A221914 Array of products of the list entries of the nonempty combinations of n, ordered in a standard way.

1, 1, 2, 2, 1, 2, 3, 2, 3, 6, 6, 1, 2, 3, 4, 2, 3, 4, 6, 8, 12, 6, 8, 12, 24, 24, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 24, 30, 40, 60, 120, 120, 1, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 6, 8, 10, 12, 12, 15, 18, 20, 24, 30, 6, 8, 10, 12, 12, 15, 18, 20, 24, 30, 24, 30, 36, 40, 48, 60, 60, 72, 90, 120, 24, 30, 36, 40, 48

Offset: 1

Author

Wolfdieter Lang, Feb 06 2013

Keywords

• nonn
• tabf

Comments

The sequence of the row lengths is 2^n-1 = A000225(n), n >= 1.

The combination choose(n,0) is the empty list [], appearing first in the list of combinations of n each taken as a list, called choose(n). The empty combination is not considered in this array. The list of lists choose(n) is ordered according to increasing m from choose(n,m), m = 0, ..., n, and for each m from 1, ..., n the lists are ordered lexicographically. E.g., choose(3) = [[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]]. The corresponding list of the products of these list entries is then (without the empty list) the row n=3 of the present array [1, 2, 3, 2, 3, 6, 6].

The row sums are (n+1)! - 1 = A033312(n+1).

One could add a column k=0 (for the empty combination) and a row n=0 by defining a(n,0) = 1, n >= 0. This enlarged array produces then after summation of the entries deriving from the choose(n,m) list the triangle T(n,m), n >= m >= 0, which coincides with the unsigned Stirling1 triangle if the rows are read backwards: T(n,m) = |S1(n+1,n+1-m)|. See A130534 for |Strirling1(n+1,m+1)|. Proof by showing the recurrence.

Examples

			The array a(n,k) begins:
n\k  1  2  3  4  5  6  7  8  9  10 11 12  13  14  15 ...
1:   1
2:   1  2  2
3:   1  2  3  2  3  6  6
4:   1  2  3  4  2  3  4  6  8  12  6  8  12  24  24
...
Row n=5:  1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 24, 30, 40, 60, 120;
Row n=6: 1, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 6, 8, 10, 12, 12, 15, 18, 20, 24, 30, 6, 8, 10, 12, 12, 15, 18, 20, 24, 30, 24, 30, 36, 40, 48, 60, 60, 72, 90, 120, 24, 30, 36, 40, 48, 60, 60, 72, 90, 120, 120, 144, 180, 240, 360, 120, 144, 180, 240, 360, 720, 720;
Row n=3: from the combinations list choose(3) (without the first entry []) [[1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]] leading to [1, 2, 3, 2, 3, 6, 6].
a(3,4) = 2 is the product of the entries of the combination list choose(3,1,1) = [1, 2], the first combination from choose(3,1).
|Stirling1| connection from like m summation: T(3,0) := 1 = |Stirling1(4,4)|, T(3,1) = 1 + 2 + 3 = 6 = |Stirling1(4,3)|,
  T(3,2) = 2 + 3 + 6 = 11 = |Stirling1(4,2)|, T(3,3) = 6 =
  |Stirling1(4,1)|.
		

Formula

a(n,k) = product(choose(n,m,j)[l],l=1..m) if the k-th entry of the choose(n) list (without the leading empty set and ordered as explained in a comment above), is choose(n,m,j), the j-th list member of the list choose(n,m), for n >= 1, 1 <= m <= n, 1 <= j <= binomial(n,m).