A221916 Array of products of the list entries of the combinations of n, taken in reverse standard order.

1, 1, 1, 2, 2, 1, 1, 6, 6, 3, 2, 3, 2, 1, 1, 24, 24, 12, 8, 6, 12, 8, 6, 4, 3, 2, 4, 3, 2, 1, 1, 120, 120, 60, 40, 30, 24, 60, 40, 30, 24, 20, 15, 12, 10, 8, 6, 20, 15, 12, 10, 8, 6, 5, 4, 3, 2, 5, 4, 3, 2, 1, 1, 720, 720, 360, 240, 180, 144, 120, 360, 240, 180, 144, 120, 120, 90, 72, 60, 60, 48, 40, 36, 30, 24, 120, 90, 72, 60, 60, 48, 40, 36, 30, 24, 30, 24

Offset: 0

Author

Wolfdieter Lang, Feb 08 2013

Keywords

• nonn
• tabf

Comments

The sequence of the row lengths is 2^n = A000079(n), n >= 0.

As a preliminary some notation. The list choose(n) consists of the 2^n combinations of n, written as lists. The first entry of choose(n) is the empty combination []. The other entries are ordered lexicographically within entries of the same length, called m, and m increases from 1 to n (m=0 is reserved for the empty combination). E.g., choose(3) = [[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]].

The array entry a(n,k=0) := n!, n >= 0, and for k = 1, 2, ... , 2^n-1 one takes a(n,k) = n!/product(comb(n,k,l), l = 1..m(n,k)), with m(n,k) the length of the (k+1)-th entry comb(n,k) of choose(n), and comb(n,k,l) is the l-th entry of comb(n,k). E.g., comb(3,5) = [1, 3], comb(3,5,1) = 1 and comb(3,5,2) = 3, hence a(3,5) = 3!/(1*3) = 2.

This array a(n,k) is the row reversed array A(n,k) = A221914(n,k) if one adds there A(n,0) = 1 for n >= 0.

If all entries of the present array which belong to the same m= m(n,k) value are summed one obtains the unsigned Stirling1(n+1,m+1) triangle A130534(n,m) because this is sigma_{n-m}(1,2,...,n) with the (n-m)-th elementary symmetric function of 1,2, ..., n.

For row No. n the sum of entries is (n+1)! = A000142(n+1), like for the triangle A130534.

Examples

			The array a(n,k) begins:
n\k  0   1   2  3  4   5  6  7  8  9  10  11  12  13  14  15
0:   1
1:   1   1
2:   2   2   1  1
3:   6   6   3  2  3   2  1  1
4:  24  24  12  8  6  12  8  6  4  3   2   4   3   2   1   1
...
Row n=5: 120, 120, 60, 40, 30, 24, 60, 40, 30, 24, 20, 15, 12, 10, 8, 6, 20, 15, 12, 10, 8, 6, 5, 4, 3, 2, 5, 4, 3, 2, 1, 1.
Row n=6: 720, 720, 360, 240, 180, 144, 120, 360, 240, 180, 144, 120, 120, 90, 72, 60, 60, 48, 40, 36, 30, 24, 120, 90, 72, 60, 60, 48, 40, 36, 30, 24, 30, 24, 20, 18, 15, 12, 12, 10, 8, 6, 30, 24, 20, 18, 15, 12, 12, 10, 8, 6, 6, 5, 4, 3, 2, 6, 5, 4, 3, 2, 1, 1.
The combinations for row n are choose(4) = [[], [1], [2], [3], [4], [1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4], [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4], [1, 2, 3, 4]]. For k=0 one takes 4! = 24. For k >= 1 one obtains 4!/1, 4!/2, 4!/3, 4!/4; 4!/(1*2), 4!/(1*3), 4!/(1*4), 4!/(2*3), 4!/(2*4), 4!/(3*4); 4!/(1*2*3),  4!/(1*2*4), 4!/(1*3*4), 4!/(2*3*4);  4!/(1*2*3*4) giving row n=4. The semicolons separate the binomial(4,m) entries with m values from 1 to 4. The example in the comment above was k=13 leading to 4!/(1*3*4) = 2 = a(4,13).
		

Crossrefs

Cf. A221914, A130534, |A008275|.

Formula

a(n,k) := n! for k=0, and for k =1,2, ..., 2^n-1 it is n!/product(comb(n,k,l),l=1..|comb(n,k)|) with |comb(n,k)| the number of entries of comb(n,k) which is the (k+1)-th entry of the list of combinations choose(n) (starting with the empty combination for k=0), and comb(n,k,l) is the l-th entry of the list comb(n,k). See a comment above how |comb(n,k)| = m(n,k) is determined.