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Re^2: 2, 2, 2, 3, 7, 22, 83, ...

by xorl (Deacon)
on Feb 06, 2006 at 20:39 UTC ( #528328=note: print w/ replies, xml ) Need Help??


in reply to Re: 2, 2, 2, 3, 7, 22, 83, ...
in thread 2, 2, 2, 3, 7, 22, 83, ...

All of the above answers are correct; it's trivial to construct a simple eigth order function which will map any solution to every single value listed.
Please show me how to do this? I haven't the slightest clue how to go about it.


Comment on Re^2: 2, 2, 2, 3, 7, 22, 83, ...
Re^3: 2, 2, 2, 3, 7, 22, 83, ...
by gsiems (Chaplain) on Feb 06, 2006 at 21:22 UTC
    The way that I seem to remember learning it way back in college looked like this
      Ah that vaguly looks familar (high school was a long time ago). Now I don't think it can be applied to this series as part of the definition says "where no two xj are the same." Or am I misssing something?
        Yes, you are missing something :-)
        In this case, the x's are 1,2,3,4,5,...,n and the y's are 2,2,2,3,7,...m
        And the method gives a function f so that f(1) = 2, f(2) = 2, f(3) = 2, f(4) = 3, and so on...

        -- 6x9=42

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