A180281 Triangle read by rows: T(n,k) = number of arrangements of n indistinguishable balls in n boxes with the maximum number of balls in any box equal to k.
1, 1, 2, 1, 6, 3, 1, 18, 12, 4, 1, 50, 50, 20, 5, 1, 140, 195, 90, 30, 6, 1, 392, 735, 392, 147, 42, 7, 1, 1106, 2716, 1652, 672, 224, 56, 8, 1, 3138, 9912, 6804, 2970, 1080, 324, 72, 9, 1, 8952, 35850, 27600, 12825, 4950, 1650, 450, 90, 10, 1, 25652, 128865, 110715, 54450, 22022, 7865, 2420, 605, 110, 11
Offset: 1
Examples
The T(4,2)=18 arrangements are {0022, 0112, 0121, 0202, 0211, 0220, 1012, 1021, 1102, 1120, 1201, 1210, 2002, 2011, 2020, 2101, 2110, 2200}. Triangle starts 1 1 2 1 6 3 1 18 12 4 1 50 50 20 5 1 140 195 90 30 6 ...
Links
- Alois P. Heinz, Rows n = 1..200, flattened (first 59 rows from R. H. Hardin)
Crossrefs
Programs
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Maple
b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i=0, 0, add(b(n-j, i-1, k), j=0..min(n, k)))) end: T:= (n, k)-> b(n$2, k)-b(n$2, k-1): seq(seq(T(n,k), k=1..n), n=1..12); # Alois P. Heinz, Aug 16 2018 # second Maple program: T:= (n, k)-> coeff(series(((x^(k+1)-1)/(x-1))^n -((x^k-1)/(x-1))^n, x, n+1), x, n): seq(seq(T(n, k), k=1..n), n=1..12); # Alois P. Heinz, Aug 17 2018
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Mathematica
T[n_,k_]:=Select[Tuples[Range[0,k],n],And[Max[#]===k,Total[#]===n]&]; (* Gus Wiseman, Sep 22 2016 *) SequenceForm@@@T[4,2] (* example *) Join@@Table[Length[T[n,k]],{n,1,6},{k,1,n}] (* sequence *) (* Second program: *) b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i == 0, 0, Sum[b[n-j, i-1, k], {j, 0, Min[n, k]}]]]; T[n_, k_] := b[n, n, k] - b[n, n, k-1]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten (* Jean-François Alcover, Aug 28 2022, after Alois P. Heinz *)
Formula
Empirical: right half of table, T(n,k) = n*binomial(2*n-k-2,n-2) for 2*k > n; also, T(n,2) = Sum_{j=1..n} binomial(n,j)*binomial(n-j,j) = 2*A097861(n). - Robert Gerbicz in the Sequence Fans Mailing List
From Alois P. Heinz, Aug 17 2018: (Start)
T(n,k) = [x^n] ((x^(k+1)-1)/(x-1))^n - ((x^k-1)/(x-1))^n.
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