2
$\begingroup$

Some while ago i sat with my professor Adrian Stern of the Ben-Gurion University for a little more than half an hour, we talked about a lot of open question and cutting edge methods in math to tackle these problems.

So he told me about "Erdos Fat sequences conjecture" to be found in Wikipedia Erdős conjecture on arithmetic progressions

The conjecture simply says that if $A$ is a set of integers (sequence) and $A$ is fat meaning $\sum \limits_{n \in A} \frac{1}{n} = +\infty$ then $A$ have a.p. of any length.

Since then i thought about many sequences that "disproof" the conjecture, i am not trying to solve the conjecture or anything, i am just baffled, that these sequences i came up with seems to falsify the conjecture.

For instance let $a_n = \lfloor 16 n \ln (16 n) \ln \ln (16 n) \rfloor$ starting with $A =\{45,137,251,379,517,665,819,980,1147,1319,\cdots \}$

The set $A$ obviously is fat since $\sum \limits_{n=1}^{\infty} \frac{1}{\lfloor 16 n \ln (16 n) \ln \ln (16 n) \rfloor}$ diverges.

Also the set $A$ have no a.p. of length $3$ or more.

So where did i get wrong, what mistakes have i done ?!

Any help is more appreciated.

Last thing, because in Israel we are in (Islamic,Jew) holidays, i could not ask my professor about the thing that baffled me, and the vacations in Israel continues till 26.10.2017, so its very long time to wait.

$\endgroup$
10
  • $\begingroup$ Proof of no arithmetic progression of length 3? $\endgroup$ – GEdgar Sep 5 '17 at 12:28
  • 1
    $\begingroup$ How do you know it has no a.p. of length $>3$ ? $\endgroup$ – Yves Daoust Sep 5 '17 at 12:28
  • 1
    $\begingroup$ @Ahmad: the theorem doesn't say contiguous a.p. $\endgroup$ – Yves Daoust Sep 5 '17 at 12:30
  • 2
    $\begingroup$ @Ahmad: that't for the case of length $3$. $\endgroup$ – Yves Daoust Sep 5 '17 at 12:36
  • 1
    $\begingroup$ How do you prove, that this has no a.p. of arbitrary length? Seems like for large enough n, the ln terms should "not change" (modulo rounding). This gives a a.p. $\endgroup$ – user369397 Sep 5 '17 at 12:39
1
$\begingroup$

To summarize the comments:

The key point is that the conjecture does not require the fat sequence to have consecutive terms in progression.

As it happens, this sequence has progressions of length $3$, even if you require consecutive terms. For example, $$(a_{92},a_{93},a_{94})=(21336,21336+280,21336+2\times 280)$$

It is, however, easy to find fat sequences which do not have consecutive $3-$term progressions. For example $$A=(1,2,11,12,21,22,\cdots)$$

Thus $A$ consists of $n\in \mathbb N$ such that $a\equiv 0,1\pmod {10}$. Of course this sequence contains infinitely long progressions, but the terms are not consecutive.

$\endgroup$
2
  • $\begingroup$ also to name few that are not consecutive entries $(9,18,26),(6,19,30),(7,26,42) , \cdots$ , oh boy, some times intuitive thinking could mislead you big time, i don't know how i did not see these obvious facts before. $\endgroup$ – Ahmad Sep 5 '17 at 13:38
  • 1
    $\begingroup$ Oh, this stuff is highly non-intuitive. Kind of explains the basis for making the conjecture though...the obvious ways to block progressions is to put in increasing gaps, but that method more or less guarantees convergence of the sum. Very hard to get a handle on the problem. $\endgroup$ – lulu Sep 5 '17 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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