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I did a little test today looking for all the numbers such as their even divisors are exactly all of them a prime number minus 1, to verify possible properties of them. These are the first terms, it is not included at OEIS:

2, [2]

4, [2, 4]

6, [2, 6]

10, [2, 10]

12, [2, 4, 6, 12]

18, [2, 6, 18]

22, [2, 22]

30, [2, 6, 10, 30]

36, [2, 4, 6, 12, 18, 36]

46, [2, 46]

58, [2, 58]

I tried to look for the one with the longest list of even divisors, but it seems that the longest one is $36$, at least up to $10^6$:

$36$, even divisors $[2, 4, 6, 12, 18, 36]$, so the primes are $[3, 5, 7, 13, 19, 37]$.

For instance, for the same exercise for the even divisors being exactly all of them a prime number plus 1 (except $1$ in the case of the even divisor $2$) it seems to be $24$

$24$, $[2, 4, 6, 8, 12, 24]$, so the primes are $[3, 5, 7, 11, 23]$.

And for instance for the case in which both minus and plus one are a prime (or $1$ for the even divisor $2$) the longest one seems to be $12$: $[2, 4, 6, 12]$.

I would like to ask the following question:

These are heuristics, but I do not understand why it seems impossible to find a greater number than those small values such as all the even divisors comply with the property and that list of divisors is longer than the list of $36$. Is there a theoretical reason behind that or should it be possible to find a greater number (maybe very big) complying with the property? The way of calculating such possibility is related somehow with Diophantine equations?

Probably the reason is very simple, but I can not see it clearly. Thank you very much in advance!

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3 Answers 3

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Sieving with small primes reveals the following.

Assume that $$ n=2^a\cdot3^bp_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}\qquad(*) $$ has the property that $d+1$ is a prime whenever $d$ is an even factor of $n$. Here $(*)$ gives the prime factorization of $n$, so $p_i$ are all distinct primes $>3$. Without loss of generality we can assume that $a>0$ and that $a_i>0$ for all $i$.

If any of the primes $p_i$ satisfies the congruence $p_i\equiv1\pmod 3$, then $2p_i+1$ is divisible by three, and $d=2p_i\mid n$ is in violation of the assumption. We can conclude that $p_i\equiv-1\pmod 3$ for all $i$.

If $k\ge 2$ then $2p_1p_2+1$ is divisible by three, so $d=2p_1p_2$ is in violation. Therefore $k\le1$. But also if $a_1\ge2$, then $2p_1^2+1$ is divisible by three and $d=2p_1^2$ is a violating factor.

At this point we know that either (call this case A) $$ n=2^a\cdot3^bp $$ for some prime $p\equiv-1\pmod3$, or (call this case B) $$ n=2^a3^b. $$

In case A we make the following further observations. First we see that $4p+1$ is again divisible by three, so $d=4p$ is in violation. Therefore in case A we must have $a=1$. Also, $2\cdot 3^3+1$ is a multiple of five, so we similarly conclude that $b\le 2$.

In case B we observe that $2^3+1$ and $2\cdot 3^3+1$ are no primes, and therefore $a\le2$ and $b\le2$.

So we are left with the possibilities

  • $n=2\cdot 3^b p$ with $b\le2$. This $n$ has $2(b+1)$ even factors, so there are at most six of them.
  • $n=2^a\cdot 3^b$ with $a,b\le2$. This $n$ has $a(b+1)$ even factors, so again there are at most six of them.
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  • $\begingroup$ very appreciated! thank you very much for this kind explanation. $\endgroup$
    – iadvd
    Jan 4, 2016 at 12:30
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Suppose that $n$ has the property that $d+1$ is prime for each even divisor $d$ of $n$, and let $k$ be the number of powers of $2$ that divide $n$.

  • If $k=0$, there is nothing to say.

  • $k\geq 3$ is impossible, since $2^3 + 1$ is not prime.

  • Suppose that $k=2$, and let $p$ be an odd prime divisor of $n$. Then $2p+1$ and $4p+1$ are prime. But working modulo $3$, we can see that this is only possible if $p=3$. So $n=4\cdot 3^l$ for some $l\geq 0$. But $2\cdot 3^3 + 1$ is not prime, so the only possibilities are $n=4,12,36$.

The remaining case is $k=1$. If $p\geq 5$ is a prime factor of $n$, then $p^2$ does not divide $n$, because $2p^2+1$ is divisible by $3$. As above, $3^3$ cannot divide $n$, so $n=2\cdot 3^l\cdot \prod_{i=1}^t q_i$, where $0\leq l \leq 2$ and the $q_i$ are distinct primes $\geq 5$.

  • If $t=0$, we have the cases $n=2,6,18$.

  • If $t=1$, we have the cases $n=2p, 6p, 18p$, none of which have more than six even factors.

  • $t\geq 2$ is impossible: at least one of $2p+1$, $2q+1$, and $2pq+1$ must be divisible by $3$.

We conclude that $n$ cannot have more than six even factors, and it has precisely six only when $n=18p$ for some prime $p$.

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    $\begingroup$ @Slade thank you very much for your time! It is easy to follow, answers like this are a good reason to love MSE. $\endgroup$
    – iadvd
    Jan 4, 2016 at 12:29
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    $\begingroup$ @Slade: I checked $n = 18p$ for the first $10^5$ primes and we have $p = 2, 11, 1451, 1931, 2741, 3761,\dots$ Is it true that for $p>2$, all have form $10m+1$? (It was for the range that I checked.) $\endgroup$ Jan 7, 2016 at 5:12
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    $\begingroup$ @TitoPiezasIII Yes, it's true. If $p\equiv 2\pmod{5}$, then $5$ divides $2p+1$. If $p\equiv 3\pmod{5}$, then $5$ divides $18p+1$. If $p\equiv 4\pmod{5}$, then $5$ divides $6p+1$. So if $p\neq 2,5$, we must have $p\equiv 1\pmod{5}$ and $p\equiv 1\pmod{2}$, so $p\equiv 1\pmod{10}$. $\endgroup$ Jan 7, 2016 at 5:29
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    $\begingroup$ @TitoPiezasIII We also have $p\equiv 2\pmod{3}$ for all such primes, otherwise $p=3$ or $3$ divides $2p+1$. So $p\equiv 11\pmod{30}$. I believe that this should be the best we can say, since our conditions can only eliminate three residue classes for higher moduli. $\endgroup$ Jan 7, 2016 at 5:32
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    $\begingroup$ @Slade: Ah, so $p \equiv 11(mod\, 30)$. And I was just going to ask was else these $p$ have in common. :) There's quite a lot of them. For the first $10^5$ primes, there's $141$ that qualify. And they're not in the OEIS. $\endgroup$ Jan 7, 2016 at 5:37
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I made a program that calculates numbers, that you are searching, but only found $$198,26118,347\color{brown}58,49338,67698,79038,109818,...$$ which has only $6$ '$prime-1$' divisors like $36$. The range was $10^8$.

It shows that a number with even, $7$ '$prime-1$' divisors like these does not exist or they are very big to calculate...

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  • $\begingroup$ thanks for the review and confirmation! $\endgroup$
    – iadvd
    Jan 4, 2016 at 13:32
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    $\begingroup$ I found the typo. The digit should have been a $5$. :) $\endgroup$ Jan 7, 2016 at 5:01

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