A333143 Triangle read by rows: T(n, k) = qStirling2(n, k, q) for q = 3, with 0 <= k <= n.
1, 1, 1, 1, 5, 1, 1, 21, 18, 1, 1, 85, 255, 58, 1, 1, 341, 3400, 2575, 179, 1, 1, 1365, 44541, 106400, 24234, 543, 1, 1, 5461, 580398, 4300541, 3038714, 221886, 1636, 1, 1, 21845, 7550635, 172602038, 371984935, 83805218, 2010034, 4916, 1
Offset: 0
Examples
[0] 1 [1] 1, 1 [2] 1, 5, 1 [3] 1, 21, 18, 1 [4] 1, 85, 255, 58, 1 [5] 1, 341, 3400, 2575, 179, 1 [6] 1, 1365, 44541, 106400, 24234, 543, 1 [7] 1, 5461, 580398, 4300541, 3038714, 221886, 1636, 1 [8] 1, 21845, 7550635, 172602038, 371984935, 83805218, 2010034, 4916, 1
Programs
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Maple
qStirling2 := proc(n, k, q) option remember; with(QDifferenceEquations): if n = 0 then return 0^k fi; if k = 0 then return n^0 fi; qStirling2(n-1, k-1, p) + QBrackets(k+1, p)*qStirling2(n-1, k, p); subs(p = q, expand(%)) end: seq(seq(qStirling2(n, k, 3), k=0..n), n=0..9);
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Mathematica
qStirling2[n_, k_, q_] /; 1 <= k <= n := (* q^(k-1) *) qStirling2[n - 1, k - 1, q] + Sum[q^j, {j, 0, k - 1}] qStirling2[n - 1, k, q]; qStirling2[n_, 0, _] := KroneckerDelta[n, 0]; qStirling2[0, k_, _] := KroneckerDelta[0, k]; qStirling2[, , _] = 0; Table[qStirling2[n + 1, k + 1, 3], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 11 2020 *)
Formula
qStirling2(n, k, q) = qStirling2(n-1, k-1, q) + qBrackets(k+1, q)*qStirling2(n-1, k, q) with boundary values 0^k if n = 0 and n^0 if k = 0.
Note that also a second definition is used in the literature which has an additional factor q^k attached to the first term in the equation above. The two versions differ by a factor of q^binomial(k,2).