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Topic: An Astute Sequence Puzzle (Read 517 times) |
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K Sengupta
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An Astute Sequence Puzzle
« on: Feb 16th, 2007, 12:05am » |
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Consider the sequence {a]n; n>0} defined by a0 = 0 and :
an+1 = [(an+ n)1/3 ]3 for n>=0; [x] is the greatest integer less than or equal to x.
(A) Determine an explicit formula representing an as a function of n only.
(B) Determine all n such that an = n
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Eigenray
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Re: An Astute Sequence Puzzle
« Reply #2 on: Feb 24th, 2007, 8:32pm » |
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To finish this off:
Say an = k3. Then an+1 = k3 as well, unless
(k3+n)1/3  k+1.
i.e., n f(k) = (k+1)3 - k3. So if n<f(k), we have an+1 = k3 = an, and once n=f(k), we have an+1 = (k+1)3.
It follows that for k>0, if f(k-1) < n  f(k), then an = k3.
So in order to have n = an = k3 >0, we need f(k-1) < k3 f(k). But k3 (k+1)3 - k3 when 2k3 (k+1)3, or 1+1/k 21/3 ~ 1.26, which requires k 3. Checking these, we find only n = 0, 8, and 27 work.
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(
(8 
(n+4)/6
+ 1) - 1)/2
3