A170942 - OEIS

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

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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..]