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
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
Programs
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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,])) }
Comments