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

[jp allouche]

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
>> does not mention this property.  Someone please help!
>> 
>> Best regards
>> Neil
>> 
>> Neil J. A. Sloane, Chairman, OEIS Foundation.
>> Also Visiting Scientist, Math. Dept., Rutgers University,
>> Email: njasloane at gmail.com
>> 
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>> 
> 
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