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 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 (list; table; graph; refs; listen; history; text; internal format)
 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 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. 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

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Last modified April 3 01:31 EDT 2020. Contains 333195 sequences. (Running on oeis4.)