A019576 Place n distinguishable balls in n boxes (in n^n ways); let f(n,k) = number of ways that max in any box is k, for 1<=k<=n; sequence gives triangle of numbers f(n,k)/n.
1, 1, 1, 2, 6, 1, 6, 45, 12, 1, 24, 420, 160, 20, 1, 120, 4800, 2450, 375, 30, 1, 720, 65520, 43050, 7560, 756, 42, 1, 5040, 1045170, 858480, 167825, 19208, 1372, 56, 1, 40320, 19126800, 19208000, 4110120, 516096, 43008, 2304, 72, 1
Offset: 1
Examples
: 1; : 1, 1; : 2, 6, 1; : 6, 45, 12, 1; : 24, 420, 160, 20, 1; : 120, 4800, 2450, 375, 30, 1; : 720, 65520, 43050, 7560, 756, 42, 1;
Links
- Alois P. Heinz, Rows n = 1..141, flattened
Programs
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Maple
b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0, add(b(n-j, i-1, k)/j!, j=0..min(k, n)))) end: T:= (n, k)-> (n-1)!* (b(n$2, k) -b(n$2, k-1)): seq(seq(T(n,k), k=1..n), n=1..12); # Alois P. Heinz, Jul 29 2014
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Mathematica
b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i<1, 0, Sum[b[n-j, i-1, k]/j!, {j, 0, Min[k, n]}]]]; T[n_, k_] := (n-1)!*(b[n, n, k]-b[n, n, k-1]); Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten (* Jean-François Alcover, Jan 15 2015, after Alois P. Heinz *)
Formula
T(n,k) = A019575(n,k)/n.
Comments