A131106 Rectangular array read by antidiagonals: k objects are each put into one of n boxes, independently with equal probability. a(n, k) is the expected number of boxes with exactly one object (n, k >= 1). Sequence gives the numerators.

1, 1, 0, 1, 1, 0, 1, 4, 3, 0, 1, 3, 4, 1, 0, 1, 8, 27, 32, 5, 0, 1, 5, 48, 27, 80, 3, 0, 1, 12, 25, 256, 405, 64, 7, 0, 1, 7, 108, 125, 256, 729, 448, 1, 0, 1, 16, 147, 864, 3125, 6144, 5103, 1024, 9, 0, 1, 9, 64, 343, 6480, 3125, 28672, 2187, 256, 5, 0, 1, 20, 243, 2048, 12005

Offset: 1

Author

David Wasserman, Jun 15 2007

Keywords

• easy
• frac
• nonn
• tabl

Comments

Problem suggested by Brandon Zeidler. To motivate this sequence, suppose that when objects are placed in the same box, they mix and the information they contain is lost. The sequence tells us how much information we can expect to recover.

Examples

			Array begins:
  1 0 0 0 0 0 ...
  1 1 3/4 1/2 5/16 3/16 ...
  1 4/3 4/3 32/27 80/81 64/81 ...
  ...
		

Crossrefs

Cf. A131107 gives the denominators. A131103, A131104 and A131105 give f(n, k, 0), f(n, k, 1) and f(n, k, 2).

Formula

a(n, k) = k*(1 - 1/n)^(k - 1). Let f(n, k, i) be the number of assignments such that exactly i boxes have exactly one object. For i > n, f(n, k, i) = 0. For i = k <= n, f(n, k, i) = n!/(n-k)!. Otherwise, f(n, k, i) = Sum_{j = 1..min(floor((k-i)/2), n-i)} A008299(k-i, j)*n!*binomial(k, i)/(n-i-j)!. Then a(n, k) = Sum_{i=1..min(n, k)} i*f(n, k, i)/n^k.