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I apologize to edit this question to request suggestions in this case for study on a sequence starting, $n\geq 2$ such that there is an integer $k>1$ that divides both $n$ and $(n/k)+1$, because it is unusual, but I think it could serve as an academic example of a case study of a sequence from The beginning.

Definition. We say that an integer $n\geq 2$ is a Seidov integer if there is an integer $k>1$ such that $k$ divides $n$, and $k$ divides $1+\frac{n}{k}$.

This sequence start as $$2, 6, 10, 12, 14, 15, 18, 20, 22, 24, 26, 28,\cdots,$$ this is Sloane sequence A139799, without known results (it seem that years ago, the author abandoned this nice definition!).

Example. The integer $6$ is Seidov since there in an integer $k=2>1$ such that $2$ divides $n=6$ and $2$ divides $\frac{6}{2}+1=3+1=4$.

Fact. The graph of this sequence is nice and seem regular.

You can see the graph of the sequence A139799, if you push the link graph in [1]. Too there is a table. My question is

Question. Can you provide to me suggestions (if is possible clain what are stardard to leave these as examples to use) to study some aspects of this sequences with methods that you believe can be employed? If you see that you want provide us a statement, feel free.

Thanks in advance.

References:

[1] https://oeis.org/A139799 ,The On-line Encyclopedia of integer sequences.

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  • $\begingroup$ My goal is to win somewhere reputation points and remove this nice definition from the hands of the genuine author (in fact has same name that I, so I take advantage of it!) $\endgroup$ – user243301 Oct 2 '15 at 7:37
  • $\begingroup$ If $d$ and $k$ are any integers, $n=k(kd-1)$ satisfies this property. $\endgroup$ – Bernard Oct 2 '15 at 7:44
  • $\begingroup$ I don't understand @Bernard, thanks $\endgroup$ – user243301 Oct 2 '15 at 7:50
  • $\begingroup$ Check the property: $k$ divides $n$, and $\dfrac nk+1=kd-1+1=kd$. $\endgroup$ – Bernard Oct 2 '15 at 8:02
  • $\begingroup$ Now I understand $k\mid n=k(kd-1)$ implies a term of the sequence since $\frac{k(kd-1)}{k}=kd$ by your computations. Thanks @Bernard $\endgroup$ – user243301 Oct 2 '15 at 10:36
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If $n$ is a perfect square, any $n$ and $k$ satisfying that condition may be a solution of the pell; $n^2-k y^2=-1$, forsome integer $y=n/k$

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  • $\begingroup$ I read your answer, and I am waiting, more results, very thanks much @Adokwu $\endgroup$ – user243301 Oct 2 '15 at 7:51

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