[seqfan] Re: 999999999888888887777777666666555554444333221

[Frank Adams-Watters]

Clearly not.

There can't be any zeros in the sequence, because including a zero 
implies that the number of zeros is zero. This means there can be no 
carries between consecutive values.

So you are just incrementing a digit d to d+1. If d = 1, removing it 
 from the number is OK, but there will either have been zero or d+1 
digits d+1 in the number. Now there is either 1 or d+2 of them, neither 
of which is permissible.

I think 9 is the smallest possible difference; it occurs many times.

Franklin T. Adams-Watters

-----Original Message-----
From: Alonso Del Arte <alonso.delarte at gmail.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Sun, Sep 21, 2014 1:14 pm
Subject: [seqfan] Re: 999999999888888887777777666666555554444333221


Side question: Does that sequence (A108571) have any terms that are
consecutive integers?

Al

On Sun, Sep 21, 2014 at 9:50 AM, M. F. Hasler <oeis at hasler.fr> wrote:

> Eric,
> this is oeis.org/A108571 :
> Any digit d in the sequence says: "I am part of an integer in which
> you'll find d digits "d".
>
> by Eric Angelini, Jul 05 2005
>
> Maximilian
>
>
> On Sat, Sep 20, 2014 at 5:43 PM, Eric Angelini <Eric.Angelini at kntv.be>
> wrote:
> > (My last post of the day, Gérard!)
> >
> > Hello SeqFans,
> > Numbers showing, for any of their digit "d", exactly d copies of 
"d":
> >
> > 
S=1,22,122,212,221,1333,3133,3313,3331,14444,41444,44144,44414,44441,...
> >
> > Example:
> > 312332 belongs to S as there are exactly
> > one "1", two "2" and three "3" in 312332.
> >
> > The sequence is finite (how many
> > terms?) and the last term is, of course,
> > this message header.
> >
> > Best,
> > É.
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>



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Alonso del Arte
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<https://www.smashwords.com/profile/view/AlonsoDelarte>
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