%I
%S 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,
%T 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,
%U 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
%N Take the permutations of lengths 1, 2, 3, ... arranged lexicographically, and replace each permutation by the number of its fixed points.
%C Length of nth row = sum of nth row = n!; number of zeros in nth row = A000166(n); number of positive terms in nth row = A002467(n). [_Reinhard Zumkeller_, Mar 29 2012]
%H Reinhard Zumkeller, <a href="/A170942/b170942.txt">Rows n=1..7 of triangle, flattened</a>
%H FindStat  Combinatorial Statistic Finder, <a href="http://www.findstat.org/St000022">The number of fixed points of a permutation</a>
%e 123,132,213,231,312,321 (corresponding to 3rd row of triangle A030298) have respectively 3,1,1,0,0,1 fixed points.
%o (Haskell)
%o import Data.List (permutations, sort)
%o a170942 n k = a170942_tabf !! (n1) (k1)
%o a170942_row n = map fps $ sort $ permutations [1..n] where
%o fps perm = sum $ map fromEnum $ zipWith (==) perm [1..n]
%o a170942_tabf = map a170942_row [1..]
%o  _Reinhard Zumkeller_, Mar 29 2012
%Y Cf. A030298, A030299.
%Y Cf. A008290, A000166, A000240, A000387, A000449, A000475, A129135, A129136, A129149, A129153, A129217, A129218, A129238, A129255.
%Y Cf. A008291.
%K nonn,tabf
%O 1,2
%A Neven Juric (neven.juric(AT)apisit.hr) and _N. J. A. Sloane_, Feb 23 2010
%E a(36)a(105) from _John W. Layman_, Feb 23 2010
%E Keyword tabf added by _Reinhard Zumkeller_, Mar 29 2012
