[math-fun] Isodiametric problem
A few days ago, at the AMS meeting, I saw a neat talk by Michael Mossinghoff about "isodiametic" problems -- e.g. Fix the diameter of a conv ex n-gon and ask for the ones with maximum perimeter. He reduced the problem to finding univariate polynomials whose non-zero coefficients alternate between +/- 1 and are divisible by a particular cyclotomic polynomial.. He was able to calculate all such up to about 500. Each such polynomial is compactly encoded by th length of the gaps between the non-zero terms. For small n almost all were periodic. He showed how a certain family of periodic constrctions give solutions, but as n got bigger the "sporadic" ones swamped them. He conjectued that these are eventually all of them. He had lots of pictures. The paper is not available yet. Victor
On Fri, Jan 9, 2009 at 1:35 PM, victor miller <victorsmiller@gmail.com> wrote:
A few days ago, at the AMS meeting, I saw a neat talk by Michael Mossinghoff about "isodiametic" problems -- e.g. Fix the diameter of a conv ex n-gon and ask for the ones with maximum perimeter.
If this is an interesting problem, and calculating the result for n <= 500 is a worthwhile feat, then my intuition that the answer is always a regular polygon must be wrong. What is the smallest n for which the answer is not regular, and what does the resulting n-gon look like? My guess is that the answer is n = 5. If so, could this be related somehow to the fact that if you want to find the minimum x such that 5 discs of diameter x cover a disc of diameter 1, the configuration of these discs that exhibits the minimum is not a 5-fold symmetric one. -- Andy.Latto@pobox.com
If I remember correctly for odd n, it's regular, but for even n it gets interesting. I think there's a non-regular example for n=6. Victor On 1/9/09, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Jan 9, 2009 at 1:35 PM, victor miller <victorsmiller@gmail.com> wrote:
A few days ago, at the AMS meeting, I saw a neat talk by Michael Mossinghoff about "isodiametic" problems -- e.g. Fix the diameter of a conv ex n-gon and ask for the ones with maximum perimeter.
If this is an interesting problem, and calculating the result for n <= 500 is a worthwhile feat, then my intuition that the answer is always a regular polygon must be wrong. What is the smallest n for which the answer is not regular, and what does the resulting n-gon look like?
My guess is that the answer is n = 5. If so, could this be related somehow to the fact that if you want to find the minimum x such that 5 discs of diameter x cover a disc of diameter 1, the configuration of these discs that exhibits the minimum is not a 5-fold symmetric one.
-- Andy.Latto@pobox.com
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Ron Graham was apparently the first solver for n=6 http://mathworld.wolfram.com/GrahamsBiggestLittleHexagon.html http://mathworld.wolfram.com/BiggestLittlePolygon.html http://www.davidson.edu/math/mossinghoff/ --Ed Pegg Jr --- On Fri, 1/9/09, Andy Latto <andy.latto@pobox.com> wrote: From: Andy Latto <andy.latto@pobox.com> Subject: Re: [math-fun] Isodiametric problem To: "math-fun" <math-fun@mailman.xmission.com> Date: Friday, January 9, 2009, 1:15 PM On Fri, Jan 9, 2009 at 1:35 PM, victor miller <victorsmiller@gmail.com> wrote:
A few days ago, at the AMS meeting, I saw a neat talk by Michael Mossinghoff about "isodiametic" problems -- e.g. Fix the diameter of a conv ex n-gon and ask for the ones with maximum perimeter.
If this is an interesting problem, and calculating the result for n <= 500 is a worthwhile feat, then my intuition that the answer is always a regular polygon must be wrong. What is the smallest n for which the answer is not regular, and what does the resulting n-gon look like? My guess is that the answer is n = 5. If so, could this be related somehow to the fact that if you want to find the minimum x such that 5 discs of diameter x cover a disc of diameter 1, the configuration of these discs that exhibits the minimum is not a 5-fold symmetric one. -- Andy.Latto@pobox.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Graham's polygon haas maximal area, rather than perimeter! WFL On 1/9/09, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
Ron Graham was apparently the first solver for n=6
http://mathworld.wolfram.com/GrahamsBiggestLittleHexagon.html http://mathworld.wolfram.com/BiggestLittlePolygon.html http://www.davidson.edu/math/mossinghoff/
--Ed Pegg Jr
--- On Fri, 1/9/09, Andy Latto <andy.latto@pobox.com> wrote: From: Andy Latto <andy.latto@pobox.com> Subject: Re: [math-fun] Isodiametric problem To: "math-fun" <math-fun@mailman.xmission.com> Date: Friday, January 9, 2009, 1:15 PM
On Fri, Jan 9, 2009 at 1:35 PM, victor miller <victorsmiller@gmail.com> wrote:
A few days ago, at the AMS meeting, I saw a neat talk by Michael Mossinghoff about "isodiametic" problems -- e.g. Fix the diameter of a conv ex n-gon and ask for the ones with maximum perimeter.
If this is an interesting problem, and calculating the result for n <= 500 is a worthwhile feat, then my intuition that the answer is always a regular polygon must be wrong. What is the smallest n for which the answer is not regular, and what does the resulting n-gon look like?
My guess is that the answer is n = 5. If so, could this be related somehow to the fact that if you want to find the minimum x such that 5 discs of diameter x cover a disc of diameter 1, the configuration of these discs that exhibits the minimum is not a 5-fold symmetric one.
-- Andy.Latto@pobox.com
_______________________________________________ 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
I should have said that Mike Mossinghoff said that he had earlier trated the maxinal area problem. -- I think in a paper in Discrete and Computational Geometry, but that the maximal perimeter problem was new. Victor On 1/9/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
Graham's polygon haas maximal area, rather than perimeter! WFL
On 1/9/09, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
Ron Graham was apparently the first solver for n=6
http://mathworld.wolfram.com/GrahamsBiggestLittleHexagon.html http://mathworld.wolfram.com/BiggestLittlePolygon.html http://www.davidson.edu/math/mossinghoff/
--Ed Pegg Jr
--- On Fri, 1/9/09, Andy Latto <andy.latto@pobox.com> wrote: From: Andy Latto <andy.latto@pobox.com> Subject: Re: [math-fun] Isodiametric problem To: "math-fun" <math-fun@mailman.xmission.com> Date: Friday, January 9, 2009, 1:15 PM
On Fri, Jan 9, 2009 at 1:35 PM, victor miller <victorsmiller@gmail.com> wrote:
A few days ago, at the AMS meeting, I saw a neat talk by Michael Mossinghoff about "isodiametic" problems -- e.g. Fix the diameter of a conv ex n-gon and ask for the ones with maximum perimeter.
If this is an interesting problem, and calculating the result for n <= 500 is a worthwhile feat, then my intuition that the answer is always a regular polygon must be wrong. What is the smallest n for which the answer is not regular, and what does the resulting n-gon look like?
My guess is that the answer is n = 5. If so, could this be related somehow to the fact that if you want to find the minimum x such that 5 discs of diameter x cover a disc of diameter 1, the configuration of these discs that exhibits the minimum is not a 5-fold symmetric one.
-- Andy.Latto@pobox.com
_______________________________________________ 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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
A very belated response. Not squares, but related. Using Triangular numbers. 1 3 6 10 15 21 T[3] = 3! T[3] + T[6] = 4! T[14]+T[5] = T[15] = 5! T[45] + T[89] = 7! T[210] + T[825] = 9! T[1770] + T[2030] = 10! T[71504] + T[85680] = 13! T[213384] + T[1603064] = T[299894] + T[1589154] = 15! I don't see an easy way to extend these. The density of triangular numbers seems to be sufficient for extended solutions. --Ed Pegg Jr Date: Mon, 3 Jul 2006 11:44:20 -0600 (MDT) From: Richard Guy <rkg@cpsc.ucalgary.ca> Reply-To: math-fun <math-fun@mailman.xmission.com> To: Number Theory List <NMBRTHRY@listserv.nodak.edu>, Math Fun <math-fun@mailman.xmission.com> Subject: [math-fun] Factorial n Presumably 0! = 1! = 0^2 + 1^2. 2! = 1^2 + 1^2 6! = 12^2 + 24^2 are the only integer solutions of n! = x^2 + y^2 but is there a proof? R.
For the triangular number problem, multiply the equation by 8 and add 2. Using T(x) = x(x+1)/2, T(a) + T(b) = N! becomes (2a+1)^2 + (2b+1)^2 = 8 N! + 2. The RHS should sometimes meet the criteria for being the sum of two squares, often multiple times. The criteria is that there is no 4K+3 prime divisor to an odd power. If you have the factorization, or know the number is twice a prime, you can calculate the sum-of-two-square representations, and then the triangles. Wrt RKG's original problem, N! = A^2 + B^2: N! has a gaggle of prime divisors that occur to the first power: Any prime P in the range N/2 < P <= N will divide N! to exactly the first power. If any prime in this range is a 4K+3 prime, it prevents splitting N! into two squares. I assume there's a quantitative version of PNT for 4K+3 primes which will give a calculable bound on N. Maybe there's a clever way to adopt the proof of Bertrand's postulate (there's always a prime between J and 2J) for 4K+3 primes. The final tidying step is to provide a sequence of 4K+3 primes, each less than twice the previous, to cover the bottom end of the range. 7, 11, 19, 31, 59, ... This suggests some related puzzles: Three squares; and whats the mod2,4,8 pattern of the odd-part of N!? Rich ________________________________________ From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Ed Pegg Jr [ed@mathpuzzle.com] Sent: Tuesday, January 20, 2009 8:14 AM To: math-fun Subject: [math-fun] Triangular+Triangular = Factorial A very belated response. Not squares, but related. Using Triangular numbers. 1 3 6 10 15 21 T[3] = 3! T[3] + T[6] = 4! T[14]+T[5] = T[15] = 5! T[45] + T[89] = 7! T[210] + T[825] = 9! T[1770] + T[2030] = 10! T[71504] + T[85680] = 13! T[213384] + T[1603064] = T[299894] + T[1589154] = 15! I don't see an easy way to extend these. The density of triangular numbers seems to be sufficient for extended solutions. --Ed Pegg Jr Date: Mon, 3 Jul 2006 11:44:20 -0600 (MDT) From: Richard Guy <rkg@cpsc.ucalgary.ca> Reply-To: math-fun <math-fun@mailman.xmission.com> To: Number Theory List <NMBRTHRY@listserv.nodak.edu>, Math Fun <math-fun@mailman.xmission.com> Subject: [math-fun] Factorial n Presumably 0! = 1! = 0^2 + 1^2. 2! = 1^2 + 1^2 6! = 12^2 + 24^2 are the only integer solutions of n! = x^2 + y^2 but is there a proof? R. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Just cleaning up old email, so this is a very, very belated response, which may have been taken care of by someone long ago. Copied to seqfans, in case someone wants to make a sequence or two out of it. Solutions of x(x+1)/2 + y(y+1)/2 = z! are solutions of (2x+1)^2 + (2y+1)^2 = 8(z!) + 2 and the first few values of z for which there are solutions can easily be ascertained: (x,y,z) = (0,1,0), (0,1,1), (1,1,2), (0,3,3), (2,2,3), (2,6,4), (0,15,5), (5,14,5), (45,89,7), (89,269,8), (210,825,9), (760,2610,10), (1770,2030,10), none for z = 11, 12, one for z = 13 (see Ed's solution below), none for z = 14, two for z = 15 (see below), one for z = 16, two for z = 17, none for z = 18, 19, 20, two for z = 21, none for z = 22, 23, two for z = 24, none for z = 25, 26, eight for z = 27, one for z = 28, two for z = 29, none for z = 30, 31, four for z = 32, 33, sixteen for z = 34, none for z = 35, 36, ..., 41, two for z = 42, none for z = 43, 44, ..., 48, sixteen for z = 49, none for z = 50, 51, 52, 53, one for z = 54, none for z = 55, 56, ..., 65, two for z = 66, none for z = 67, sixteen for z = 68, ... (E&OE, and PARI is slowing down a bit now: AND it would take rather longer to find the actual solutions!) R. On Tue, 20 Jan 2009, Ed Pegg Jr wrote:
A very belated response.
Not squares, but related. Using Triangular numbers. 1 3 6 10 15 21
T[3] = 3! T[3] + T[6] = 4! T[14]+T[5] = T[15] = 5! T[45] + T[89] = 7! T[210] + T[825] = 9! T[1770] + T[2030] = 10! T[71504] + T[85680] = 13! T[213384] + T[1603064] = T[299894] + T[1589154] = 15!
I don't see an easy way to extend these. The density of triangular numbers seems to be sufficient for extended solutions.
--Ed Pegg Jr
Date: Mon, 3 Jul 2006 11:44:20 -0600 (MDT) From: Richard Guy <rkg@cpsc.ucalgary.ca> Reply-To: math-fun <math-fun@mailman.xmission.com> To: Number Theory List <NMBRTHRY@listserv.nodak.edu>, Math Fun <math-fun@mailman.xmission.com> Subject: [math-fun] Factorial n
Presumably 0! = 1! = 0^2 + 1^2. 2! = 1^2 + 1^2 6! = 12^2 + 24^2
are the only integer solutions of
n! = x^2 + y^2
but is there a proof? [Later: for n > 6, n! will always contain a prime factor of shape 4k - 1 raised to an odd power, in fact raised to the first power, by a suitable form of the prime number theorem.] R. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I've added the factors of numbers of the form 4*z!+1 that I recently found to the factor database: http://factordb.com/index.php?query=4*z%21%2B1 In particular, 4*69!+1 = 96493309986243088621721365030167853723206604771 * P52, so there is no solution for z = 69. Feel free to use them to compute related sequences for OEIS (enough for z up to 80). I'll continue factoring as my computer time permits. Warut On Sat, Sep 11, 2010 at 2:29 AM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Just cleaning up old email, so this is a very, very belated response, which may have been taken care of by someone long ago. Copied to seqfans, in case someone wants to make a sequence or two out of it. Solutions of
x(x+1)/2 + y(y+1)/2 = z!
are solutions of (2x+1)^2 + (2y+1)^2 = 8(z!) + 2 and the first few values of z for which there are solutions can easily be ascertained:
(x,y,z) = (0,1,0), (0,1,1), (1,1,2), (0,3,3), (2,2,3), (2,6,4), (0,15,5), (5,14,5), (45,89,7), (89,269,8), (210,825,9), (760,2610,10), (1770,2030,10), none for z = 11, 12, one for z = 13 (see Ed's solution below), none for z = 14, two for z = 15 (see below), one for z = 16, two for z = 17, none for z = 18, 19, 20, two for z = 21, none for z = 22, 23, two for z = 24, none for z = 25, 26, eight for z = 27, one for z = 28, two for z = 29, none for z = 30, 31, four for z = 32, 33, sixteen for z = 34, none for z = 35, 36, ..., 41, two for z = 42, none for z = 43, 44, ..., 48, sixteen for z = 49, none for z = 50, 51, 52, 53, one for z = 54, none for z = 55, 56, ..., 65, two for z = 66, none for z = 67, sixteen for z = 68, ... (E&OE, and PARI is slowing down a bit now: AND it would take rather longer to find the actual solutions!) R.
On Tue, 20 Jan 2009, Ed Pegg Jr wrote:
A very belated response.
Not squares, but related. Using Triangular numbers. 1 3 6 10 15 21
T[3] = 3! T[3] + T[6] = 4! T[14]+T[5] = T[15] = 5! T[45] + T[89] = 7! T[210] + T[825] = 9! T[1770] + T[2030] = 10! T[71504] + T[85680] = 13! T[213384] + T[1603064] = T[299894] + T[1589154] = 15!
I don't see an easy way to extend these. The density of triangular numbers seems to be sufficient for extended solutions.
--Ed Pegg Jr
Date: Mon, 3 Jul 2006 11:44:20 -0600 (MDT) From: Richard Guy <rkg@cpsc.ucalgary.ca> Reply-To: math-fun <math-fun@mailman.xmission.com> To: Number Theory List <NMBRTHRY@listserv.nodak.edu>, Math Fun <math-fun@mailman.xmission.com> Subject: [math-fun] Factorial n
Presumably 0! = 1! = 0^2 + 1^2. 2! = 1^2 + 1^2 6! = 12^2 + 24^2
are the only integer solutions of
n! = x^2 + y^2
but is there a proof? [Later:
for n > 6, n! will always contain a prime factor of shape 4k - 1 raised to an odd power, in fact raised to the first power, by a suitable form of the prime number theorem.] R.
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participants (7)
-
Andy Latto -
Ed Pegg Jr -
Fred lunnon -
Richard Guy -
Schroeppel, Richard -
victor miller -
Warut Roonguthai