

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
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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 + (2k1) with 1*k*(2k1) = 2k*(2k1)/2 a triangular number, and 3k+2 = 1 + k + (2k+1) with 1*k*(2k+1) = 2k(2k+1)/2 a triangular number.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000


EXAMPLE

a(10) = 1 since 10 = 3 + 3 + 4, and 3 = (5^21)/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 = (5^21)/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]^21)/8*y*(n(Prime[i]^21)/8y)], 1, 0], {i, 2, PrimePi[Sqrt[8n/3+1]]}, {y, (Prime[i]^21)/8, (n(Prime[i]^21)/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

ZhiWei Sun, Oct 24 2013


STATUS

approved



