|
|
A055775
|
|
a(n) = floor(n^n / n!).
|
|
25
|
|
|
1, 1, 2, 4, 10, 26, 64, 163, 416, 1067, 2755, 7147, 18613, 48638, 127463, 334864, 881657, 2325750, 6145596, 16263866, 43099804, 114356611, 303761260, 807692034, 2149632061, 5726042115, 15264691107, 40722913454, 108713644516
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
Stirling's approximation for n! suggests that this should be about e^n/sqrt(pi*2n). Bill Gosper has noted that e^n/sqrt(pi*(2n+1/3)) is significantly better.
There are n^n distinct functions from [n] to [n] or sequences on n symbols of length n, the number of those sequences having n distinct symbols is n!. So the probability P(n) of bijection is n!/n^n. The expected value of the number of functions that we pick until we found a bijection is the reciprocal of P(n), or n^n/n!. - Washington Bomfim, Mar 05 2012
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
a(5)=26 since 5^5=3125, 5!=120, 3125/120=26.0416666...
|
|
MATHEMATICA
|
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|