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
KEYWORD
nonn
AUTHOR
Washington Bomfim, Mar 16 2012
STATUS
approved