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A170942
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Take the permutations of lengths 1, 2, 3, ... arranged lexicographically, and replace each permutation with the number of its fixed points.
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19
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1, 2, 0, 3, 1, 1, 0, 0, 1, 4, 2, 2, 1, 1, 2, 2, 0, 1, 0, 0, 1, 1, 0, 2, 1, 0, 0, 0, 1, 1, 2, 0, 0, 5, 3, 3, 2, 2, 3, 3, 1, 2, 1, 1, 2, 2, 1, 3, 2, 1, 1, 1, 2, 2, 3, 1, 1, 3, 1, 1, 0, 0, 1, 2, 0, 1, 0, 0, 1, 1, 0, 2, 1, 0, 0, 0, 1, 1, 2, 0, 0, 2, 0, 1, 0, 0, 1, 3, 1, 2, 1, 1, 2, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0
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OFFSET
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1,2
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COMMENTS
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Length of n-th row = sum of n-th row = n!; number of zeros in n-th row = A000166(n); number of positive terms in n-th row = A002467(n). [Reinhard Zumkeller, Mar 29 2012]
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LINKS
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EXAMPLE
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123,132,213,231,312,321 (corresponding to 3rd row of triangle A030298) have respectively 3,1,1,0,0,1 fixed points.
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PROG
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(Haskell)
import Data.List (permutations, sort)
a170942 n k = a170942_tabf !! (n-1) (k-1)
a170942_row n = map fps $ sort $ permutations [1..n] where
fps perm = sum $ map fromEnum $ zipWith (==) perm [1..n]
a170942_tabf = map a170942_row [1..]
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CROSSREFS
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Cf. A008290, A000166, A000240, A000387, A000449, A000475, A129135, A129136, A129149, A129153, A129217, A129218, A129238, A129255.
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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