[seqfan] Re: Last term?

[M. F. Hasler]

Dear SeqFans,
Eric agreed with the correction I suggested,
S = 1, 21, 2, 11, 3, 4, 111, 1122, 5, 6, 7, 8, 9, 22, ...
cf. http://oeis.org/draft/A261299

Have a nice week,

Maximilian

On Sun, Oct 25, 2015 at 2:16 AM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
> Hello SeqFans,
> What is the last term of the zeroless seq S?
>
> S=1,21,2,22,23,24,25,111,26,1111,27,22222,3,4,5,28,222222,6,7,8,9,29,...
>
> The length of the successive integers
> is given by the successive digits of S
> itself:
> Slenghts = 1,2,1,2,2,2,2,3,2,4,2,5,1,1,1,...
>
> S was always extended with the smallest
> integer not occuring earlier and not
> leading to a contradiction.
>
> Note that S will show only nine 1's
> as there are only nine 1-digit integers
> in Z (all zeroes are forbidden in S as
> a zero would produce an integer of
> length zero later -- which has no sense).
> (we could have decided that zero produces
> a 10-digit integer later in the seq. --
> but we didn't select this option here).
>
> Best,
> É.
>
> P.-S.
> The seq. P obeys the same rules -- but
> all terms of P must be primes: what
> is the last term of P?