Observe that: 0 < a_n = m^2 - n^2 - k^2 = n + k - m < n for all data of A082184, or if you want, this identity can also be written: 0 < m - k < n, 0 < m < n + k. I did not prove, but would guess that this identity, a generalization of the triangle inequality, holds throughout. The sequence of a_n = n + k - m, a_n: 1, 2, 3, 2, 3, 6, 5, 3, 5, 7, 8, 6, 4, 10, 15, 8, 9, 14, 5, 7 . . . (NaN) seems a preferable choice for characterizing and analyzing irregularity. In plain language, it can be described as the distance of (n,k,m) from a Pythagorean triple. Maybe there is some way to relate this to Pythagorean Triples? --Brad On Wed, Feb 19, 2020 at 11:46 AM Allan Wechsler <acwacw@gmail.com> wrote:
I think that k = T(n) - 1 is an upper bound. T(k) makes a huge triangle; all the elements of the T(n) triangle can be thinly plated onto the side of the big one as a single additional row, producing T(k+1), so m = k+1. I think m-k would also make an interesting sequence.
On Wed, Feb 19, 2020 at 12:35 PM Neil Sloane <njasloane@gmail.com> wrote:
Let T(i) = i(i+1)/2. Given n, let k be smallest number such that T(n) + T(k) = T(m) for some m. The k and m values are in A082183 and A082184. It must be classical that k and m always exist. - can someone supply a reference or a proof?
The graph of the k values is quite irregular. Is there an upper bound? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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