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