[seqfan] Re: Shouldn't A158624 be hard and base?
israel at math.ubc.ca
israel at math.ubc.ca
Mon Apr 2 03:26:38 CEST 2012
Yes, the explanation could be made a lot clearer. Since the 3rd last digit
of 5^n is either 1 or 6, we choose the largest (6). Now when the last 3
digits are 625, the 4th last is 0 or 5, so we choose 5. When the last 4
digits are 5625, the 5th last is 1 or 6, so we choose 6...
Robert Israel
University of British Columbia
On Apr 1 2012, D. S. McNeil wrote:
>> I don't understand the sequence. 5^2 = 1 mod 8, so 5^n ends in ...625
>> if n is even and ...125 if n is odd (for n > 2). So it doesn't
>> stabilize to the indicated value (or any value!).
>
>As I read it, it doesn't claim that it's the value that it stabilizes
>at, it's saying that it's the (presumably least) upper bound, i.e. the
>supremum.
>
>
>Doug
>
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