Here, apparently, is the missing computer code (Mathematica): RichTriples[TNn_] := Sort[Select[{TNn, k, m} /. Solve[MapThread[#1 == #2 &, {{TNn (TNn + 1)/#, #}, {(m + k + 1), (m - k)}}] ][[1]] & /@ Complement[Divisors[TNn (TNn + 1)][[2 ;; -2]], {TNn}], And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]] SymTriples[sqNn_] := Sort[Select[({sqNn, k, m} - 1)/2 /. Solve[MapThread[#1 == #2 &, {{(sqNn^2 - 1)/#, #}, {(m - k), (m + k)}}] ][[1]] & /@ Divisors[sqNn^2 - 1][[2 ;; -2]], And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]] Position[RichTriples /@ Range[2, 200], {}][[All, 1]] /. x_Integer :> {x + 1, x + 2} SymTriples[2 # + 1] & /@ Range[2, 20]; %[[All, 1, 2]] Out[1] = {{2, 3}, {3, 4}, {4, 5}, {7, 8}, {16, 17}, {31, 32}, {127, 128}} Out[2] = {2, 5, 9, 3, 5, 27, 10, 4, 8, 14, 17, 9, 5, 21, 135, 12, 14, 35, 6} Absence of fail cases in symmetric form follows from odd + 1 = even, not prime. I did not prove if the algorithm generates all of S2, nor if it includes all terms from OEIS, but it looks that way. Also interesting: Complement[SymTriples[2 # + 1], RichTriples[#]] & /@ Range[2, 50] Complement[RichTriples[#], SymTriples[2 # + 1]] & /@ Range[2, 50] --Brad On Wed, Feb 19, 2020 at 7:22 PM Brad Klee <bradklee@gmail.com> wrote:
Elegant proof: The two sets,
S1 = { (n,k,m) in Z^3 : n*(n+1) + k*(k+1) - m*(m+1) = 0 } S2 = { (x,y,z) in (2Z+1)^3 : x^2 + y^2 - z^2 - 1 = 0 }