OFFSET
1,6
COMMENTS
Another way to obtain this sequence (with offset 0): Form the infinite array U(n,k) = number of labeled partitions of (n,k) into pairs (i,j), for n >= 0, k >= 0 and read it by antidiagonals. In other words, U(n,k) = number of partitions of n black objects labeled 1..n and k white objects labeled 1..k. Each block must have at least one white object.
Then T(n,k)=U(n+k,k+1). Thus the two versions are related like "multichoose" to "choose". - Augustine O. Munagi, Jul 16 2007
LINKS
Alois P. Heinz, Rows n = 1..141, flattened
FORMULA
T(n,1)=0 for n>=2; T(n,2)=1 for n>=2; T(n,3)=1+2^(n-3) for n>=3; T(n,n)=B(n-1), T(n,n-1)=B(n-1)-B(n-2), where B(q) are the Bell numbers (A000110).
Double e.g.f.: exp(exp(x)*(exp(y)-1)).
U(n,k) = Sum_{i=0..k} i^(n-k)*Stirling2(k,i). - Vladeta Jovovic, Jul 12 2007
EXAMPLE
Triangle T(n,k) starts:
1;
0,1;
0,1,2;
0,1,3,5;
0,1,5,10,15;
T(5,3)=5 because we have 1245|3, 145|2|3, 14|25|3, 15|24|3 and 1|245|3.
The arrays U(n,k) starts:
1 0 0 0 0 ...
1 1 1 1 1 ...
2 3 5 9 17 ...
5 10 22 52 130 ...
15 37 99 283 855 ...
MATHEMATICA
T[n_, k_] := Sum[If[n == k, 1, i^(n-k)]*StirlingS2[k, i], {i, 0, k}];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 10 2024, after Vladeta Jovovic *)
KEYWORD
AUTHOR
Christian G. Bower, Jun 03 2005; Emeric Deutsch, Nov 14 2006
EXTENSIONS
Edited by N. J. A. Sloane, May 22 2008, at the suggestion of Vladeta Jovovic. This entry is a composite of two entries submitted independently by Christian G. Bower and Emeric Deutsch, with additional comments from Augustine O. Munagi.
STATUS
approved