login
Triangle read by rows: T(n,k) = number of partitions of n with k circular successions (n>=0, 0 <= k <= n).
4

%I #17 Jan 22 2015 13:38:17

%S 0,1,0,1,0,1,1,3,0,1,4,4,6,0,1,11,20,10,10,0,1,41,66,60,20,15,0,1,162,

%T 287,231,140,35,21,0,1,715,1296,1148,616,280,56,28,0,1,3425,6435,5832,

%U 3444,1386,504,84,36,0,1,17722,34250,32175,19440,8610,2772,840,120,45,0,1

%N Triangle read by rows: T(n,k) = number of partitions of n with k circular successions (n>=0, 0 <= k <= n).

%H T. Mansour, A. O. Munagi, <a href="http://dx.doi.org/10.1016/j.ejc.2014.06.008">Set partitions with circular successions</a>, European Journal of Combinatorics, 42 (2014), 207-216.

%e Triangle begins:

%e 0

%e 1, 0,

%e 1, 0, 1,

%e 1, 3, 0, 1,

%e 4, 4, 6, 0, 1,

%e 11, 20, 10, 10, 0, 1,

%e 41, 66, 60, 20, 15, 0, 1,

%e 162, 287, 231, 140, 35, 21, 0, 1,

%e 715, 1296, 1148, 616, 280, 56, 28, 0, 1,

%e 3425, 6435, 5832, 3444, 1386, 504, 84, 36, 0, 1,

%e 17722, 34250, 32175, 19440, 8610, 2772, 840, 120, 45, 0, 1

%e ...

%t t[n_, k_] := Binomial[n, k]*((-1)^(n-k)+Sum[(-1)^(j-1)*BellB[n-k-j], {j, 1, n-k}]); t[0, 0]=0; t[1, 0]=1; t[1, 1]=0; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 09 2014 *)

%Y A124323 is an essentially identical triangle, differing only in row 0 and 1.

%Y For columns see A000296, A250105 - A250107.

%K nonn,tabl

%O 0,8

%A _N. J. A. Sloane_, Nov 16 2014