[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.
Allan, You should look at this paper by Ono, Robbins and Wahl http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/006.pdf . In it, among other things, they give an explicit formula for the number of representations of n as a sum of 3 triangular numbers. Victor On Wed, Jan 15, 2014 at 2:13 PM, Allan Wechsler <acwacw@gmail.com> wrote:
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
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
Thank you for the pointers, Victor and Rich. With these clues, I looked in OEIS for the values of 8K+3 for the K's that I gave earlier, and sure enough, I found A095811. Note that the comment deprecates these values as conjectural, though we know that at least A(1) is correct. On Wed, Jan 15, 2014 at 2:51 PM, Schroeppel, Richard <rschroe@sandia.gov>wrote:
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
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related stuff ... Done OEIS offer Superseeker, and does it find A095811 when you ask about your K sequence? Every positive integer is also the sum of three generalized pentagonal numbers. Ordinary pentagonal numbers are (3n^2-n)/2 for n>=0. GPNs allow n<0 as well, or can be taken as (3n^2+-n)/2 with n>=0. These are the exponents in the non-zero terms of the (Euler-)Ramanujan product of (1-z^i). The series begins 1 - z - z^2 + z^5 + z^7 - z^12 - ... This result follows from the 3-square representations of 24N+3. The individual squares must be odd (easy). I also need that there's a rep where at least one of the squares is not a multiple of 3 (and therefore all three squares are non-multiples of 3). This is easy unless 24N+3 = T happens to be a multiple of 9. For this case, we assume (by induction) that T/9 has a 3-square rep A2+B2+C2 with C a non-multiple-of-3. Then T = (2A+2B-C)^2 + (2A-B+2C)^2 + (-A+2B+2C)^2. If 2A+2B-C is a multiple of 3, then 2A+2B+C is not, so we flip the sign of C in the T formula. This gives us T as the sum of 3 squares of numbers in the shape 6K+-1, which we force to 6K-1 by negating K where required. Then feed the three Ks into the (3n^2-n)/2 formula to get the pentagonal numbers. 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 2:59 PM To: math-fun Subject: Re: [math-fun] [EXTERNAL] Largest n-fold tri-triangular numbers Thank you for the pointers, Victor and Rich. With these clues, I looked in OEIS for the values of 8K+3 for the K's that I gave earlier, and sure enough, I found A095811. Note that the comment deprecates these values as conjectural, though we know that at least A(1) is correct. On Wed, Jan 15, 2014 at 2:51 PM, Schroeppel, Richard <rschroe@sandia.gov>wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (3)
-
Allan Wechsler -
Schroeppel, Richard -
Victor Miller