

A221876


T(n,k) is the number of orderpreserving full contraction mappings (of an nchain) with exactly k fixed points.


9



1, 2, 1, 5, 2, 1, 12, 5, 2, 1, 28, 12, 5, 2, 1, 64, 28, 12, 5, 2, 1, 144, 64, 28, 12, 5, 2, 1, 320, 144, 64, 28, 12, 5, 2, 1, 704, 320, 144, 64, 28, 12, 5, 2, 1, 1536, 704, 320, 144, 64, 28, 12, 5, 2, 1, 3328, 1536, 704, 320, 144, 64, 28, 12, 5, 2, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

The matrix inverse starts
1;
2,1;
1,2,1;
0,1,2,1;
1,0,1,2,1;
2,1,0,1,2,1;
3,2,1,0,1,2,1;
4,3,2,1,0,1,2,1;
5,4,3,2,1,0,1,2,1;
6,5,4,3,2,1,0,1,2,1;
7,6,5,4,3,2,1,0,1,2,1;  R. J. Mathar, Apr 12 2013
...
T(n,k) is also the total number of occurrences of parts k in all compositions (ordered partitions) of n, see example. The equivalent sequence for partitions is A066633. Omar E. Pol, Aug 26 2013


REFERENCES

A. D. Adeshola, V. Maltcev and A. Umar, Combinatorial results for certain semigroups of orderpreserving full contraction mappings of a finite chain, (submitted 2013).


LINKS



FORMULA

T(n,n) = 1, T(n,k) = (nk+3)*2^(nk2) for n>=2 and n > k > 0.
T(2*n+1,n+1) = T(n+1,1) = A045623(n) for n>=0.


EXAMPLE

T(5,3) = 5 because there are exactly 5 orderpreserving full contraction mappings (of a 5chain) with exactly 3 fixed points, namely: (12333), (12334), (22344), (23345), (33345).
Triangle begins:
1,
2, 1,
5, 2, 1,
12, 5, 2, 1,
28, 12, 5, 2, 1,
64, 28, 12, 5, 2, 1,
144, 64, 28, 12, 5, 2, 1,
320, 144, 64, 28, 12, 5, 2, 1,
704, 320, 144, 64, 28, 12, 5, 2, 1,
1536, 704, 320, 144, 64, 28, 12, 5, 2, 1,
3328, 1536, 704, 320, 144, 64, 28, 12, 5, 2, 1,
...
Note that column k is column 1 shifted down by k positions.
Row 4 is [12, 5, 2, 1]: in the compositions of 4
[ 1] [ 1 1 1 1 ]
[ 2] [ 1 1 2 ]
[ 3] [ 1 2 1 ]
[ 4] [ 1 3 ]
[ 5] [ 2 1 1 ]
[ 6] [ 2 2 ]
[ 7] [ 3 1 ]
[ 8] [ 4 ]
there are 12 parts=1, 5 parts=2, 2 part=3, and 1 part=4.


MATHEMATICA

T[n_, n_] = 1; T[n_, k_] := (n  k + 3)*2^(n  k  2);


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



