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 Over fourteen (Posted on 2011-12-02)
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.

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 Some thoughts | Comment 1 of 4

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(n-1)^2+b(n-1)^2, b(n)=a(n-1)*b(n-1), 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 2011-12-04 03:38:00

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