A221878 Number of order-preserving or order-reversing full contraction mappings (of an n-chain) with exactly k fixed points.

1, 0, 1, 1, 2, 1, 2, 8, 2, 1, 6, 22, 5, 2, 1, 14, 57, 12, 5, 2, 1, 34, 136, 28, 12, 5, 2, 1, 78, 315, 64, 28, 12, 5, 2, 1, 178, 710, 144, 64, 28, 12, 5, 2, 1, 398, 1577, 320, 144, 64, 28, 12, 5, 2, 1, 882, 3460, 704, 320, 144, 64, 28, 12, 5, 2, 1

Offset: 1

Author

Abdullahi Umar, Feb 28 2013

Keywords

• nonn
• tabl

Comments

Its row sum is A221882.

Examples

			T (4,0) = 6 because there are exactly 6 order-preserving or order-reversing full contraction mappings (of a 4-chain) with no fixed point, namely: (2111), (3321), (3322), (4321), (4322), (4443).
Triangle:
1,
0, 1,
1, 2, 1,
2, 8, 2, 1,
6, 22, 5, 2, 1,
14, 57, 12, 5, 2, 1,
34, 136, 28, 12, 5, 2, 1,
78, 315, 64, 28, 12, 5, 2, 1,
178, 710, 144, 64, 28, 12, 5, 2, 1,
398, 1577, 320, 144, 64, 28, 12, 5, 2, 1,
882, 3460, 704, 320, 144, 64, 28, 12, 5, 2, 1
		

Links

• A. D. Adeshola, V. Maltcev and A. Umar, Combinatorial results for certain semigroups of order-preserving full contraction mappings of a finite chain, arXiv:1303.7428 [math.CO], 2013.
• A. D. Adeshola, A. Umar, Combinatorial results for certain semigroups of order-preserving full contraction mappings of a finite chain, JMCC 106 (2017) 37-49

Crossrefs

Cf. A221876, A221877, A221879, A221880, A221881, A221882.

Formula

T(n,0) = T(n-1,1), T(n,1) = A059570(n) + A221876(n,1) - n and T(n,k) = A221876 if k > 1.