[seqfan] Re: wanted: solution to a recurrence
Roland Bacher
Roland.Bacher at ujf-grenoble.fr
Thu Oct 21 19:30:52 CEST 2010
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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