26 Sep
2013
26 Sep
'13
10:07 p.m.
Oh, well. Then I guess that simplex (whose volume is always (n-1)/n!) is a maximum configuration only for those values. --Dan Charles wrote: ----- It's (n-1)/n! only for n = 2, 3, 4. ----- On 2013-09-26, at 3:57 PM, Dan Asimov wrote:
The nth item on the list below (starting from n=1) is (n-1)/n! (though I haven't checked all of them).
This corresponds for example to the volume of the n-simplex in the n-cube [0,1]^n whose vertices are the origin and the n points whose coordinates are one 0 and the rest 1's.
--Dan
On 2013-09-26, at 11:47 AM, Charles Greathouse wrote:
So this is basically the Hadamard problem, and the maximum volume of an n-simplex is A003432 (n)/n!:
1 1 2 1/2 3 1/3 4 1/8 5 1/24 6 1/80 7 2/315 8 1/720 9 1/2520 10 1/11340