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A277855
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Irregular triangle read by rows: T(n,k) is the maximum length of the longest common subsequence of k distinct permutations of n items with n>=1 and 1<=k<=n!
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1
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1, 2, 1, 3, 2, 2, 1, 1, 1, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1
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OFFSET
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1,2
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COMMENTS
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The formulas given below are correct. The sequence can be used to normalize the length of the longest common subsequence of a set of k full preference orderings relative to the maximum attainable length. This normalized number is a measure of concordance in the set of preference orderings.
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LINKS
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FORMULA
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T(n,1)=n.
For n>1, 1<=k<=n! and 1<=j<=n, T(n,k)=n-j if binomial(n,n-j+1)*(j-1)!+1<=k<=binomial(n,n-j)*j!.
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EXAMPLE
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The permutations {abc, acb} have 2 longest common subsequences of length 2: ab and ac. The permutations {abc, acb, cab} have one longest common subsequence: ab of length 2. The formula above yields T(3,3)= 2.
The triangle begins:
1
2,1
3,2,2,1,1,1
4,3,3,3,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1
5,4,4,4,4,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,...
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MATHEMATICA
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Flatten[Table[(n - Select[Range@ n, Function[j, Binomial[n, n - j + 1] (j - 1)! + 1 <= k <= Binomial[n, n - j] j!]]) /. {} -> {n}, {n, 5}, {k, n!}], {3}] // Flatten (* Michael De Vlieger, Nov 04 2016 *)
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CROSSREFS
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A277517: the maximum number of common subsequences of k distinct permutations of n items.
A152072: the maximum number of length-k longest common subsequences of a pair of length-n strings.
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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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