%I #9 Feb 11 2013 12:33:47
%S 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,
%T 120,120,60,40,30,24,60,40,30,24,20,15,12,10,8,6,20,15,12,10,8,6,5,4,
%U 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
%N Array of products of the list entries of the combinations of n, taken in reverse standard order.
%C The sequence of the row lengths is 2^n = A000079(n), n >= 0.
%C 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]].
%C The array entry a(n,k=0) := n!, n >= 0, and for k = 1, 2, ... , 2^n1 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 lth 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.
%C 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.
%C 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_{nm}(1,2,...,n) with the (nm)th elementary symmetric function of 1,2, ..., n.
%C For row No. n the sum of entries is (n+1)! = A000142(n+1), like for the triangle A130534.
%F a(n,k) := n! for k=0, and for k =1,2, ..., 2^n1 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 lth entry of the list comb(n,k). See a comment above how comb(n,k) = m(n,k) is determined.
%e The array a(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
%e 0: 1
%e 1: 1 1
%e 2: 2 2 1 1
%e 3: 6 6 3 2 3 2 1 1
%e 4: 24 24 12 8 6 12 8 6 4 3 2 4 3 2 1 1
%e ...
%e 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.
%e 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.
%e 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).
%Y Cf. A221914, A130534, A008275.
%K nonn,tabf
%O 0,4
%A _Wolfdieter Lang_, Feb 08 2013
