fun with primorials and prime numbers

Does x range over all non-negative reals?

What's the smallest value of b for which { mx + b | x in [0,oo) } contains all the primes?

Hi,

x ranges over the positive integers (starting at zero).

Maybe your formula with x as a real would work, I hadn't thought of that, but the formula I am using has x as a positive integer, and there are multiple b values for a given m (also a positive integer).

cheers, Jamie

Those are not the positive integers; they're the non-negative integers.

Answer my question. What's the smallest value of b for which { mx + b | x non-negative integer } contains all the primes?

That isn't the correct formula, to get all the primes you need to use multiple values of b.

For the simple example of primorial 6 for m, there are two values of b required, 1 and 5.

y = 6x + 1 y = 6x + 5

for all the non negative integers x, all primes will be in y, half in 6x + 1 and half in 6x + 5

For bigger primorials m, the values of b get more interesting, and seem to just be all the primes below m, as well as numbers with prime factors larger than m's prime factors or something.

cheers, Jamie

Why? Chose m = 1, b = 0, and you'll get all primes.

On Tue, 08 Mar 2016 03:02:41 -0800, Jamie M wrote: ...

Instead of "half in 6x + 1 and half in 6x + 5" you should say "about half in 6x + 1 and about half in 6x + 5". As noted in an answer at

Are you claiming that if m is a primodial integer, then there's a subset K of [1, m-1] /\ N for which { mx + b | x non-negative integer, b in K }, will include all of the primes?

In what half are the primes 2 and 3?

Any rational that is not some integer divided by a power of ten is represented by a recurring decimal.

Sylvia.

I think the count of primes in the two formulas has to be defined as equal, if they weren't equal then how can the primes have a pseudorandom distribution?

cheers, Jamie

Hi,

Yes that is true, ie

these formulas produce all the primes except for 2,3,5:

y = 30x + 1 y = 30x + 7 y = 30x + 11 y = 30x + 13 y = 30x + 17 y = 30x + 19 y = 30x + 23 y = 30x + 29

cheers, Jamie

Not unless you allow x = 0, which is often NOT regarded as a natural number,

But then they produce every natural number not divisible by any of 2, 3 or 5,so they naturally also include every prime not divisible by any of 2, 3 or 5!

A similar list of equations can be constructed to produce all primes but the first n primes for any natural number n.

You are mixing up terms again - certainly "defined" does not come into it. And there is nothing "pseudorandom" about the distribution of primes - as far as we know, the distribution is /random/, with certain constraints. (To be pseudorandom, there would need to be a reasonable algorithm for generating the primes.)

And you are also misunderstanding what random means, and what James meant by "the difference can be arbitrarily large".

To explain, lets move to something clearly random - coin tosses. If you keep tossing a coin, and record the number of heads and the number of tails, you will not expect them to be equal - you will expect them to be /about/ equal. But you will also expect the difference between the counts to vary, and there will be no fixed limit to that difference - a limit would imply that the number of heads influenced the number of tails and vice-versa. So the difference between the counts can be arbitrarily large.

The distribution of 6x+1 primes and that of 6x+5 primes are independent in the same way. Thus you would expect roughly the same numbers of each, but there can be arbitrary differences between them.

(Terms like "about the same" can be quantified probabilistically - you can calculate that there is a 90% chance of being at least as close as one limit, 70% chance of being at least as close as another limit, and so on, and figure out a probability density function. But that is left as an exercise for the reader :-)

Hi,

I used the term pseudorandom interchangeably with random, as the distribution of the both random and pseudorandom is considered random for most purposes, and as you say the prime digits aren't exactly randomly distributed, so I hesitate to say it is a random distribution just because I don't know the pattern if one exists.

The number of primes in 6x + 5 vs 6x + 1 can be arbitrarily large, however at the same time as x approaches infinity, the difference in the number of primes in 6x + 5 vs 6x + 5 approaches zero percent, although for ranges of x there may be (arbitrarily) many more primes in

cheers, Jamie

"pseudorandom" and "random" mean very different things - they only /look/ similar. Don't use the terms interchangeably - look them up, and use them correctly.

It's okay to be ignorant about a topic - we all start out ignorant. It's easy to cure ignorance with learning and knowledge. But don't be proud or wilful about your ignorance. When you have a gap in your knowledge and someone tries to fill it, the proper reaction is "thanks for that explanation" or "that's interesting, I'll read up on some more details". You don't write "I prefer to remain ignorant and continue to make the same mistakes".

And I didn't same that "prime digits aren't exactly randomly distributed". Written like that, the sentence doesn't even make sense.

That is roughly correct, although I don't think that the difference between the two counts will "approach zero percent". It certainly will not converge to zero percent - that is contradicted by the arbitrarily large differences between them. It may be that by taking some other measure than the usual convergence, you can argue that it tends towards zero. But I don't for sure. (While I know a good deal more mathematics than most people, I am not an expert in number theory.)

Hi,

I am not proud of not knowing whether the prime numbers have a random or a pseudorandom distribution, I just don't know, but I tend to think it is a pseudorandom distribution, at least until proven otherwise, as then it is more interesting to think of some unknown formula that can define the prime number placements.

Also I don't know if the arbitrarily long sequences of primes applies to the formulas 6n+1 and 6n+5 as I think that formula is for coprimes maybe, not sure:

cheers, Jamie

What do you mean by random distribution and pseudorandom distribution?

According to Dirichlet's theorem there are infinitely many primes in each of the sequences {6n+1} and {6n+5}.

It is fine for you not to know if the primes are random or pseudorandom. That does not mean it is fine to use the terms "random" and "pseudorandom" interchangeably. Do your best to learn the terms and use them appropriately - you have no way to discuss whether they are random or pseudorandom until you get the terms correct.

And by saying "I tend to think it is a pseudorandom distribution", you are begging the question. Your investigations are trying to find out if there is a formula (other than trial and error) for generating primes - you can't start off by assuming that one exists and working from there. (You /can/ say "/if/ it exists, what would it look like?).

You won't get arbitrarily long sequences of primes of the form 6n+1 or

Hi,

Basically there is no noticeable difference between a random distribution and a pseudorandom distribution in the meaning I am using them in, except that the pseudorandom distribution has a known or unknown formula that describes the sequence.

But 6 isn't prime, that wiki page says for the formula there needs to be two primes, so 6n+1 and 6n+5 doesn't work.

cheers, Jamie

Oops I was confused about the meaning of coprime:

"two integers a and b are said to be relatively prime, mutually prime, or coprime if the only positive integer that divides both of them is 1"

cheers, Jamie