

A213910


Irregular triangle read by rows: T(n,k) is the number of involutions of length n that have exactly k inversions; n>=0, 0<=k<=binomial(n,2).


2



1, 1, 1, 1, 1, 2, 0, 1, 1, 3, 1, 2, 1, 1, 1, 1, 4, 3, 3, 4, 2, 4, 1, 3, 0, 1, 1, 5, 6, 5, 9, 5, 10, 5, 9, 4, 7, 3, 3, 2, 1, 1, 1, 6, 10, 9, 16, 13, 19, 17, 19, 19, 17, 19, 13, 17, 7, 13, 3, 8, 1, 4, 0, 1, 1, 7, 15, 16, 26, 29, 34, 43, 39, 54, 41, 61, 40, 62, 36, 58, 28, 47, 21, 34, 15, 21, 10, 11, 6, 4, 3, 1, 1
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OFFSET

0,6


COMMENTS

Row sums are A000085.
Sum_{k>=0} T(n,k)*k = A211606(n).
Diagonal is A214086.


LINKS

Alois P. Heinz, Rows n = 0..40, flattened


FORMULA

T(n,k) = T(n1,k) + Sum_{j=1..n1} T(n2,k2*(nj)+1) for n>=0, k>0; T(n,k) = 0 for n<0 or k<0; T(n,0) = 1 for n>=0.  Alois P. Heinz, Mar 07 2013


EXAMPLE

T(4,3) = 2 because we have: (3,2,1,4), (1,4,3,2).
Triangle T(n,k) begins:
1;
1;
1, 1;
1, 2, 0, 1;
1, 3, 1, 2, 1, 1, 1;
1, 4, 3, 3, 4, 2, 4, 1, 3, 0, 1;
1, 5, 6, 5, 9, 5, 10, 5, 9, 4, 7, 3, 3, 2, 1, 1;


MAPLE

T:= proc(n) option remember; local f, g, j; if n<2 then 1 else
f, g:= [T(n1)], [T(n2)]; for j to 2*n3 by 2 do
f:= zip((x, y)>x+y, f, [0$j, g[]], 0) od; f[] fi
end:
seq(T(n), n=0..10); # Alois P. Heinz, Mar 05 2013


MATHEMATICA

Table[Distribution[Map[Inversions, Involutions[n]], Range[0, Binomial[n, 2]]], {n, 0, 9}]//Flatten


CROSSREFS

Cf. A008302 (permutations of [n] with k inversions).
Sequence in context: A124035 A204184 A157897 * A288002 A140129 A029347
Adjacent sequences: A213907 A213908 A213909 * A213911 A213912 A213913


KEYWORD

nonn,tabf


AUTHOR

Geoffrey Critzer, Mar 04 2013


STATUS

approved



