[seqfan] Re: Help with a(n) and a cumulative sum

[israel at math.ubc.ca]

Yes, it will continue. E.g. from the recurrence relation, a(n) mod 9 is 
determined by a(n-3),a(n-2),a(n-1) mod 9. So once the 4-tuple 1,6,4,3 
occurs a second time, the rest must continue with a(n) == a(n-6) mod 9.

Cheers,
Robert

On Jul 15 2019, Ali Sada via SeqFan wrote:

> Dear Dr. Israel,
>In your sequence, A308900, a(n) mod 9 is 
>1,6,4,3,7,0,1,6,4,3,7,0,1,6,4,3,7,0
>Will this pattern continue? And if so why? 
>I really appreciate your response in advance.
>Best,
>Ali 
>
>
>
>    On Monday, July 15, 2019, 3:57:05 PM EDT, jean-paul allouche 
> <jean-paul.allouche at imj-prg.fr> wrote:
> 
> Dear Neil, dear all
>
>Yes it looks like a topological argument should make it. (But
>it is not really clear to me [yet] how to do this.)
>
>best wishes
>jean-paul
>
>Le 15/07/2019 à 05:53, Neil Sloane a écrit :
>> Robert,  I was just looking at A309151, as it happens.  Could you add 
>> your sequence 1, 6, 4, 66, 34, 666, 334, ... to the OEIS please?  It is 
>> a useful data point.
>>
>> Is A309151 going to need an argument from topology to show that it exists
>> (thus proving I was wrong when I said it was not deep!) ?  Jean-Paul?
>>
>> Best regards
>> Neil
>>
>> Neil J. A. Sloane, President, OEIS Foundation.
>> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
>> Phone: 732 828 6098; home page: http://NeilSloane.com
>> Email: njasloane at gmail.com
>>
>>
>>
>> On Sun, Jul 14, 2019 at 11:29 PM <israel at math.ubc.ca> wrote:
>>
>>> There are infinite sequences such that a(n) doesn't share any digit 
>>> with the cumulative sum. For example: 1, 6, 4, 66, 34, 666, 334, ... 
>>> with cumulative sums 1, 7, 11, 77, 111, 777, 1111, ... So your sequence 
>>> A309151 can indeed be infinite.
>>>
>>> Cheers,
>>> Robert
>>>
>>> On Jul 14 2019, Éric Angelini wrote:
>>>
>>>> Hello SeqFans, Here is the draft of a strange sequence:
>>>> https://oeis.org/draft/A309151 It says: Lexicographically earliest
>>>> sequence of different terms starting with a(1) = 1 such that a(n)
>>> doesn't
>>>> share any digit with the cumulative sum a(1) + a(2) + a(3) + ... +
>>> a(n-1)
>>>> + a(n). And in the Comments section: As this sequence needs a lot of 
>>>> backtracking, we don't guarantee the accuracy of the last 79 integers 
>>>> of the 1079-term b-file. Indeed, the problem comes from the fact that 
>>>> some cumulative sums quickly block the extension of the sequence. This 
>>>> is the case with 10 (or any other sum ending in zero). But this is the 
>>>> case too with 301 after 
>>>> 1,2,3,5,4,6,7,8,9,10,11,12,14,22,20,23,24,30,13,15,16,18,28. So, we 
>>>> bumped quite often in a "bad sum" (on the average, the sequence was 
>>>> extended by 100 terms for every backtrack). To make a prior list of 
>>>> "bad sums" is difficult (meaning impossible, I guess): 258002 is such 
>>>> a "bad sum" if you have previously used {1,3,4,6,7,9,39,49,69,79} else 
>>>> 258002 + 79 = 258081 would be ok. So my questions are: could the 
>>>> sequence be infinite? Could a list of "bad sum numbers" be easely 
>>>> defined and used? Post-scriptum: I am working, together with 
>>>> Jean-Marc, on the variants: "Lexicographically [...] such that a(n) 
>>>> shares exacly k digit with the cumulative sum a(1) + a(2) + a(3) + ... 
>>>> + a(n-1) + a(n)", with k = 1, 2, 3, ... Best, É.
>>>>
>>>> --
>>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>>
>>>>
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