OFFSET
1,7
COMMENTS
Equivalently, a(n) is the number of triples [n,k,m] with k>0 satisfying the Diophantine equation n*(3*n-1) + k*(3*k-1) - m*(3*m-1) = 0. Any such triple satisfies a triangle inequality, n+k > m. The n for which there is a triple [n,n,m] are listed in A137694. Solutions of the form [n,m-1,m] appear only when n=3*z+1, z > 0. The n for which a(n)=0 are listed in A135768.
REFERENCES
N. J. A. Sloane et al., "sum of 2 triangular numbers is a triangular number", math-fun mailing list, Feb. 19-29, 2020.
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
B. Klee, Pentathagorean triples: two easy proofs, seqfan mailing list, Mar. 20, 2020.
M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.
EXAMPLE
Isosceles case, n=5: 2*5*(3*5-1) - 7*(3*7-1) = 0.
MATHEMATICA
PentaTriples[PNn_] := Sort[Select[{PNn,
(-PNn + 3 PNn^2 + # - 3 #^2)/(6 #),
(-PNn + 3 PNn^2 + # + 3 #^2)/(6 #)
} & /@ Divisors[PNn*(3*PNn - 1)],
And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]]
Length[PentaTriples[#]] & /@ Range[100]
a[n_] := Length@FindInstance[n > 0 && y > 0 && z > 0 &&
n (3 n - 1) + y (3 y - 1) == z (3 z - 1), {y, z}, Integers, 10^9];
a /@ Range[100]
CROSSREFS
KEYWORD
nonn
AUTHOR
Bradley Klee, Mar 01 2020
STATUS
approved