# [seqfan] Re: correcting A181539

Klaus Brockhaus klaus-brockhaus at t-online.de
Thu Nov 4 20:02:50 CET 2010

```  Max,

my version has to be corrected in agreement with your previous posting.
We need the condition m>1 and then have a(1)=9, and the first square in
the comment is 81.

Thanks and best regards,
Klaus

Am 04.11.2010 16:17, schrieb Max Alekseyev:
> Klaus,
>
> According to your definition, a(2) must be equal 1. Why it is not?
>
> Your terms from a(2) onward match those from my purported update to
> A181539 that I sent to SeqFan earlier.
>
> Regards,
> Max
>
> On Thu, Nov 4, 2010 at 2:39 AM, Klaus Brockhaus
> <klaus-brockhaus at t-online.de>  wrote:
>>
>> Sequence: A181539
>> Sender: Klaus Brockhaus
>> Date of submission: Nov 01 2010
>>
>> Description of changes, start:
>> a(2) through a(4), a(7) through a(11) corrected, comment added, example replaced by
>> Description of changes, end
>>
>>    %I A181539
>> ! %S A181539 1,49,249,1249,18751,218751,781249,24218751,74218751,1425781249,
>> ! %T A181539 13574218751,163574218751
>>    %N A181539 a(n) = smallest number m such that m^2 == 1 mod 10^n.
>> ! %C A181539 All terms have last digit 1 or 9. Squares of terms are 1, 2401, 62001, 1560001, 351600001, 47852000001, 610350000001, 586547900000001, 5508423000000001, 2032852170000000001, 184259414700000000001, 26756525040000000000001.
>> ! %e A181539 1249^2 = 1560001 == 1 mod 10^4, and there is no smaller m such that m^2 == 1 mod 10^4. Hence a(4) = 1249.
>>    %K A181539 more,nonn,new
>>    %O A181539 1,2
>>    %A A181539 Kevin Batista (kevin762401(AT)yahoo.com), Oct 29 2010
>>

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