To say that the problem is "completely solved" may be an overstatement, since we haven't answered a fundamental question about the general quadratic case. If you count the output of SymTriples, it should equal the number of odd divisors of n*(n+1), less one: A309507[n_] := Subtract[Length[Select[Divisors[n + n^2], OddQ[#] &]], 1] A309507 /@ Range[2, 50] This code would be a good addition to OEIS; however, it is easy to doubt theorem five of the Tripathi paper ( https://www.fq.math.ca/Papers1/46_47-4/Tripathi.pdf ), because we already have two similar algorithms that give different output. The proof given for Theorem 5 doesn't say anything about (x-y)*(x+y) being, in some sense, a "complete factorization". If we adapted this style of proof to another less-symmetric algorithm, we could end up with a wrong theorem. By changing coordinates, we can always transform a similar quadratic problem to the form (x-y)*(x+y)=n, with some extra conditions on the divisors of (x,y,n). Is it always the case that the symmetric factorization is complete, i.e. that it's solution over divisors of n gives all possible solutions? I wouldn't be surprised if this question took more than two days to answer... --Brad On Fri, Feb 21, 2020 at 10:59 PM Neil Sloane <njasloane@gmail.com> wrote:
Thanks to everyone who replied. Wow! In two days the problem was completely solved. I've prepared a summary of the upper bounds, which I will add to A082183 tomorrow.
The original question was to find a bound for A082813. New sequences arising from the suggestions and comments are A332547, -548, -549, -552, -553, -554. Thanks, Allan, Brad, Michael, Rich.
Brad, thanks also for the pointer to the "squares" version, A055527. It suggests that one should look at a general quadratic form (instead of n^2 or n*(n+1)) to really understand what is going on ...!
Neil
On Thu, Feb 20, 2020 at 10:34 AM Brad Klee <bradklee@gmail.com> wrote:
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 }
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