[seqfan] Numbers of the Form 2^j (mod 10^k)

[FRANCISCO SALINAS]

Dear Paul,

You asked for the numbers of the form: 2^j (mod 10^k) where j and k are any 
positive integers. Given k, this gives a sequence for j of period 4*5^(k-1) 
whose values for the periodic part, when ordered, are the multiples (non 
divisible by 5) of 2^k (which are >= 2^k and <= 2^k*(5^k-1)) or 
equivalently: 2^j (mod 10^k) is the same as 2^k*A047201(n) = 
2^k*(n+floor((n-1)/4)) where n = (j-1) (mod 4*5^(k-1)) + 1. So, for example, 
if we have only the last 5  digits of a large integer N, you find that N is 
NOT some power of 2 if and only if the number given by those digits is 
divisible by 5 OR not a multiple of 2^k.

B. Regards,

        Fco.


>From: "Paul D. Hanna" <pauldhanna at juno.com>
>To: seqfan at ext.jussieu.fr
>Subject: Re: [seqfan] Numbers of the Form 2^j (mod 10^k)
>Date: Fri, 27 Aug 2004 11:54:38 -0400
>
>The statements below need correction (from prior email).
>
>The final digits (5 digits in the example) that are given of the unknown
>integer N,
>are not to begin with leading zeros.
>
>And the sequence of numbers of the form 2^j (mod 10^k)
>will not list numbers with leading zeros, like 008 for example.
>These numbers will be listed elsewhere, such as 1008.
>
>And, obviously, I should have typed a '5' for the power of 10, not '6',
>when I said:
>"That is, which 5-digit numbers are of the form:
>     2^j (mod 10^5)
>where j is any positive integer?"
>
>Sorry to trouble you, should you find these matters trivial.
>     Paul
>
>On Fri, 27 Aug 2004 11:19:49 -0400 "Paul D. Hanna" <pauldhanna at juno.com>
>writes:
>      Given only the last 5 (say) digits of a large integer N,
>how can you determine if N is NOT some power of 2?
>That is, which 5-digit numbers are of the form:
>     2^j (mod 10^6)
>where j is any positive integer?
>So, if the last 5 digits of N were not in this sequence,
>then N would not be a power of 2.

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