# [seqfan] PS Period of final n digits of 2^k

jp allouche jean-paul.allouche at imj-prg.fr
Fri Jul 8 10:19:56 CEST 2022

```PS When I mistakenly wrote "Z/p^nZ cyclic", I meant of course
"(Z/p^nZ)^× cyclic" (i.e., the multiplicatively invertible elements)
Sorry for the mistake
jean-paul

Le 08/07/2022 09:22, Jean-Paul Allouche a écrit :
> Dear Neil, dear all
>
> Quick hint: it suffices to look at 2^k mod 5^n. Mimicking a classical
> proof
> that Z/p^aZ is cyclic for p odd prime, one proves by induction that
> 2^{4.5^{b-1}}  = 1 + r_b 5^b where r_b is not divisible by 5. This
> gives
> period 4.5{b-1} for mod 5^b. Hence the result.
>
> best
> jp
>
> Le ven. 8 juil. 2022 à 08:05, Neil Sloane <njasloane at gmail.com> a écrit
> :
>
>> Dear Seq Fans,
>> Consider the sequence {2^k, k >= 0} mod 10^n. Let a(n) be its period.
>>
>> 2^k mod 10 is 1 2 4 8 6 2 4 8 6 ... which has period 4 (see A000689).
>> So
>> a(1) = 4.
>> a(2) = 20 (A000855).
>> a(3) = 100 (A126605) according to Zak Seidov, who also says that
>> a(4) = 500.
>>
>> So 4, 20, 100, 500, ? : how does it continue? Is it A005054? But
>> A005054
>>
>> Best regards
>> Neil
>>
>> Neil J. A. Sloane, Chairman, OEIS Foundation.
>> Also Visiting Scientist, Math. Dept., Rutgers University,
>> Email: njasloane at gmail.com
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/

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