A210024 Floor of the expected value of number of trials until all cells are occupied in a random distribution of 2n balls in n cells.

1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 5, 6, 7, 9, 11, 13, 16, 19, 23, 27, 33, 39, 47, 57, 68, 81, 97, 116, 139, 167, 199, 239, 286, 342, 409, 489, 585, 700, 838, 1002, 1199, 1434, 1716, 2053, 2456, 2938, 3515, 4205, 5030, 6018, 7199, 8612, 10302, 12325, 14744, 17638

Offset

1,6

Comments

Also floor of the expected value of number of trials until we have n distinct symbols in a random sequence on n symbols of length 2n.

From (2.3), see first reference,

p_0(2n,n)=Sum_{v=0..n-1}((-1)^v * binomial(n,v) * (n-v)^(2n)/n^(2n))

= 1/n^(2n).Sum_{v=0..n-1}( (-1)^v * binomial(n,v) * (n-v)^(2n)), so

the expected value 1/p_0(2n, n) =

1/(1/n^(2n).Sum_{v=0..n-1}( (-1)^v * binomial(n,v)*(n-v)^(2n)))

= n^(2n)/Sum_{v=0..n-1}( (-1)^v * binomial(n,v)*(n-v)^(2n) )

References

W. Feller, An Introduction to Probability Theory and its Applications, 2nd ed, Wiley, New York, 1968, (2.3) p. 92. (Occupancy problems)

Links

Washington Bomfim and T. D. Noe, Table of n, a(n) for n = 1..1000 (Washington Bomfim computed the first 100 terms)

Formula

a(n) = floor(n^(2n)/Sum_{v=0..n-1}( (-1)^v * binomial(n,v) * (n-v)^(2n) ))

Example

For n=2, with symbols 0 and 1, the 2^4 sequences on 2 symbols of length 4 can be represented by 0000, 0001, 0010, 0011, 0100, 0101,0110, 0111, 1000, 1001, 1010, 1011, 1100, 1110, and 1111. We have 2 sequences with a unique symbol, and 14 sequences with 2 distinct symbols, so a(2) = floor(16/14) = floor(8/7) = 1.

Mathematica

Table[Floor[n^(2 n)/Sum[((-1)^v*Binomial[n, v]*(n - v)^(2 n)), {v, 0, n - 1}]], {n, 100}] (* T. D. Noe, Mar 16 2012 *)

Crossrefs

Cf. A209899, A209900.

Sequence in context: A185325 A125890 A067661 * A052839 A125894 A091493

Adjacent sequences: A210021 A210022 A210023 * A210025 A210026 A210027

Keyword

nonn

Author

Washington Bomfim, Mar 16 2012

Status

approved