A361950 Array read by antidiagonals: T(n,k) = n! * Sum_{s} 2^(Sum_{i=1..k-1} s(i)*s(i+1))/(Product_{i=1..k} s(i)!) where the sum is over all nonnegative compositions s of n into k parts.

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 6, 1, 0, 1, 4, 13, 26, 1, 0, 1, 5, 22, 81, 162, 1, 0, 1, 6, 33, 166, 721, 1442, 1, 0, 1, 7, 46, 287, 1726, 9153, 18306, 1, 0, 1, 8, 61, 450, 3309, 24814, 165313, 330626, 1, 0, 1, 9, 78, 661, 5650, 50975, 494902, 4244481, 8488962, 1, 0

Offset: 0

Author

Andrew Howroyd, Mar 31 2023

Keywords

• nonn
• tabl

Comments

T(n,k) corresponds to c(k,n) in the Klarner reference. This is an intermediate step in the computation of the number of labeled weakly graded (ranked) posets. The number of elements in the poset is n and the rank k.

Examples

			Array begins:
======================================================
n/k| 0 1      2       3        4        5        6 ...
---+--------------------------------------------------
0  | 1 1      1       1        1        1        1 ...
1  | 0 1      2       3        4        5        6 ...
2  | 0 1      6      13       22       33       46 ...
3  | 0 1     26      81      166      287      450 ...
4  | 0 1    162     721     1726     3309     5650 ...
5  | 0 1   1442    9153    24814    50975    91866 ...
6  | 0 1  18306  165313   494902  1058493  1957066 ...
7  | 0 1 330626 4244481 13729846 29885567 55363650 ...
  ...
T(3,2) = 26: the nonnegative integer compositions of 3 with 2 parts are (0,3), (1,2), (2,1), (3,0). These contribute, respectively 2^0*3!/(0!*3!) = 1, 2^2*3!/(1!*2!) = 12, 2^2*3!/(2!*1!) = 12, 2^0*3!/(0!*3!) = 1, so T(3,2) = 1 + 12 + 12 + 1 = 26.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1325 (first 51 antidiagonals).
• D. A. Klarner, The number of graded partially ordered sets, J. Combin. Theory, 6 (1969), 12-19.

Crossrefs

Columns k=0..7 are A000007, A000012, A047863, A000684, A001827, A001828, A001829, A001830.

Rows 0..2 are A000012, A000027, A028872(n+1).

The unlabeled version is A361952.

Cf. A361951.

Programs

PARI

S(M)={matrix(#M, #M, i, j, sum(k=0, i-j, 2^((j-1)*k)*M[i-j+1,k+1])/(j-1)! )}
C(n, m=n)={my(M=matrix(n+1, n+1), c=vector(m+1), A=O(x*x^n)); M[1, 1]=1; c[1]=1+A; for(h=1, m, M=S(M); c[h+1]=sum(i=0, n, vecsum(M[i+1, ])*x^i, A)); c}
R(n)={Mat([Col(serlaplace(p)) | p<-C(n)])}
{ my(A=R(6)); for(i=1, #A, print(A[i,])) }