-4
$\begingroup$

Following this, let's define the notion of perfect sequence as follows:

$(u_{i})_{i}$ is a perfect sequence if and only if it is the sequence of divisors of an even perfect number in increasing order or the infinite sequence $(u_{i})_{i}, \ \ u_{i}=2^{-i}$.

Let's now list the set of finite perfect sequences by increasing number of terms, so that $(v_{1,i})=(1, 2, 3, 6)$, $(v_{2,i})=(1, 2, 4, 7, 14, 28)$, And so on.

My question is: is the diagonal sequence $(v_{i,i})_{i}$ such that $\sum_{i}{v_{i,i}}^{-1}=2$?

Many thanks in advance.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

You are asking if a sum of $i$ numbers equals $2$.
The partial sum
$S_k=\frac{1}{u_{1,1}}+...+\frac{1}{u_{k,k}}$ does not change as $i$ grows for every $k<i$ so, if you want to reach $2$ there are these possibilities :
1. $i=\infty$
2. $i<\infty$
We don't know if there are infinitely or finitely many even perfect numbers so we cannot expect to answer your question.
It is like you are asking if the sum of the reciprocals of even perfect numbers converges to a rational.
If it does, there are finitely many,but we do not have any idea at all about how many exist.
But in your comment above, it seems that you understand what I am saying.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Indeed. I was hoping for methods outside number theory, like maybe functional analysis (say, consider a Hilbert space generated by finite perfect sequences in which such sequences would converge to the infinite perfect sequence, or something like that) to shed some light on the problem. $\endgroup$
    – Sylvain JULIEN
    Dec 8, 2015 at 21:09
1
$\begingroup$

The $i$th even perfect number is $2^{n_i-1}(2^{n_i}-1)$, where the $n_i$ are an enumeration of those $n$ such that $2^n-1$ is prime. In particular, the $n_i$ are prime themselves, so that $n_i>i$ for each $i$. Now you can see that the $i$th factor (ordered small to large) of the $i$th even perfect number is $2^{-i}$, so that the sum of the reciprocals of the $i$th factors is 2, as conjectured.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ But that would mean there are infinitely many even perfect numbers, which has been unknown so far, am I wrong? $\endgroup$
    – Sylvain JULIEN
    Dec 8, 2015 at 19:08
  • 1
    $\begingroup$ Right. I was assuming that there are infinitely many even perfect numbers. So the conjecture is equivalent to the existence of infinitely many even perfect numbers. It's almost trivial to deduce from the existence of infinitely many even perfect numbers (as above), but it seems unlikely that this would provide a useful method of attacking the question of whether there are infinitely many even perfect numbers - how could you control the sum without knowing how many terms it has? $\endgroup$
    – Anthony Quas
    Dec 8, 2015 at 19:48
  • $\begingroup$ I was thinking of the equality of the sum of the reciprocals as some kind of property that would remain true while considering the diagonal, hence my question. $\endgroup$
    – Sylvain JULIEN
    Dec 8, 2015 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.