 Adjacent sequences: A230593 A230594 A230595 * A230597 A230598 A230599 KEYWORD nonn AUTHOR Zhi-Wei Sun, Oct 24 2013 STATUS approved

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Last modified January 28 05:03 EST 2023. Contains 359850 sequences. (Running on oeis4.)