[seqfan] Re: pentathagorean triples

[Neil Sloane]

Brad, Olivier - excellent, please submit it (or them) (the pentagonal +
pentagonal = pentagonal version)

Call a triple that works, [n,k,m]. Then the sequences would be something
like:
the list of n's that work
for given n , how many solutions for k
for an n that works, the smallest k and the corresponding m




Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com



On Sun, Mar 1, 2020 at 12:02 PM Brad Klee <bradklee at gmail.com> wrote:

> 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
>
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