### Abstract

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) |
---|---|

Pages (from-to) | 139-147 |

Number of pages | 9 |

Journal | Ars Combinatoria |

Volume | 71 |

State | Published - Apr 2004 |

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### All Science Journal Classification (ASJC) codes

- Mathematics(all)

### Cite this

*Ars Combinatoria*,

*71*, 139-147.

}

*Ars Combinatoria*, vol. 71, pp. 139-147.

**Prime power divisors of the number of n × n alternating sign matrices.** / Frey, Darrin D.; Sellers, James Allen.

Research output: Contribution to journal › Article

TY - JOUR

T1 - Prime power divisors of the number of n × n alternating sign matrices

AU - Frey, Darrin D.

AU - Sellers, James Allen

PY - 2004/4

Y1 - 2004/4

N2 - 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.

AB - 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.

UR - http://www.scopus.com/inward/record.url?scp=26844456482&partnerID=8YFLogxK

UR - http://www.scopus.com/inward/citedby.url?scp=26844456482&partnerID=8YFLogxK

M3 - Article

VL - 71

SP - 139

EP - 147

JO - Ars Combinatoria

JF - Ars Combinatoria

SN - 0381-7032

ER -