[seqfan] Last term? (correction)

[Eric Angelini]

I guess S starts more like this:

S=1,21,2,11,3,4,111,1122,5,6,7,8,9,22,23,22222,
Slenghts=1  2  1  2 1  1   3    4    1  1  1 1 1  2,  2,  5

Thank you, Maximilian!
Best,
É.
Catapulté de mon aPhone

Le 25 oct. 2015 à 07:17, Eric Angelini <Eric.Angelini at kntv.be<mailto:Eric.Angelini at kntv.be>> a écrit :

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?

Catapulté de mon aPhone

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