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   An Interesting Recursion
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   Author  Topic: An Interesting Recursion  (Read 639 times)
ThudnBlunder
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An Interesting Recursion  
« on: Nov 26th, 2008, 9:30am »
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Consider the recursive formula a(n) = a(a(n-1)) + a(n - a(n-1))
 
1) What is a(3)?
 
2) When does a(n)/n equal 0.5?
 
3) What is the least value of n such that a(n)/n - 0.5 < 0.05 for all greater n?
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towr
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Re: An Interesting Recursion  
« Reply #1 on: Nov 26th, 2008, 10:08am »
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For a(n)=0
1) 0
2) never,  
3) doesn't exist
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SMQ
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Re: An Interesting Recursion  
« Reply #2 on: Nov 26th, 2008, 10:12am »
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What's the base case for the recursion?
 
My first thoughts are that in order for the recursion to be evaluated straightforwardly, 0 < a(n) <= n.  If a(n-1) <= 0 for some n, then n - a(n-1) >= n, and a(n - a(n-1)) is unknown; likewise if a(n-1) > n-1 for some n then a(a(n-1)) is unknown.
 
Moreover, assuming the field of integers, 0 < a(n) <= n implies a(1) = 1; but if a(1) = 1, than a(2) = 2 and a(n) = a(n-1) + a(1) = n for all positive integer n.
 
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ThudnBlunder
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Re: An Interesting Recursion  
« Reply #3 on: Nov 26th, 2008, 10:14am »
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on Nov 26th, 2008, 10:08am, towr wrote:
For a(n)=0

Oh yes, a(1) = a(2) = 1
 
I always forget something.  Roll Eyes
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Re: An Interesting Recursion  
« Reply #4 on: Nov 26th, 2008, 10:28am »
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Ah, well, in that case.
1) 2
2) for any n=2k
3) no idea there
« Last Edit: Nov 26th, 2008, 10:28am by towr » IP Logged

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ThudnBlunder
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Re: An Interesting Recursion  
« Reply #5 on: Dec 4th, 2008, 4:09pm »
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Mathworld has a page about this batrachion.
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