Partial step: N is the sum of three triangular numbers when and only when (whh?) 8N+3 is the sum of three squares (from 1 + 8 * x(x+1)/2 = (2x+1)^2 ). The count of three-square reps is intimately connected to the class number of the quadratic number field for sqrt( - (8N+3)). 8*53+3 = 427, which is the largest imaginary quadratic field with class number 2. The class numbers, and therefore the number of 3-square reps, grow erratically, very roughly as sqrt(N). This establishes that the number of tri-tri reps grows to infinity, so the count of N with a particular # of tri-tri reps is finite. But it might be 0 reps, which would spoil a "well-defined" maximum N. It seems likely that every class number occurs, since for class # K there are roughly O(K) candidate fields. But I know of no proof. Rich -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Wednesday, January 15, 2014 12:14 PM To: math-fun Subject: [EXTERNAL] [math-fun] Largest n-fold tri-triangular numbers For today's foray into Waring theory ... Consider, please, OEIS sequence A002636, the number of ways of expressing n as the sum of 3 triangular numbers. (For the present purposes, 0 counts as a triangular number.) By inspecting the list, I conjecture that 53 is the largest number to have only one tri-triangular representation: 53 = 28 + 15 + 10. Is this very very hard to prove? The largest number I could find with only two tri-triangular representations is 194. The entries (53,194) are enough to see easily that this sequence -- if it is well-defined -- is not in OEIS. I conjecture that for every n in N1, there is a largest k such that A002636(k) = n. How hard is *this* to prove? It looks like when n = 3, k = 470; when n = 4, k = 788; and when n = 5, k = 1730. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun