[seqfan] Re: a(n+1) is the multiple of at least one digit of a(n)

[Lars Blomberg]

Looking at the first 35 million items I found no counter-example to Hans 
Havermann's conjecture.

Furthermore, several other conjectures where a(k)==k, can be made:

- m*10^n, n >= 0, m=19,127..199,217..220,1012,1063..1999
- 12*10^n + a, n >=1, a=7,8,9
- m*10^n + a, n >=1, m=13..19, a=0..9
- 22*10^n, n >= 1
- m*10^n + a, n >=2, m=11..19, a=0..99
and probably many more.

-----Ursprungligt meddelande----- 
From: Hans Havermann
Sent: Tuesday, May 28, 2013 9:03 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: a(n+1) is the multiple of at least one digit of a(n)

Conjecture: For nonnegative integer n, 2*10^n appears at index 2*10^n.


On May 28, 2013, at 10:46 AM, Eric Angelini <Eric.Angelini at kntv.be> wrote:

> I guess this is a permutation of the Naturals:
>
> a(n+1) is the smallest integer not yet in the sequence such that
> a(n+1) is the multiple of at least one digit of a(n):
>
> S=1,2,4,8,16,3,6,12,5,10,7,14,9,18,11,13,15,17,19,20,22,24,...

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