Some information a brother of mine found with respect to your question.
If f(0)=0, the sequence is known as the Hofstadter G-sequence, see
http://mathworld.wolfram.com/HofstadterG-Sequence.html or
http://www.research.att.com/~njas/sequences/A005206
solved by
HG=: (-:<:%:5) <[EMAIL PROTECTED] >:@i.
HG 100
0 1 1 2 3 3 4 4 5 6 6 7 8 8 9 9 10 11 11 12 12 13 14 14 15 16 16 17 17 18 19
19 20 21 21 22 22 23 24 24 25 25 26 27 27 28 29 29 30 30 31 32 32 33 33 34
35 35 36 37 37 38 38 39 40 40 41 42 42 43 43 44 45 45 46 46 47 48 48 49 50
50 51 51 52 53 53 54 55 55 56...
R.E. Boss
> -----Oorspronkelijk bericht-----
> Van: [EMAIL PROTECTED] [mailto:programming-
> [EMAIL PROTECTED] Namens Jacobs, Jan
> Verzonden: donderdag 17 april 2008 19:19
> Aan: [email protected]
> CC: Elsen, Jack van der
> Onderwerp: [Jprogramming] problem
>
> ls,
> I heard of the following problem from a friend.
> Given:
> f(x) = x - f(f(x-1))
> f(0) = 1
>
> I can compute f(1) = 1 - f(f(1-1)) = 1 - f(f(0)) = 1 - f(1) <=> f(1) = 1
> - f(1) <=> 2f(1) = 1 <=> f(1) = 1/2. But this does not help much: f(2) =
> 2 - f(f(2-1)) = 2 - f(f(1)) = 2 - f(1/2).
>
> Questions:
> 1. Do functions f exists on R -> R that are continuous?
> 2. Can J help him with the question? Could memoization help?
>
> Thanks for any help.
> Jan.
>
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