A221879 Triangle T(n,k) read by rows: Number of order-reversing full contraction mappings (of an n-chain) with 1 fixed point and height exactly k.
1, 2, 0, 3, 2, 1, 4, 6, 4, 0, 5, 12, 12, 4, 1, 6, 20, 28, 18, 6, 0, 7, 30, 55, 52, 27, 6, 1, 8, 42, 96, 120, 88, 36, 8, 0, 9, 56, 154, 240, 230, 136, 48, 8, 1, 10, 72, 232, 434, 516, 400, 200, 60, 10, 0, 11, 90, 333, 728, 1036, 996, 650, 280, 75, 10, 1
Offset: 1
Examples
T (4,6) = 6 because there are exactly 6 order-reversing full contraction mappings (of a 4-chain) with 1 fixed point and of height exactly 2, namely: (3222), (2221), (2211), (4433), (4333), (3332). Triangle starts: 1, 2, 0, 3, 2, 1, 4, 6, 4, 0, 5, 12, 12, 4, 1, 6, 20, 28, 18, 6, 0, 7, 30, 55, 52, 27, 6, 1, 8, 42, 96, 120, 88, 36, 8, 0, 9, 56, 154, 240, 230, 136, 48, 8, 1, 10, 72, 232, 434, 516, 400, 200, 60, 10, 0, 11, 90, 333, 728, 1036, 996, 650, 280, 75, 10, 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
Programs
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Maple
A221879 := proc(n,k) option remember ; if n<1 then 0 ; elif n=1 then if k = 1 then 1; else 0 ; end if; else if n = 2 and k=2 then 0; else (n-k+1)*binomial(n-2,k-1)+procname(n-2,k-2) ; end if; end if; end proc: seq(seq( A221879(n,k),k=1..n),n=1..20) ; # R. J. Mathar, Aug 15 2025
Formula
T(n, 1) = 1, T(2,2) = 0 and T(n,k) = (n-k+1)*C(n-2,k-1) + T(n-2,k-2) for k > 0.
Sum_{k=1..n} T(n,k) = A059570(n).
Comments