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        riddles >> medium >> An Astute Sequence Puzzle
        
(Message started by: K Sengupta on Feb 16^th, 2007, 12:05am)

Title: An Astute Sequence Puzzle
Post by K Sengupta on Feb 16^th, 2007, 12:05am

Consider the sequence {a]_n; n>0} defined by a₀ = 0 and :
a_n+1 = [(a_n+ n)^1/3 ]³ for n>=0; [x] is the greatest integer less than or equal to x.

(A) Determine an explicit formula representing a_n as a function of n only.

(B) Determine all n such that a_n = n

Title: Re: An Astute Sequence Puzzle
Post by towr on Feb 16^th, 2007, 12:56am

A: http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lceil.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(8 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lfloor.gif(n+4)/6http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/rfloor.gif+ 1) - 1)/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/rceil.gif³

Title: Re: An Astute Sequence Puzzle
Post by Eigenray on Feb 24^th, 2007, 8:32pm

To finish this off:

Say a_n = k³. Then a_n+1 = k³ as well, unless

(k³+n)^1/3 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif k+1.

i.e., n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif f(k) = (k+1)³ - k³. So if n<f(k), we have a_n+1 = k³ = a_n, and once n=f(k), we have a_n+1 = (k+1)³.

It follows that for k>0, if f(k-1) < n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif f(k), then a_n = k³.

So in order to have n = a_n = k³ >0, we need f(k-1) < k³ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif f(k). But k³ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif (k+1)³ - k³ when 2k³ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif (k+1)³, or 1+1/k http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 2^1/3 ~ 1.26, which requires k http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 3. Checking these, we find only n = 0, 8, and 27 work.