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%I A131106 #3 Mar 30 2012 17:38:08
%S A131106 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 A131106 256,405,64,7,0,1,7,108,125,256,729,448,1,0,1,16,147,864,3125,6144,
%U A131106 5103,1024,9,0,1,9,64,343,6480,3125,28672,2187,256,5,0,1,20,243,2048,12005
%N 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.
%C A131106 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 A131106 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 A131106 Array begins:
%e A131106 1 0 0 0 0 0
%e A131106 1 1 3/4 1/2 5/16 3/16
%e A131106 1 4/3 4/3 32/27 80/81 64/81
%Y A131106 Cf. A131107 gives the denominators. A131103, A131104 and A131105 give f(n, k, 0), f(n, k, 1) and f(n, k, 2).
%K A131106 easy,frac,nonn,tabl
%O A131106 1,8
%A A131106 _David Wasserman_, Jun 15 2007

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