# [seqfan] wanted: solution to a recurrence (corrected)

N. J. A. Sloane njas at research.att.com
Thu Oct 21 21:31:07 CEST 2010

```Dear Seq Fans (and especially Eric Angelini)

Corrected version - I forgot one condition!

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, but
a(i) != a(n+1) for i = n+2, ..., a(n)+n.

In other words, a(a(n)+n+1) is the next occurrence of a(n+1) (after a(n+1)).

For example,
1 2 3 4 5 6 7 8 9 ...
1 2 2 2 2 2 2 2 2 ...

is not a solution, since (with n=2) we have
a(3)=a(5)=2, but we also have a(4)=2, which is not allowed.

I would like the earliest such sequence.

To put it another way, say a(n) = P, a(n+1) = Q.
Then the rule says that the next time we see Q is P steps down the road!

The entries should be positive integers, not all 1's.

The only nontrivial solution I know is

1 3 3 1 3 3 1 3 3 1 3 3 ...

Are there others?

Neil

```

More information about the SeqFan mailing list