# [seqfan] Re: Shouldn't A158624 be hard and base?

Maximilian Hasler maximilian.hasler at gmail.com
Mon Apr 2 16:03:44 CEST 2012

```This is another instance which shows how weird the current indexing
convention for constants is. We have to write
"
The first digit of the backward value of 5^n is a(0) = 5 for all n.
The second digit is a(-1) = 2 for all n >= 2.
The third digit is a(-2) = 6 for all even n >= 4, etc.
"

If we had a more natural indexing scheme with increasing indices like e.g.

c = sum( i=offset..infinity, a(i)/10^i )

then we would have, in the above, a(1)=5, a(2)=2, a(3)=6, etc.

At present, adding the "cons" kw reverses the indexing, making b-files
invalid when they were conceived for increasing indices.
For example, since the 'all ones" sequence also is the expansion of
1/9, they are indexed
by 0,-1,-2,-3,..., cf http://oeis.org/A000012/list

Maximilian

On Sun, Apr 1, 2012 at 9:26 PM,  <israel at math.ubc.ca> wrote:
> 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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>
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```