[seqfan] Re: Digital question; a question about the palindromes
Vladimir Shevelev
shevelev at bgu.ac.il
Fri Mar 4 18:01:48 CET 2011
Thanks, Robert, for complete result on 3-digit (2n+1,2n+2)-palindromes. I think that it is worth to publish this sequence: a(n)=(n+1)*((2*n+1)*(2*n+2)-1): 22,87,220,445,786,1267, etc.
I was not able to generalize your result. What is about nontrivial 4-digit (b,c)-palindromes? I know
none of them.
Regards,
Vladimir
----- Original Message -----
From: Robert Israel <israel at math.ubc.ca>
Date: Thursday, March 3, 2011 19:36
Subject: [seqfan] Re: Digital question; a question about the palindromes
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>
>
> On Thu, 3 Mar 2011, Robert Israel wrote:
>
> > For example, two-digit (b,c)-palindromes satisfy b a_1 + a_2 =
> c a_2 + a_1,
> > i.e. (b - 1) a_1 = (c - 1) a_2, or a_1/a_2 = (c-1)/(b-
> 1). Nontrivial
> > examples will require gcd(c-1,b-1) > 1.
> > For example, 45_11 = 54_9 = 49.
> >
> > Three-digit (b,c) palindromes satisfy
> > (b^2-1) a_1 + (b-c) a_2 + (c^2-1) a_3 = 0.
>
> Of course that should have been
> (b^2-1) a_1 + (b-c) a_2 - (c^2 - 1) a_3 = 0.
> This typo doesn't affect the rest of the message.
>
> > If c=b+1 this becomes
> > a_2 = (b^2-1) a_1 - (b^2 + 2b) a_3. But then a_2 == -a_1
> mod b, so
> > a_2 = b - a_1, and then a_1 = a_3 + (1 + 2 a_3)/b. With
> 0 <= a_3 < b,
> > this can only happen if b = 1 + 2 a_3. Thus we have
> three-digit
> > (2n+1, 2n+2) palindromes (n+1,n,n)_{2n+1} = (n,n,n+1)_{2n+2}, e.g.
> > 544_9 = 445_10.
> >
> >
> > Robert
> Israel israel at math.ubc.ca
> > Department of
> Mathematics http://www.math.ubc.ca/~israel University of
> > British
> Columbia Vancouver, BC, Canada
> >
> > On Thu, 3 Mar 2011, Vladimir Shevelev wrote:
> >
> >> Let base-b of n is
> [a_1...a_k]_b. Consider base-c reverse: [a_k...a_1]
> >> _c.
> >> If [a_1...a_k]_b= [a_k...a_1] _c, then n is called
> (b,c)-palindromic.
> >> E.g., n=0 and n=1 are (b,c)-palindromic for every b,c. On the
> other hand,
> >> nontrivial (b,c)-palindromes could exist when the ratio b/c
> is closed to 1,
> >> e.g., c=b+1 for sufficiently large b.
> >> I ask anyone to find some of them.
> >>
> >> This problem arose from David's digit question and the
> following its
> >> generalization:
> >> d | n= [a_1...a_k]_b <=>d | [a_k...a_1] _c <=>d | bc-1
> >> which one can prove by same way.
> >> The case b=c corresponds to David's observation.
> >> Palindromes arose in my last message on this topic.
> >>
> >> Regards,
> >> Vladimir
> >>
> >> ----- Original Message -----
> >> From: Vladimir Shevelev <shevelev at bgu.ac.il>
> >> Date: Friday, February 25, 2011 22:34
> >> Subject: [seqfan] Re: Digital question
> >> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> >>
> >>> 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,
> >>> Vladimir
> >>>
> >>>
> >>> ----- 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,
> >>>>>> Vladimir
> >>>>>
> >>>>>
> >>>>>
> >>>>> -----
> >>>>> No virus found in this message.
> >>>>> Checked by AVG - www.avg.com
> >>>>> Version: 10.0.1204 / Virus Database: 1435/3459 - Release
> >>> Date:
> >>>> 02/22/11>
> >>>>>
> >>>>> _______________________________________________
> >>>>>
> >>>>> Seqfan Mailing list - http://list.seqfan.eu/
> >>>>
> >>>>
> >>>> _______________________________________________
> >>>>
> >>>> Seqfan Mailing list - http://list.seqfan.eu/
> >>>>
> >>>
> >>> Shevelev Vladimir
> >>>
> >>> _______________________________________________
> >>>
> >>> Seqfan Mailing list - http://list.seqfan.eu/
> >>>
> >>
> >> Shevelev Vladimir
> >>
> >> _______________________________________________
> >>
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >
Shevelev Vladimir
More information about the SeqFan
mailing list