[seqfan] Re: Period of final n digits of 2^k

[Jean-Paul Allouche]

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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