[seqfan] Re: wanted: solution to a recurrence

[Roland Bacher]

Dear Neil,

What's wrong with  1,2,2,2,2,2,2,2,2,2,......?

Roland

On Thu, Oct 21, 2010 at 12:14:10PM -0400, N. J. A. Sloane wrote:
> Dear Seq Fans (and especially Eric Angelini)
> 
> I'm looking for a sequence a(1), a(2), ..., not all 1's, 
> that satisfies the forward-looking recurrence
> a(n+1) = a(a(n)+n+1) for n>= 1.
> 
> In fact I would like the earliest such sequence.
> 
> Here is an unfinished attempt at constructing one:
> 
>  n:= 1 2 3 4 5 6 7 8 91011121314 ...
> a(n) 2 4 2 4 6 4 2 x 6 x 2 4 y 6 ... 
> 
> where x and y are still to be chosen.
> 
> This does satisfy the recurrence so far:
> 
> n=1: we need a(2) = a(a(1)+2) = a(4), and indeed 
> a(2)=a(4)=4
> 
> n=2: we need a(3) = a(a(2)+3) = a(7), and indeed
> a(3)=a(7)=2
> 
> and so on.
> 
> Can someone produce an explicit sequence satisfying the recurrence?
> The first term might be 1. The entries should be positive
> integers, not all 1's.
> I do not have a solution.
> 
> Neil
> 
> 
> 
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