

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.


4



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
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OFFSET

1,8


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.


LINKS

Table of n, a(n) for n=1..71.


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!/(nk)!. Otherwise, f(n, k, i) = sum_{j = 1..min(floor((ki)/2), ni) A008299(ki, j)*n!*binomial(k, i)/(nij)!. Then a(n, k) = sum_{i=1..min(n, k)} i*f(n, k, i)/n^k.


EXAMPLE

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).
Sequence in context: A274819 A106646 A056969 * A298739 A325011 A294188
Adjacent sequences: A131103 A131104 A131105 * A131107 A131108 A131109


KEYWORD

easy,frac,nonn,tabl


AUTHOR

David Wasserman, Jun 15 2007


STATUS

approved



