[seqfan] wanted: solution to a recurrence

[N. J. A. Sloane]

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