[seqfan] Re: A sequence and a related array

[Ali Sada]

 Hi Olivier and Charles,
Thank you very much for your very informative responses. I really appreciate them. I know this is not a sequence related email, but it's just a funny story I want to share with the seqfans if possible. 

I served one year in a military prison during Saddam days (for not joining the compulsory military service after graduation.) To fill the long empty days, I went back to my old friend, numbers. I was always fascinated by number theory but unfortunately, all the math I studied in college was calculus and applied mathematics. 

Pens and papers were not permitted in the prison, so I did all the calculations in my head and I started to see these incredible patterns and formulas. When I was released I wanted to share my groundbreaking knowledge with the world. I contacted a couple of math institutions to tell them about my "discoveries" which included things like the sieve of Eratosthenes and the Euclid algorithm! The "most modern" was what I learned later to be Fermat's small theorem! I can imagine now how much fun people in those institutions had while reading my messages.
So now, Monsieur  Gerard, you are telling me that I reached Gauss days? That is a significant time leap for me! 
Thank you again and again.
Best,
Ali 






    On Monday, July 5, 2021, 4:53:36 AM GMT, Olivier Gerard <olivier.gerard at gmail.com> wrote:  
 
 Dear Ali,

I should have added a few things about what you are looking into.

Your sequence ideas lead to cyclotomic polynomials, which are at least as
old as Gauss' first book.

Your example for n=7 is the factorization of  x^7-x : in that case you have
5 factors, of total degree 7
(so you don't really gain any simplicity for prime numbers)

x^7 - x = x (x-1) (x^5 + x^4 + x^3 + x^2 + x + 1)

This should lead you to A032741, which gives the number of terms of the
factorization.

A032741(5+1) = 3 so you will have 3+2 = 5 terms.

Then you can explore the links with finite fields, integer partitions and
binomial coefficients.

Regards,

Olivier



On Sun, Jul 4, 2021 at 10:17 AM Olivier Gerard <olivier.gerard at gmail.com>
wrote:

>
> The sequence you are looking for is
> "Kempner Numbers", A002034
>
> 1, 2, 3, 4, 5, 3, 7, 4, 6, 5, 11, 4, 13, 7, 5, 6, 17, 6, 19, 5, 7, ...
>
> Olivier Gérard
>
>
> On Sun, Jul 4, 2021 at 10:15 AM Ali Sada via SeqFan <seqfan at list.seqfan.eu>
> wrote:
>
>> Hi everyone,
>>
>> If we want to make sure that we have a multiple of a certain positive
>> integer n we can simply multiply n consecutive integers. For example,
>> multiplying 11 consecutive numbers will certainly give us a multiple of 11.
>> However, some numbers don’t need n terms to get that multiple. For
>> example, we can get a multiple of 6 by multiplying only three integers
>> m(m+1)(m+2), which means that a(6) = 3. Or if we want a multiple of 7 we
>> need only five terms m(m+1) (m-1)(m^2+m+1)(m^2-m+1). So, what is the least
>> number of these non-factor-able terms we need to multiply in order to get a
>> multiple of n? I would really appreciate your help with this sequence if
>> you thought it’s suitable for the OEIS.
>>
>> The related array is “the largest common factor of m^k-m^n, where m > 1,
>> n = 1,2,3,.., and k > n."
>> For example, the largest common factor of m^8-m^2 is 252. I would
>> appreciate your help with this one too.
>>
>> Best,
>>
>> Ali
>>
>>
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>

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