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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
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.
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: A106646 A056969 A335532 * A298739 A346366 A325011
KEYWORD
easy,frac,nonn,tabl
AUTHOR
David Wasserman, Jun 15 2007
STATUS
approved