# [seqfan] Re: Digital question; a question about the palindromes

Mon Mar 14 20:06:27 CET 2011

  In connection with the 4-digit (b,c) palindromes, I obtained an  identity:

(2n+2)*(14n+9)^3+(3n+3)*(14n+9)^2+(5n+3)*(14n+9)+(2n+1))=
(2n+1)*(14n+11)^3+(5n+3)*(14n+11)^2+(3n+3)*(14n+11)+(2n+2).
This proves that we have infinitely many 4-digit (b,c)-palindromes.

So, for n=1, we have [4,6,8,3]_23=[3,8,6,4]_25=52029; for n=2,
[6,9,13,5]_37=[5,13,9,6]_39=316725, etc.

Moreover, it makes more plausible a general conjecture:  for a given k>=2,  there are infinitely many k-digit (b,c)-palindromes (in which leading 0's are not allowed).

Best regards,

----- Original Message -----
From: Robert Israel <israel at math.ubc.ca>
Date: Friday, March 4, 2011 21:37
Subject: [seqfan] Re: Digital question; a question about the palindromes
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Here are a few, generated by a brute-force search:
>
> 2331_9 = 1332_11 = 1729
> 2651_13 = 1562_16 = 5474
> 3831_11 = 1383_16 = 4995
> 4051_11 = 1504_16 = 5380
> 4211_7 = 1124_11 = 1478
> 4521_7 = 1254_11 = 1632
> 4701_9 = 1074_15 = 3484
> 4781_11 = 1874_16 = 6260
> 4811_11 = 1184_18 = 6304
> 4662_9 = 2664_11 = 3458
> 5141_10 = 1415_16 = 5141
> 5241_7 = 1425_11 = 1842
> 5271_11 = 1725_17 = 6975
> 5431_5 = 1345_8 = 741
> 5551_7 = 1555_11 = 1996
> 5581_8 = 1855_12 = 2945
> 5701_9 = 1075_16 = 4213
> 5942_11 = 2495_15 = 7790
> 5503_9 = 3055_11 = 4053
> 6271_7 = 1726_11 = 2206
> 6401_7 = 1046_13 = 2255
> 6112_7 = 2116_10 = 2116
> 6202_11 = 2026_16 = 8230
> 6352_8 = 2536_11 = 3306
> 6932_11 = 2396_16 = 9110
> 6143_9 = 3416_11 = 4494
> 6723_7 = 3276_9 = 2418
> 6993_9 = 3996_11 = 5187
> 7152_11 = 2517_16 = 9495
> 7702_7 = 2077_11 = 2746
> 7882_11 = 2887_16 = 10375
> 7273_11 = 3727_14 = 9639
> 7453_7 = 3547_9 = 2635
> 7834_9 = 4387_11 = 5782
> 8071_9 = 1708_16 = 5896
> 8014_8 = 4108_10 = 4108
> 8194_14 = 4918_17 = 22278
> 8474_9 = 4748_11 = 6223
> 8945_11 = 5498_13 = 11786
> 9471_10 = 1749_19 = 9471
> 9571_9 = 1759_17 = 7030
> 9303_11 = 3039_16 = 12345
> 9315_9 = 5139_11 = 6818
> 9495_11 = 5949_13 = 12567
>
> Cheers,
> Robert
>
>
>
> On Fri, 4 Mar 2011, Vladimir Shevelev wrote:
>
> >    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,
> >
> >
> > ----- 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
> >>
> >>>
> >>> 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,
> >>>>
> >>>> ----- 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,
> >>>>>
> >>>>>
> >>>>> ----- 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,
> >>>>>>>
> >>>>>>>
> >>>>>>>
> >>>>>>> -----
> >>>>>>> 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>
> >>>>>>>
> >>>>>>> _______________________________________________
> >>>>>>>
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> >>>>>>
> >>>>>>
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> >>>>>>
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> >>>>>>
> >>>>>
> >>>>>
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> >>>>>
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> >>>>>
> >>>>
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> >>>
> >