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A131106
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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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4
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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; internal format)
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OFFSET
| 1,8
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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.
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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.
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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
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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: A019984 A106646 A056969 * A152151 A152148 A198261
Adjacent sequences: A131103 A131104 A131105 * A131107 A131108 A131109
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KEYWORD
| easy,frac,nonn,tabl
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AUTHOR
| David Wasserman (dwasserm(AT)earthlink.net), Jun 15 2007
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