A219585 Number A(n,k) of k-partite partitions of {n}^k into distinct k-tuples; square array A(n,k), n>=0, k>=0, read by antidiagonals.
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 5, 2, 1, 1, 15, 40, 17, 2, 1, 1, 52, 457, 364, 46, 3, 1, 1, 203, 6995, 14595, 2897, 123, 4, 1, 1, 877, 136771, 937776, 407287, 21369, 323, 5, 1, 1, 4140, 3299218, 88507276, 107652681, 10200931, 148257, 809, 6, 1
Offset: 0
Examples
A(1,3) = 5: [(1,1,1)], [(1,1,0),(0,0,1)], [(1,0,1),(0,1,0)], [(1,0,0),(0,1,0),(0,0,1)], [(0,1,1),(1,0,0)]. A(3,2) = 17: [(3,3)], [(3,0),(0,3)], [(3,2),(0,1)], [(2,3),(1,0)], [(3,1),(0,2)], [(2,2),(1,1)], [(1,3),(2,0)], [(2,1),(1,2)], [(2,1),(1,1),(0,1)], [(3,0),(0,2),(0,1)], [(2,2),(1,0),(0,1)], [(2,1),(0,2),(1,0)], [(1,2),(2,0),(0,1)], [(1,2),(1,1),(1,0)], [(0,3),(2,0),(1,0)], [(2,0),(1,1),(0,2)], [(2,0),(0,2),(1,0),(0,1)]. Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, 1, ... 1, 1, 2, 5, 15, 52, 203, ... 1, 1, 5, 40, 457, 6995, 136771, ... 1, 2, 17, 364, 14595, 937776, 88507276, ... 1, 2, 46, 2897, 407287, 107652681, ... 1, 3, 123, 21369, 10200931, 10781201973, ... 1, 4, 323, 148257, 233051939, ... 1, 5, 809, 970246, 4909342744, ...
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..209
Crossrefs
Programs
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Mathematica
f[n_, k_] := f[n, k] = 1/2 Product[Sum[O[x[j]]^(n+1), {j, 1, k}]+1+ Product[x[j]^i[j], {j, 1, k}], Evaluate[Sequence @@ Table[{i[j], 0, n}, {j, 1, k}]]]; a[0, ] = a[, 0] = 1; a[n_, k_] := SeriesCoefficient[f[n, k], Sequence @@ Table[{x[j], 0, n}, {j, 1, k}]]; Table[Print[a[n-k, k]]; a[n-k, k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 11 2013, updated Sep 16 2019 *)
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PARI
EulerT(v)={Vec(exp(x*Ser(dirmul(v,vector(#v,n,1/n))))-1, -#v)} D(p, n, k)={my(v=vector(n)); for(i=1, #p, v[p[i]]++); EulerT(v)[n]^k/prod(i=1, #v, i^v[i]*v[i]!)} T(n, k)={my(m=n*k+1, q=Vec(exp(intformal(O(x^m) - x^n/(1-x)))/(1+x))); if(n==0, 1, (-1)^m*sum(j=0, m, my(s=0); forpart(p=j, s+=(-1)^#p*D(p, n, k), [1, n]); s*q[#q-j])/2)} \\ Andrew Howroyd, Dec 16 2018
Formula
A(n,k) = [(Product_{j=1..k} x_j)^n] 1/2 * Product_{i_1,...,i_k>=0} (1+Product_{j=1..k} x_j^i_j).
Comments