Re^2: 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, ...

Replies are listed 'Best First'.
Re^3: 2, 2, 2, 3, 7, 22, 83, ... by gsiems (Deacon) on Feb 06, 2006 at 21:22 UTC
The way that I seem to remember learning it way back in college looked like this	[reply]
Re^4: 2, 2, 2, 3, 7, 22, 83, ... by xorl (Deacon) on Feb 07, 2006 at 13:35 UTC
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 x_j are the same." Or am I misssing something?	[reply]
Re^5: 2, 2, 2, 3, 7, 22, 83, ... by Articuno (Beadle) on Feb 07, 2006 at 21:03 UTC
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	[reply]