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A221914
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Array of products of the list entries of the nonempty combinations of n, ordered in a standard way.
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3
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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
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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FORMULA
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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).
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EXAMPLE
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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)|.
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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