[seqfan] Re: Digital question

Vladimir Shevelev shevelev at bgu.ac.il
Fri Feb 25 13:40:06 CET 2011

Let us separate a nontrivial part of your proof,

If inv(d)<d, then it is sufficient to choose n=d ; if inv(d)>d and not multiple of d, then n=d is required as well. What is the left? d could be either PALINDROMIC (inv(d)=d)  or a number wth the reversal  multiple of d (cf. A031877).  Denote the set of such numbers R_b.

Thus from your proof we conclude that, if d is not a divisor of b^2-1, then, for every d \in R_b, there exists  m such that m*d is not only in R_b but even inv(m*d) is not multiple of d.

Maybe, there exists a direct proof of this fact?

Thanks,

----- Original Message -----
From: Juan Arias de Reyna <arias at us.es>
Date: Thursday, February 24, 2011 10:59
Subject: [seqfan] Re: Digital question
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Here is the proof of
>
>  (d does not divide b^2-1) => there exists n with d | n but
> not d | n_1.
>
>
>
> Assume that (d does not divide b^2-1)   we want to
> prove that there exists n with
> d | n but not d | n_1.
>
> (a) We may assume that  gcd(d,b)=1. In other
> case   d = d_1*d_2 with  gcd(d_1,b)=1
> d_2 | b^k  for a suitable k.   Now  (d_1
> does not divide b^2-1)    and if we find
> n with  d_1 | n but not d_1 | n_1.  Take
> m=n*b^k  it is clear that d | m   and
> m_1 = n_1, so that d does not divide m_1.
>
>
> (b) Let  e the least natural number with  d | b^e-
> 1.  We will have e > 2, since
> d does not divide b^2 -1 and hence also does not divide b-1.
>
> Let x_0=1, x_1, x_2,  x_{e-1} , defined by x_j is the rest
> of b^j mod d.
> The usual criterion of divisibility gives us  d | n is
> equivalent to
> d | A_0+A_1*x_1+A_2*x_2+ ... + A_{e-1}*x_{e-1} where if
> A_j = a_j+a_{j+e}+a_{j+2e}+...
> are the sum of the digits of the representation of n in base b.
>
> Given elements (z_j)_{j=0}^{e-1} in Z/(dZ) it is easy to
> construct a number n such that
> A_j(n)\equiv z_j mod d  and  A_j(n_1) equiv z_{e-j-1}
> mod d.  Therefore we only need to construct
> elements z_j in Z/(dZ) such that  z_0+z_1*x_1+z_2*x_2+ ...
> + z_{e-1}*x_{e-1}=0 and
> z_{e-1}+z_{e-2}*x_1+z_{e-3}*x_2+ ... + z_{0}*x_{e-1} != 0
>
>
>
> Assume that e is odd. e = 2f+1  then take  z_0 =
> 0,  z_f = -b^f, z_{e-1}=1 and all other
> z_j = 0. Then
>
> z_0+z_1*x_1+z_2*x_2+ ... + z_{e-1}*x_{e-1} = -b^f x_f+x_{2f} = -
> b^f b^f + b^(2f) = 0
> and
> z_{e-1}+z_{e-2}*x_1+z_{e-3}*x_2+ ... + z_{0}*x_{e-1} =1 - b^f
> x_f = 1-b^(2f) != 0
> since b^e = b^(2f+1) is the first power of b  equal to 1.
>
> Assume now that e is even  e = 2f. Since e >2  we
> have  e >= 4.
> In this case take  z_0 = -b^f, z_{f-1} = b, z_f = 0,
> z_{2f-1}=0. Then we will have
> z_0+z_1*x_1+z_2*x_2+ ... + z_{e-1}*x_{e-1} = -b^f + b x_{f-1} = -
> b^f + b^f = 0
> and
> z_{e-1}+z_{e-2}*x_1+z_{e-3}*x_2+ ... + z_{0}*x_{e-1} = b x_f -
> b^f x_{e-1} =
> b^(f+1) - b^f b^(2f-1) = b^(f+1) - b^(f-1) = b^(f-1) (b^2-1) != 0.
>
> This ends the proof.
>
> Regards,
> Juan Arias de Reyna
>
>
> El 23/02/2011, a las 01:05, David Wilson escribió:
>
> > Very good. Now show
> >
> > (d does not divide b^2-1) => there exists n with d | n but not
> d | n_1.
> >
> >
> > ----- Original Message ----- From: "Vladimir Shevelev"
> <shevelev at bgu.ac.il>> To: "Sequence Fanatics Discussion list"
> <seqfan at list.seqfan.eu>> Sent: Tuesday, February 22, 2011
> 7:33 AM
> > Subject: [seqfan] Re: Digital question
> >
> >
> >> From b^(t-1)*n-n_1==0 (mod d) we conclude that d | n => d |
> n_1. Since {n_1 is inv. n}<=>
> >> {n is inv. n_1}, then also b^(t-1)*n_1-n==0 (mod d) and we
> conclude that d | n_1 => d | n.
> >>
> >> Regards,
> >
> >
> >
> > -----
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