RE: [math-fun] Multiplicative Magic Squares
There's a standard proof that the center of a 3x3 magic square is K/3, where K is the row sum. (Add up the four lines through the center, subtract the whole square.) Of course this works for multiplication too, so the magic product is always the cube of the center. There's a collection of similar results for a 4x4 square, such as "the sum of the four corners is K" etc. Although the arguments offered so far are plausible, I think a bit more proof is in order before claiming minimality of the various squares. Maybe a program? There aren't that many potential products < 5040 with 16 divisors, and we can also require the exponents in the prime factorization form a descending staircase. Allowing 0 as an entry doesn't seem helpful. I suspect that allowing negatives will turn out to be equivalent to the positive case with duplications allowed (but not triplications & up). But I don't see the proof that minus signs can always be attached to a positive square in a consistent way. Rich rcs@cs.arizona.edu -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Christian Boyer Sent: Thursday, September 22, 2005 9:02 AM To: 'math-fun' Subject: RE: [math-fun] Multiplicative Magic Squares Of course replace "magic sum" by "magic product", twice in my previous message... -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : jeudi 22 septembre 2005 16:30 À : 'math-fun' Objet : RE: [math-fun] Multiplicative Magic Squares In his old book on recreative mathematics, Boris Kordiemski described exactly the Michael's method (using two latin squares) for 4x4 multiplicative squares. He also described a method for 3x3 multiplicative squares: a b² a²b a²b² ab 1 b a² ab² With a=2 and b=3, you get 2 9 12 36 6 1 3 4 18 Magic sum = 216 (*). It is the smallest possible multiplicative square. In 1667 (a long time ago...), Arnauld gave a 3x3 multiplicative square. Perhaps the oldest published multiplicative square? Using powers of 2, its magic sum was bigger : 4096. Christian. (*) 216 is also the smallest cube being also sum of three cubes. 216 = 6^3 = 3^3 + 4^3 + 5^3 _______________________________________________ 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
So far, I have found the following criteria on the minimal 4x4 magic product m: m <= 5040 m must have nonincreasing exponents in its standard prime factorization m must have >= 16 divisors d for which m/d is not of the form 1, p, p^2, p^3, p^4, p^5, p^6, p^7, pq, p^2q, p^3q, pqr for primes p,q,r. I believe these criteria reduce the candidates to: 1440 1680 1800 2160 2520 2880 3360 3456 3600 3840 4320 4608 5040 ----- Original Message ----- From: "Schroeppel, Richard" <rschroe@sandia.gov> To: "math-fun" <math-fun@mailman.xmission.com> Cc: <rcs@cs.arizona.edu> Sent: Thursday, September 22, 2005 12:07 PM Subject: RE: [math-fun] Multiplicative Magic Squares There's a standard proof that the center of a 3x3 magic square is K/3, where K is the row sum. (Add up the four lines through the center, subtract the whole square.) Of course this works for multiplication too, so the magic product is always the cube of the center. There's a collection of similar results for a 4x4 square, such as "the sum of the four corners is K" etc. Although the arguments offered so far are plausible, I think a bit more proof is in order before claiming minimality of the various squares. Maybe a program? There aren't that many potential products < 5040 with 16 divisors, and we can also require the exponents in the prime factorization form a descending staircase. Allowing 0 as an entry doesn't seem helpful. I suspect that allowing negatives will turn out to be equivalent to the positive case with duplications allowed (but not triplications & up). But I don't see the proof that minus signs can always be attached to a positive square in a consistent way. Rich rcs@cs.arizona.edu -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Christian Boyer Sent: Thursday, September 22, 2005 9:02 AM To: 'math-fun' Subject: RE: [math-fun] Multiplicative Magic Squares Of course replace "magic sum" by "magic product", twice in my previous message... -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : jeudi 22 septembre 2005 16:30 à : 'math-fun' Objet : RE: [math-fun] Multiplicative Magic Squares In his old book on recreative mathematics, Boris Kordiemski described exactly the Michael's method (using two latin squares) for 4x4 multiplicative squares. He also described a method for 3x3 multiplicative squares: a b² a²b a²b² ab 1 b a² ab² With a=2 and b=3, you get 2 9 12 36 6 1 3 4 18 Magic sum = 216 (*). It is the smallest possible multiplicative square. In 1667 (a long time ago...), Arnauld gave a 3x3 multiplicative square. Perhaps the oldest published multiplicative square? Using powers of 2, its magic sum was bigger : 4096. Christian. (*) 216 is also the smallest cube being also sum of three cubes. 216 = 6^3 = 3^3 + 4^3 + 5^3 _______________________________________________ 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
----- Original Message ----- From: "David Wilson" <davidwwilson@comcast.net> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Thursday, September 22, 2005 3:47 PM Subject: Re: [math-fun] Multiplicative Magic Squares
So far, I have found the following criteria on the minimal 4x4 magic product m:
m <= 5040 m must have nonincreasing exponents in its standard prime factorization m must have >= 16 divisors d for which m/d is not of the form 1, p, p^2, p^3, p^4, p^5, p^6, p^7, pq, p^2q, p^3q, pqr for primes p,q,r.
To explain this last criterion. Let m be the magic product of a 4x4 square. Let d be a cell of that square. Let the row containing d have cells d,u,v,w. Let the column containing d have cells d,x,y,z. Then u,v,w,x,y,z are distinct positive integers with uvw = xyz = m/d. This is insoluble if m/d has any of the prime signatures listed above.
I believe these criteria reduce the candidates to:
1440 1680 1800 2160 2520 2880 3360 3456 3600 3840 4320 4608 5040
For these candidate magic products m, the allowable cell values are: m = 1440: d = (1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 30 40) m = 1680: d = (1 2 3 4 5 6 7 8 10 12 14 15 20 21 28 35) m = 1800: d = (1 2 3 4 5 6 8 9 10 12 15 18 20 25 30 50) m = 2160: d = (1 2 3 4 5 6 8 9 10 12 15 18 20 24 27 30 36 45 60) m = 2520: d = (1 2 3 4 5 6 7 8 9 10 12 14 15 18 20 21 28 30 35 42 70) m = 2880: d = (1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 30 32 36 40 48 60 80) m = 3360: d = (1 2 3 4 5 6 7 8 10 12 14 15 16 20 21 24 28 30 35 40 42 56 70) m = 3456: d= (1 2 3 4 6 8 9 12 16 18 24 32 36 48 72 96) m = 3600: d = (1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 30 36 40 45 50 60 75 100) m = 3840: d = (1 2 3 4 5 6 8 10 12 15 16 20 24 32 40 48 64 80) m = 4320: d = (1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 27 30 36 40 45 48 54 60 72 90 120) m = 4608: d = (1 2 3 4 6 8 9 12 16 18 24 32 48 64 96 128) m = 5040: d = (1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21 24 28 30 35 36 40 42 45 56 60 63 70 84 105 140) The product of all the cells in the square is m^4. This means that for a given m, - m^4 must divide the product of all d values. - The product of the smallest 16 d values must be <= m^4. - The product of the largest 16 d values must be >= m^4. The only two m above that satisfy all these criteria are 4320 and 5040. So the smallest magic product of a 4x4 square of positive integers is either 4320 or 5040. Presumably someone can check whether 4320 is a magic product.
So the smallest magic product of a 4x4 square of positive integers is either 4320 or 5040. Presumably someone can check whether 4320 is a magic product.
Nice analysis! By hand (the programming was getting nowhere), I found 1 27 8 20 10 16 9 3 24 2 15 6 18 5 4 12 Which has a constant of 4320 on the rows and columns, but not the diagonals. With further noodling, perhaps the diagonals can be fixed. I'm fairly certain Dudeney's 60,466,176 for the 5x5 can be beaten. Ed Pegg Jr Ed Pegg Jr
So the smallest magic product of a 4x4 square of positive integers is either 4320 or 5040. Presumably someone can check whether 4320 is a magic product.
Nice analysis! By hand (the programming was getting nowhere), I found 1 27 8 20 10 16 9 3 24 2 15 6 18 5 4 12 Which has a constant of 4320 on the rows and columns, but not the diagonals. With further noodling, perhaps the diagonals can be fixed. I'm fairly certain Dudeney's 60,466,176 for the 5x5 can be beaten. Ed Pegg Jr
Nice square by hand, Ed! By program, it is possible to construct other semi-magic 4x4 squares like yours with magic product 4320. But IF my program (written perhaps too quickly...) is correct, it is impossible to get at least one diagonal = 4320, when all rows and columns = 4320. The nearest possible diagonal = 4608, for example with: 16 1 10 27 5 24 18 2 6 12 3 20 9 15 8 4 It means that the smallest possible magic product for 4x4 seems to be 5040. Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de ed pegg Envoyé : vendredi 23 septembre 2005 06:24 À : math-fun Objet : Re: [math-fun] Multiplicative Magic Squares
So the smallest magic product of a 4x4 square of positive integers is either 4320 or 5040. Presumably someone can check whether 4320 is a magic product.
Nice analysis! By hand (the programming was getting nowhere), I found 1 27 8 20 10 16 9 3 24 2 15 6 18 5 4 12 Which has a constant of 4320 on the rows and columns, but not the diagonals. With further noodling, perhaps the diagonals can be fixed. I'm fairly certain Dudeney's 60,466,176 for the 5x5 can be beaten. Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
So we have my argument that 4320 is the only 4x4 magic product candidate < 5040, and Christian Boyer's argument that 4320 is not in fact a 4x4 magic product. Since both of these arguments involve computer programs, it would be a good idea to get them independently verified before being confident of our results. ----- Original Message ----- From: "Christian Boyer" <cboyer@club-internet.fr> To: "'math-fun'" <math-fun@mailman.xmission.com> Sent: Friday, September 23, 2005 6:07 AM Subject: RE: [math-fun] Multiplicative Magic Squares Nice square by hand, Ed! By program, it is possible to construct other semi-magic 4x4 squares like yours with magic product 4320. But IF my program (written perhaps too quickly...) is correct, it is impossible to get at least one diagonal = 4320, when all rows and columns = 4320. The nearest possible diagonal = 4608, for example with: 16 1 10 27 5 24 18 2 6 12 3 20 9 15 8 4 It means that the smallest possible magic product for 4x4 seems to be 5040. Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de ed pegg Envoyé : vendredi 23 septembre 2005 06:24 à : math-fun Objet : Re: [math-fun] Multiplicative Magic Squares
So the smallest magic product of a 4x4 square of positive integers is either 4320 or 5040. Presumably someone can check whether 4320 is a magic product.
Nice analysis! By hand (the programming was getting nowhere), I found 1 27 8 20 10 16 9 3 24 2 15 6 18 5 4 12 Which has a constant of 4320 on the rows and columns, but not the diagonals. With further noodling, perhaps the diagonals can be fixed. I'm fairly certain Dudeney's 60,466,176 for the 5x5 can be beaten. Ed Pegg Jr _______________________________________________ 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 paper on the minimal 4x4 magic product would be a nice addition to the JIS. I would like to see my logic checked as well.
Can a 3x3 magic square be both additively and multiplicatively magic? ----- Original Message ----- From: "David Wilson" <davidwwilson@comcast.net> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, September 23, 2005 6:45 AM Subject: Re: [math-fun] Multiplicative Magic Squares
A paper on the minimal 4x4 magic product would be a nice addition to the JIS.
I would like to see my logic checked as well.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
David Wilson wrote:
Can a 3x3 magic square be both additively and multiplicatively magic?
Not even close, even if you don't try for diagonals. There's at most one solution to {x+y=S, xy=P} for any S and P -- as we were taught when we learned to factor polynomials -- and any z in a both-ways-magic square would force its row-mates and column-mates to be two solutions to such a system. Nice work on the possible 4x4 magic products, by the way... --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Nice argument. Can we satisfy a+b+c = abc two different ways as a first step towards a 4x4? ----- Original Message ----- From: "Michael Kleber" <michael.kleber@gmail.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, September 23, 2005 10:02 AM Subject: Re: [math-fun] Multiplicative Magic Squares David Wilson wrote:
Can a 3x3 magic square be both additively and multiplicatively magic?
Not even close, even if you don't try for diagonals. There's at most one solution to {x+y=S, xy=P} for any S and P -- as we were taught when we learned to factor polynomials -- and any z in a both-ways-magic square would force its row-mates and column-mates to be two solutions to such a system. Nice work on the possible 4x4 magic products, by the way... --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
David Wilson wrote:
A paper on the minimal 4x4 magic product would be a nice addition to the JIS.
It seems clear to me that the logic we've been using will be enough to completely describe (with only finitely much work) the sequence "All possible magic products of 4x4 multiplicative magic squares". Surely that's the desired paper. As we've made clear, the property only depends on the exponents in the prime factorization of the potential magic product. Moreover any multiple of a magic product is again one, since we can multiply any magic square pointwise by any of eight rotations or reflections of n 1 1 1 1 1 n 1 1 1 1 n 1 n 1 1 Oh, wait, it's not quite that easy -- it's not immediately clear that you can always find one of these that still has no duplicate entries. Hmm. Ignoring that for the moment, the question boils down to what the minimal exponent sets are that allow magic squares. The 5040 square shows that signature [4,2,1,1] works, and the same construction will give lots of others, so let's make it clear: Prop 1. For any 6-tuple (a1,a2,a3;b1,b2,b3), there is a multiplicative magic square whose sixteen entries are the pairwise products ai*bj, 0<=i,j<=3, where a0=b0=1. Its magic product is a1*a2*a3*b1*b2*b3. (Of course, one still needs to check that these entires are all distinct.) The 5040 example is concisely expressed by the 6-tuple (p,q,pq; pp,r,s), where each letter represents a distinct prime. We can construct a table of signatures attainable this way: 6-tuple prod. signature (minimum magic product) (p,q,r; s,t,u) [1,1,1,1,1,1] 30030 (p,pp,q; r,s,t) [3,1,1,1,1] 9240 (p,pp,q; r,rr,s) [3,3,1,1] 7560 (p,q,pq; r,s,t) [2,2,1,1,1] 13860 (p,q,pq; r,s,rs) [2,2,2,2] 44100 (p,q,pq; pp,r,s) [4,2,1,1] 5040 (p,q,pq; pp,qq,r) [4,4,1] 6480 (p,q,r; pp,qq,rr) [3,3,3] 27000 ...etc... (p,pp,ppp; p^4,p^8,p^12) [30] 1073741824 There's no intelligent order here; I'm just doodling by hand. But a computer could easily take any tentative product signature and try to factor it as a 6-tuple with 16 distinct pairwise products, and fill this table out completely This gives us lots of magic squares that do exist. Contrarywise, to find ones that don't exist, we implement David's "needs 16 factors with plausible quotients" rule, and start ruling out prime signatures that are impossible. This criterion can be extended a bit, too: eight of the 16 entires in the magic square participate in a row, column, and diagonal, so the quotient has to be writable as a product of three terms in three ways, with all nine terms distinct. That will rule out lots of prime signatures. Between those two techniques, there should only be finitely many signatures that are possible but not yet constructed, and all small enough that an exhaustive search will finish the problem off. So that's fuzzy, but seems a plausible plan, if there are folks with the time to play with it. (Sadly I don't count these days.) --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Amidst all this talk about multiplicative magic, somehow I had completely forgotten that, just a few years ago, a paper by Ahmed, De Loera, and Hemmecke calculated the Grobner basis for the polyhedral cone of 4x4 magic squares (math.CO/0201108, "Polyhedral Cones of Magic Cubes and Squares"). According to their result, every 4x4 magic square -- by which we mean the entries are nonnegative integers, and the rows, columns, and main diagonals have the same sum, but *not* that the entries are distint -- is a positive integer linear combination of the following 20 matrices: [ 1 0 0 0 ] [ 0 0 1 0 ] and its seven rotations and reflections; [ 0 0 0 1 ] [ 0 1 0 0 ] [ 1 0 1 0 ] [ 0 0 0 2 ] and its seven rotations and reflections; [ 0 1 1 0 ] [ 1 1 0 0 ] [ 0 0 1 1 ] [ 0 1 0 1 ] and its one rotation; and [ 1 0 1 0 ] [ 1 1 0 0 ] [ 1 0 1 0 ] [ 0 0 1 1 ] and its one rotation. [ 1 1 0 0 ] [ 0 1 0 0 ] The construction I gave originally, with ABCDabcd, generated all the magic squares you could make using only the first eight of these; the corresponding construction with 20 variables will generate every possible multiplicative magic square. (For each of these new matrices, you pick a constant, which is multiplied in each location where the matrix has a 1, and whose square is multiplied in the location of a 2, if any; the magic product is multiplied by the square of the constant.) This doesn't address the distinct entries requirement at all. But ignoring that requirement, you can now conclude, for example, that the number of multiplicative magic squares with magic product p1^e1 * p2^e2 *...* pn^en is just product( binomial( ei+19, ei ) ) over i=1...n. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
I can make up the all-2s matrix in at least two different ways. Add up [ 1 0 0 0 ] [ 0 0 1 0 ] and its seven rotations and reflections; [ 0 0 0 1 ] [ 0 1 0 0 ] or, add up two copies of [ 0 0 1 1 ] [ 0 1 0 1 ] and its one rotation. [ 1 0 1 0 ] [ 1 1 0 0 ] I think this makes the binomial product below a little generous. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Michael Kleber Sent: Sun 9/25/2005 12:53 PM To: math-fun Subject: Re: [math-fun] Multiplicative Magic Squares Amidst all this talk about multiplicative magic, somehow I had completely forgotten that, just a few years ago, a paper by Ahmed, De Loera, and Hemmecke calculated the Grobner basis for the polyhedral cone of 4x4 magic squares (math.CO/0201108, "Polyhedral Cones of Magic Cubes and Squares"). According to their result, every 4x4 magic square -- by which we mean the entries are nonnegative integers, and the rows, columns, and main diagonals have the same sum, but *not* that the entries are distint -- is a positive integer linear combination of the following 20 matrices: [ 1 0 0 0 ] [ 0 0 1 0 ] and its seven rotations and reflections; [ 0 0 0 1 ] [ 0 1 0 0 ] [ 1 0 1 0 ] [ 0 0 0 2 ] and its seven rotations and reflections; [ 0 1 1 0 ] [ 1 1 0 0 ] [ 0 0 1 1 ] [ 0 1 0 1 ] and its one rotation; and [ 1 0 1 0 ] [ 1 1 0 0 ] [ 1 0 1 0 ] [ 0 0 1 1 ] and its one rotation. [ 1 1 0 0 ] [ 0 1 0 0 ] The construction I gave originally, with ABCDabcd, generated all the magic squares you could make using only the first eight of these; the corresponding construction with 20 variables will generate every possible multiplicative magic square. (For each of these new matrices, you pick a constant, which is multiplied in each location where the matrix has a 1, and whose square is multiplied in the location of a 2, if any; the magic product is multiplied by the square of the constant.) This doesn't address the distinct entries requirement at all. But ignoring that requirement, you can now conclude, for example, that the number of multiplicative magic squares with magic product p1^e1 * p2^e2 *...* pn^en is just product( binomial( ei+19, ei ) ) over i=1...n. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Rich wrote:
I can make up the all-2s matrix in at least two different ways.
Hmm, true. Good eyes. In fact, you can make the all-1s matrix in at least three ways. I'll recheck the paper, to see if it's my typo or misunderstanding, and if not, talk to DeLoera... --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Rich wrote:
I can make up the all-2s matrix in at least two different ways.
Yes, I'm just an idiot; there's no claim of uniqueness at all. Every 4x4 magic square is a sum of those 20, and that's the minimal set of vectors with that property (and *that set* is unique). But sums can arise in multiple ways. Sorry about the confusion.
I think this makes the binomial product below a little generous.
Of course. In fact the number of 4x4 magic squares with given magic sum was calculated by Beck, Cohen, Cuomo, and Gribelyuk, in an article in the Monthly (110, no. 8 (2003); the preprint is math.CO/0201013, and came out just a week and a half before the Ahmed, De Loera, and Hemmecke paper I mentioned before). It's in the EIS, of course: A093199. It's the value of one of two degree-7 polynomials, depending on whether the magic sum is even or odd. And the ADH paper gives a generating fn. The leading term of those polynomials is 1/480 s^7, while if you just use the eight permutation matrices, you would only get 1/7! s^7. So if you only used the abcdABCD construction, then for each prime p, you'd have roughly a 2/21 chance of being able to make the desired exponents, and your overall odds are (2/21)^(# primes which appear). So abcdABCD makes a pretty small fraction of all multiplicative magic squares, as the magic product (and the exponents on each prime in its factorization) get large. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Could a subset of this bases be used to generate a cofinite subset of all relevant magic squares? ----- Original Message ----- From: "Michael Kleber" <michael.kleber@gmail.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Monday, September 26, 2005 9:28 AM Subject: Re: [math-fun] Multiplicative Magic Squares Rich wrote:
I can make up the all-2s matrix in at least two different ways.
Yes, I'm just an idiot; there's no claim of uniqueness at all. Every 4x4 magic square is a sum of those 20, and that's the minimal set of vectors with that property (and *that set* is unique). But sums can arise in multiple ways. Sorry about the confusion.
I think this makes the binomial product below a little generous.
Of course. In fact the number of 4x4 magic squares with given magic sum was calculated by Beck, Cohen, Cuomo, and Gribelyuk, in an article in the Monthly (110, no. 8 (2003); the preprint is math.CO/0201013, and came out just a week and a half before the Ahmed, De Loera, and Hemmecke paper I mentioned before). It's in the EIS, of course: A093199. It's the value of one of two degree-7 polynomials, depending on whether the magic sum is even or odd. And the ADH paper gives a generating fn. The leading term of those polynomials is 1/480 s^7, while if you just use the eight permutation matrices, you would only get 1/7! s^7. So if you only used the abcdABCD construction, then for each prime p, you'd have roughly a 2/21 chance of being able to make the desired exponents, and your overall odds are (2/21)^(# primes which appear). So abcdABCD makes a pretty small fraction of all multiplicative magic squares, as the magic product (and the exponents on each prime in its factorization) get large. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
From: "Michael Kleber" <michael.kleber@gmail.com>
So abcdABCD makes a pretty small fraction of all multiplicative magic squares, as the magic product (and the exponents on each prime in its factorization) get large.
Here is an interesting family of 4x4 multiplicative squares, using only three variables a, b, c: 1 abbb bc aaa ac aab ab bb aaab c abb b bbb a aa abc Magic product = a^4 * b^4 * c, for the 4 rows, 4 columns, 2 diagonals and 2 broken diagonals. And also for other sets of four cells (the corners for example, the 4 numbers of each quarter, ...). Apply a=2 and b=3, and you will get one of the various possible examples with magic product 6480. 6480 is the second smallest product after 5040, as listed in my previous email. Christian.
participants (5)
-
Christian Boyer -
David Wilson -
ed pegg -
Michael Kleber -
Schroeppel, Richard