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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.
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%I #3 Mar 30 2012 17:38:08

%S 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,

%T 256,405,64,7,0,1,7,108,125,256,729,448,1,0,1,16,147,864,3125,6144,

%U 5103,1024,9,0,1,9,64,343,6480,3125,28672,2187,256,5,0,1,20,243,2048,12005

%N 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.

%C 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.

%F 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.

%e Array begins:

%e 1 0 0 0 0 0

%e 1 1 3/4 1/2 5/16 3/16

%e 1 4/3 4/3 32/27 80/81 64/81

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

%K easy,frac,nonn,tabl

%O 1,8

%A _David Wasserman_, Jun 15 2007