Question

A sphenic number is a number that is the product of 3 unique primes only. So 2*3*5 is the smallest sphenic number. A sphenic triple is three consecutive sphenic numbers. The first sphenic triple is:

1309, 1310, and 1311, which have factorizations 7*11*17, 2*5*131, and 3*19*23.

Are there infinitely many sphenic triples or sphenic doubles?

Answer 1

Is this an open conjecture? This sounds like of comparable difficulty to the twin prime conjecture.

Answer 2

I honestly don't know, I was just curious. For instance there aren't any sphenic quadruples since that would mean one of them is divisible by 4 which means it's not sphenic.

I guess I'm just looking for some ideas about sphenic numbers and to play around with them haha.

Answer 3

Clearly at least one of the number in a sphenic triple must be even. Since 2k and 2k+1 don't share any prime factors if I remember correctly, this conjecture is definitely plausible.

Answer 4

Oh unique prime factors are not required...

Answer 5

Yeah, since they are consecutive we also know at least one of them is divisible by 3 and the middle one has to be divisible by 2, since if the first and last one are divisible by 2 then at least one of them is also divisible by 4.

Answer 6

Unique prime factors are required to be sphenic.

Answer 7

But they are not required to be coprime right?

Answer 8

Pairwise coprime that is.

Answer 9

for clarification: \(12 =2*2*3\) is not sphenic and \(60=2^2*3*5\) is not sphenic.

Only requirement is that they are the product of 3 unique primes.

Answer 10

BUT I think it turns out that a sphenic triple has to be pairwise coprime but it's because of divisibility and being consecutive that limits them. haha.

Answer 11

Since 3 consecutive numbers can't share a factor larger than 2, there's just not enough space to make like, the first and third number divisible by 17 or something, you need at least 17 consecutive numbers to do that if you get me

Answer 12

I am really tired now and apparently I spilled out something which I don't fully understand yet is apparently true?????

Answer 13

Suppose there are two even numbers in a triple, then it must be the case that there exist two semiprimes such that the difference of the two semiprimes is 1. That seemed really impossible to me.

Answer 14

For example, can more than one of these three numbers be divisible by 5? n, n+1, and n+2

No, cause numbers that are divisible by 5 happen every 5th number, so you've gotta have at least n and n+5 on the list to do that.

Answer 15

Don't worry about it if it doesn't make any sense, if you try figuring other stuff out it'll naturally pop out at you I think eventually.

Answer 16

How did they find the first sphenic triple? That seemed like way too much work to brute force... Is there an algorithm to enumerate all sphenic triples?

Answer 17

Pigeonholing principle in essence right?

Answer 18

Sounds like it could be deep number theory stuff.

Answer 19

I have no idea, the wikipedia article doesn't really say much and I don't want to "cheat" I want to just figure out as much about sphenic numbers on my own for fun first. https://en.wikipedia.org/wiki/Sphenic_number

Answer 20

Definitely sounds like deep number theory stuff. The prime decomposition of the sphenic triple must be vastly different in size or else the difference could not be 1.

Answer 21

So here's what I've figured out, they must have the divisors in one of these three setups:

\((3, 2, \cdot)\)
\((\cdot, 6, \cdot)\)
\((\cdot, 2, 3)\)

So since there's 9 primes in total, there are 7 unknown primes left haha. In that first triple, we had the third case, \((\cdot, 2, 3)\) since we can fill in the rest of the primes: \((7*11*17,2*5*131,3*19*23)\).

Even if we can't "prove this" or whatever today I'm just trying to see how far I can get for the fun of it not really doing this with the serious goal of "getting the answer".

Some other questions as examples of stuff to look into, is there a relationship between perfect numbers, twin primes, or goldbach's conjecture to this? Does one imply the other? Is it possible to generate a sphenic triple given a sphenic double?

Answer 22

What??? How did you get that?

Answer 23

I dunno I said a lot which part seems weirdest to you? xD

Answer 24

I follow you @Kainui :P

Answer 25

Basically what I've said is at least one of the numbers in the triple is divisible by 3 and the middle one has to be divisible by 2.

Here's a proof that the first and last one aren't divisible by 2:

We'd have them be \(a=n\) and \(b=n+2\) since they are 2 apart. We are saying they're divisible by 2, so \(n=2k\).
\(a=2k\) and \(b=2k+2\) But we can factor them as:
\(a=2(k)\) and \(b=2(k+1)\)

Since k and k+1 are consecutive numbers, one of these must be divisible by 2 right? Even and odd numbers come every other number after all. But if that's true, then that means \(a\) or \(b\) which is already divisible by 2 must also be divisible by 2 again, which means it can't be sphenic since sphenics are unique primes only!

Answer 26

How do you get the middle is divisible by 6 part?

Answer 27

Well, it might be that there are no sphenic triples with the middle divisible by 6. That would be quite interesting if you could prove, but at least one of 3 consecutive numbers must be divisible by 3. So since it's already divisible by 2, there's one possibility where the middle is divisible by 3.

Maybe try to find some set of 3 consecutive numbers where none of them are divisible by 3 if you are not sure. Or maybe I don't understand your question.

Answer 28

I think maybe the notation I made up just there is throwing you off.

Answer 29

Your notation threw me off indeed but I am really tired as well.

Answer 30

I figured out quite a handful more about the specific cases and forms of the primes involved but I feel like I could probably simplify it a bit before I say it I guess if anyone's thinking about this I see two people watching lol.

Answer 31

I don't particularly think sphenic triples are interesting, but I think there's some potential to find something interesting in the unknown so that's the appeal to me right now lol. Any kinda comments or suggestions are welcome.

Answer 32

Yes, there are infinite many

Answer 33

sphenic triples and sphenic doubles

Answer 34

How do you know @daniel.ohearn1 ?

Answer 35

Because there are infinite primes

Answer 36

Is the twin prime conjecture true?

Answer 37

There are infinitely many primes and there are infinitely many sphenic numbers because of this. But there isn't necessarily infinitely many sphenic triples.

A sphenic triple only happens when 3 consecutive numbers each has 3 unique prime factors. It's a pretty particular setup, so there's no guarantee that there's infinitely many triples like this.

Similarly, there are infinitely many primes, but it's unproven if there's infinitely many twin primes, which are primes that are separated by 2. It's kind of similar to this in a way.

Answer 38

Yeah, once you get up to about Grahams Number of countable primes, we can say praise Aleph there's computers.

Answer 39

lol maybe I should write a program to phind sphenic triples

Answer 40

here are the first integers of the sphenic triples below 10k

[1309,1885,2013,2665,3729,5133,6061,6213,6305,6477,6853,6985,7257,7953,8393,8533,8785,9213,9453,9821,9877]

Answer 41

\(\tau(x) = \tau(x+1)=\tau(x+2)=8\)
and
\(\mu(x)=\mu(x+1)=\mu(x+2)=-1\)

Answer 42

Funny I was just thinking about that, also they're all relatively prime to each other

\[\tau(x(x+1)(x+2))=2^9\]

Answer 43

What's \(\tau()\) and \(\mu()\)?

Answer 44

\(\tau(n)\) is the divisor counting function, it tells you the number of divisors of n. We can write it as this where the d|n means add a term for every divisor of n.
\[\tau(n)=\sum_{d|n}1\]

So for instance, since any number to the 0 power is 1, I'll go ahead and write the divisors that way to make it clear that I'm counting all of them: \(\tau(12)=1^0+2^0+3^0+4^0+6^0+12^0 = 1+1+1+1+1+1=6\)

Yeah not super exciting yet, but it turns out this is part of a fairly large class of functions called multiplicative functions which allow you to separate them apart in a peculiar way. If the numbers are relatively prime ( \(\gcd(a,b)=1\) ), you can separate them like this:

\[\tau(a*b)=\tau(a)*\tau(b)\]

So alternatively we can calculate:
\[\tau(12)=\tau(4)*\tau(3) = (1+1+1)*(1+1)=6\]

\(\mu(n)\) is called the Mobius function and is also one of these types of multiplicative functions and it is a very strange function on its own. Really all the multiplicative functions are directly related to the Riemann Zeta Function by something called Dirichlet Series and Dirichlet convolutions, and one way to define it is as these coefficients:

\[\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s}\]

It ends up being one of the most useful ones because it allows you to do something called Mobius inversion on the Abelian group formed by the multiplicative functions, and there are some surprising and weird identities.