[seqfan] Re: Digital question
David Wilson
davidwwilson at comcast.net
Tue Feb 22 03:35:56 CET 2011
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
>
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