[seqfan] Re: How many terms?

[Eric Angelini]

Clever, Frank!

-----Message d'origine-----
De : SeqFan [mailto:seqfan-bounces at list.seqfan.eu] De la part de Frank Adams-Watters
Envoyé : lundi 8 décembre 2014 10:47
À : seqfan at list.seqfan.eu
Objet : [seqfan] Re: How many terms?

3,6,12,13,14,15,16,... goes 30.

Franklin T. Adams-Watters

-----Original Message-----
From: Eric Angelini <Eric.Angelini at kntv.be>
To: Sequence Discussion list <seqfan at list.seqfan.eu>
Sent: Mon, Dec 8, 2014 3:02 am
Subject: [seqfan] How many terms?



Hello SeqFans,
How many terms could the longest such sequence S contain?
1) No term of S has a repeated digit (ex. 232 is forbidden)
2) a(n+1)=a(n)+[one digit of a(n)]
3) a(n+1)>a(n)

This S has 28 terms, for instance, and is impossible to extend:

S=1,2,4,8,16,17,18,19,28,36,39,42,46,52,54,58,63,69,75,82,84,92,94,98,106
,107,108,109.

Start S' with any integer and try to beat 28.
Best,
É.


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