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

Robert Israel israel at math.ubc.ca
Fri Mar 4 20:36:42 CET 2011 

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,
> 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
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

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