The sequence a(1),a(2),a(3),..., is formed according to the recursive rule
a(1)=1, a(2)=a(1)+1/a(1),..., a(n+1)=a(n)+1/a(n), ...
Prove that a(100) > 14.
No direct evaluation, of course.
This looks pretty hard. After trying a few ideas unsuccessfully, I started reading around and came to this:
http://mathworld.wolfram.com/FractionalChromaticNumber.html
and this:
http://mathworld.wolfram.com/MycielskiGraph.html
The successive terms can be rewritten as a(n)=a(n1)^2+b(n1)^2, b(n)=a(n1)*b(n1), c=a(n)/b(n). The first few a,b, are:
a 1 2 5 29 941 969581 1014556267661
b 1 1 2 10 290 272890 264588959090
i.e. a,b quickly become gigantic. The pattern of additional terms is also quite interesting:
a=1, b=(1+1/1), c=(1+1/1)+1/(1+1/1), d=(1+1/1)+1/(1+1/1)+1/((1+1/1)+1/(1+1/1)), e=(1+1/1)+1/(1+1/1)+1/((1+1/1)+1/(1+1/1))+1/((1+1/1)+1/(1+1/1)+1/((1+1/1)+1/(1+1/1))), etc; it seems that simply writing out the sum for the 100th member of the series in this form would require 10^30 '1's, which is a bit daunting.
So I will be interested to see how the stipulation is proved, as required, without direct evaluation.

Posted by broll
on 20111204 03:38:00 