We let A(n) equal the number of n × n alternating sign matrices. From the work of a variety of sources, we know that A(n) =∏ l=0 n-1(3l+1)!/(n+l)! We find an efficient method of determining ord p(A(n)), the highest power of p which divides A(n), for a given prime p and positive integer n, which allows us to efficiently compute the prime factorization of A(n). We then use our method to show that for any nonnegative integer k, and for any prime p > 3, there are infinitely many positive integers n such that ord p(A(n)) =k. We show a similar but weaker theorem for the prime p = 3, and note that the opposite is true for p = 2.
|Original language||English (US)|
|Number of pages||9|
|State||Published - Apr 2004|
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