A170942 Take the permutations of lengths 1, 2, 3, ... arranged lexicographically, and replace each permutation with the number of its fixed points.

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

Offset

1,2

Comments

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]

Links

Reinhard Zumkeller, Rows n=1..7 of triangle, flattened

FindStat - Combinatorial Statistic Finder, The number of fixed points of a permutation

Example

123,132,213,231,312,321 (corresponding to 3rd row of triangle A030298) have respectively 3,1,1,0,0,1 fixed points.

Prog

(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..]
-- Reinhard Zumkeller, Mar 29 2012

Crossrefs

Cf. A030298, A030299.

Cf. A008290, A000166, A000240, A000387, A000449, A000475, A129135, A129136, A129149, A129153, A129217, A129218, A129238, A129255.

Cf. A008291.

Sequence in context: A238347 A263322 A353496 * A325660 A342657 A002187

Adjacent sequences: A170939 A170940 A170941 * A170943 A170944 A170945

Keyword

nonn,tabf

Author

Neven Juric (neven.juric(AT)apis-it.hr) and N. J. A. Sloane, Feb 23 2010

Extensions

a(36)-a(105) from John W. Layman, Feb 23 2010

Keyword tabf added by Reinhard Zumkeller, Mar 29 2012

Status

approved