[seqfan] Re: Digital question

[David Wilson]

d | b^2 - 1 => forall n (d | n => d | n') is easy.

Perhaps the converse is indeed evident, in which case a proof should be no 
problem. So where is it?

----- Original Message ----- 
From: "Vladimir Shevelev" <shevelev at bgu.ac.il>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Sent: Sunday, February 20, 2011 4:49 PM
Subject: [seqfan] Re: Digital question


> Let us prove that, if d | b^2-1, then d | n iff d | base-b reverse of n.
>
> Indeed, let n=sum{i=1,...,t}a_i*b^(t-i), n_1=sum{i=1,...,t}a_i*b^{i-1}. 
> Then b^(t-1)*n-n_1=
> sum{i=1,...,t-1}a_i*(b^(2*t-i-1)-b^(i-1))=sum{i=1,...,t-1}a_i*((b^2)^(t-i)-1)*b^(i-1)==0 
> (mod d).
> Quite analogously we obtain that n-n_1* b^(t-1)==0 (mod d). Now the 
> statement follows.
>
> The inverse statement is evident: there exist many examples when d divides 
> n and
> divide neither b^2-1 nor n_1.
>
> Regards,
> Vladimir
> 


-----
No virus found in this message.
Checked by AVG - www.avg.com
Version: 10.0.1204 / Virus Database: 1435/3458 - Release Date: 02/21/11