[seqfan] pentathagorean triples

[Brad Klee]

After A046079 and A309507, Olivier Gérard suggested that
the same sort of thing could be done with pentagonal numbers,
and I calculated the following using Mathematica:

PentTriples[PNn_] :=
 Sort[Select[{PNn, k, m} /.
   Solve[MapThread[#1 == #2 &, {{PNn (3 PNn - 1)/#, #}, {(3 m +
             3 k - 1), (m - k)}}]][[1]] & /@ Divisors[PNn (3 PNn - 1)],
   And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]]

Length[PentTriples[#]] & /@ Range[30]
Out[]= {0, 0, 0, 1, 1, 0, 2, 0, 0, 1, 0, 1, 1, 1, 0, 1, 2, 0, 2, 1,
1, 3, 1, 0, 2, 3, 0, 3, 1, 1}  (* nAn *)

These numbers are also supported by brute force calculation
of solutions to n*(3*n-1) + y*(3*y-1) = z*(3*z-1).

--Brad