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

Charles Greathouse charles.greathouse at case.edu
Mon Apr 2 16:38:04 CEST 2012

```b-files for constants are always 'wrong' -- the server doesn't let
people upload b-files with decreasing indexes.

The negative ordering is implied by two conditions: a given decimal
place has the same index in two different sequences, and the offset is
the number of digits before the decimal place.  To change it we must
lose one or both of these.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Mon, Apr 2, 2012 at 10:03 AM, Maximilian Hasler
<maximilian.hasler at gmail.com> wrote:
> 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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```