A221914 - OEIS

%I #6 Feb 11 2013 12:33:18

%S 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,

%T 3,4,5,6,8,10,12,15,20,6,8,10,12,15,20,24,30,40,60,24,30,40,60,120,

%U 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

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

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

%C 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].

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

%C 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.

%F 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).

%e The array a(n,k) begins:

%e n\k  1  2  3  4  5  6  7  8  9  10 11 12  13  14  15 ...

%e 1:   1

%e 2:   1  2  2

%e 3:   1  2  3  2  3  6  6

%e 4:   1  2  3  4  2  3  4  6  8  12  6  8  12  24  24

%e ...

%e 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;

%e 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;

%e 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].

%e 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).

%e |Stirling1| connection from like m summation: T(3,0) := 1 = |Stirling1(4,4)|, T(3,1) = 1 + 2 + 3 = 6 = |Stirling1(4,3)|,

%e   T(3,2) = 2 + 3 + 6 = 11 = |Stirling1(4,2)|, T(3,3) = 6 =

%e   |Stirling1(4,1)|.

%K nonn,tabf

%O 1,3

%A _Wolfdieter Lang_, Feb 06 2013