A351580 a(n) is the number of multisets of size n-1 consisting of permutations of n elements.

1, 2, 21, 2600, 9078630, 1634935320144, 22831938997720867560, 34390564970975286088924022400, 7457911916650283082000186530740981347120, 300682790088737748950725540713718365319268411170195200, 2830053444386286847574443631356044745870287426798365860653876609636480

Offset: 1

Author

Dan Eilers, Feb 13 2022

Keywords

• nonn

Comments

a(n) is the number of reduced men's ranking tables in the stable marriage problem of order n. In the SMP (as noted in A351409), relabeling men or women has no effect on the number of stable matchings. So the women can be relabeled to normalize the order of man #1's rankings (with woman #1 as his first choice and woman n as his last choice), and then the men except man #1 can be relabeled to normalize the lexicographic order of those men's rankings. Since man #1's rankings end up fixed in natural order, they do not contribute to the number of possibilities, leaving n! multichoose (n-1) ways to arrange the rankings of the other n-1 men.

The number of unreduced men's ranking tables is given by A036740. Relabeling just the women reduces this to A134366. Alternately, relabeling just the men reduces A036740 to A344690. Relabeling both men and women reduces the men's relabeling reduction, A344690, by a factor of (n!+n-1)/n to a(n).

It might be tempting to try to reduce A344690 by a factor of n!, but that doesn't work because not all of man #1's rankings are equally likely after relabeling all the men to give man #1 the lexicographically least rankings.

There is room for further relabeling reduction from a(n), given by A263921. The reduction from a(n) to A263921 is analogous to the reduction from reduced latin squares, A000315, to A123234.

Each of the a(n) reduced men's ranking tables can be combined with the A036740 possible unreduced women's ranking tables to form complete instances, but these instances have more possibilities than A351409. For example, a(3)*A036740(3)=21*216=4536 > A351409(3)=3888. However, fewer possibilities result from using A263921 in place of a(n), although the men's ranking tables of A263921 may not be as straightforward to generate. With A263921(3)=10, 10*216=2160 < 3888.

Examples

			Starting with the following men's ranking table of order 3, where row k represents man k's rankings, the 1 in the 2nd position of row 3 means that man #3 ranks woman #2 as his 1st choice.
  213
  321
  213
Step 1: reorder columns so row 1 is in natural order:
  123
  231
  123
Step 2: reorder rows 2 to n so rows are in lexical order:
  123
  123
  231
a(3)=21 because there are 1+2+3+4+5+6 = 21 possibilities for the last two rows in lexical order, with 3!=6 possible permutations for each row.
The 21 tables for a(3) are the following:
  123   123   123   123   123   123   123
  123   123   123   123   123   123   132
  123   132   213   231   312   321   132
.
  123   123   123   123   123   123   123
  132   132   132   132   213   213   213
  213   231   312   321   213   231   312
.
  123   123   123   123   123   123   123
  213   231   231   231   312   312   321
  321   231   312   321   312   321   321
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..31
• Wikipedia, Counting multisets

Crossrefs

Cf. A036740, A071919, A134366, A344690, A263921, A351409, A000315, A185141, A030495.

Programs

Mathematica

Table[Binomial[n!+n-2,n-1],{n,15}] (* Harvey P. Dale, Jun 02 2023 *)

PARI

a(n) = binomial(n! + n - 2, n - 1) \\ Andrew Howroyd, Feb 13 2022

Formula

a(n) = binomial(n! + n - 2, n - 1).

a(n) = n*A344690(n)/A030495(n-1).

a(n) = A344690*n/(n! + n - 1).

a(n) = A071919(n-1,n!). - Alois P. Heinz, Feb 16 2022

Extensions

Erroneous Mathematica program deleted by N. J. A. Sloane, Jun 02 2023