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

[Benoît Jubin]

This sequence is very easy in base 2 (it is the identity) and also
manageable in base 3 (it might be worth entering it in the OEIS). As
often with these sequences, one sees better what happens in small
bases (base 2 when it is not trivial, or base 3, like here). For
instance, you can prove that it is a permutation of the integers (this
is equivalent to having an infinite number of digits "1" in the
sequence, which is the case), and you see the fixed points (it might
be possible to characterize them). Once you see the pattern, you might
be able to extend it to arbitrary bases...

On Wed, May 29, 2013 at 2:54 PM, Hans Havermann <gladhobo at teksavvy.com> wrote:
> Likely an infinite number of conjectures because of the fractal-like nature of the sequence where at each successive power of ten, new "detail" emerges. I had grabbed the most blatant one.
>
> On May 29, 2013, at 5:53 AM, Lars Blomberg <lars.blomberg at visit.se> wrote:
>
>> ...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.
>
>
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