what is this pattern in the primes?

Write the numbers from 1 to your current primorial. If that's 6, then you have

1 2 3 4 5 6

Now remove the numbers that are divisible by the factors

1 X X X 5 X

Now consider your next primorial, which is given by multiplying the current by the next prime, p.

Create p - 1 new lines by adding your current primorial to each number, successively:

1 X X X 5 X 7 X X X 11 X 13 X X X 17 X 19 X X X 23 X 25 X X X 29 X

Now, this contains all the numbers from 1 to the new primorial, with the multiples of the current primorial's factors already removed.

Since the previous primorial is coprime with the new factor p, each column that hasn't already been removed contains exactly one number that is divisible by p. We can rearrange the columns to put that number last, then remove the entirety of the last row.

1 X X X 11 X 7 X X X 17 X 13 X X X 23 X 19 X X X 29 X

So the count of numbers remaining (being the coprime count for the next primorial) is (p-1) * the count of numbers in the first row (being the coprime count for the current primorial).

Sylvia.

Hi,

Thanks, seems to make sense.

I noticed this just now too:

primorial 210 is the smallest primorial that has more coprimes than the count of primes smaller than it (48 coprimes vs 46 primes smaller than

210), then for larger primorials the two numbers converge I think at infinity.

primorial coprimes countOfPrimesLessThanPrimorial ratio

2 1 0 0 6 2 3 1.5 30 8 10 1.25 210 48 46 0.9583... 2310 480 343 0.7146... 30030 5760 3248 0.5638... 510510 92160 42331 0.4593...

If the count of primes and coprimes for a primorial converges at infinity, then the formula:

coprimes=(nextprimefactor-1)*(coprimes of next smallest primorial)

should give an accurate count of how many primes are below an infinite sized primorial as it is the same as the number of coprimes.

Also it would be interesting to find the error term between the count of primes below a primorial and the count of the primorial's coprimes for smaller primorials, to get an accurate count of primes in different ranges of numbers.

cheers, Jamie

I just checked and the sequence 0,3,10,46,343... is in oeis.org:

"Number of primes

Your ratio is (primes less than n) / (coprimes of n less than n). It is reducing, which is consistent with my intuition that the coprimes would rise faster than primes. After all, each higher primordial loses exactly one prime as a coprime, but gets a bunch of other primes as coprimes.

Given the above - not at all.

Sylvia.

Hi,

Thanks for the correction, let me restate it:

for all numbers below an infinite primorial, the number of primes are infinitely less than the number of coprimes, and the number of coprimes is infinitely less than the magnitude of the infinite primorial.

So there are infinitely many more coprimes than primes, but at the same time both primes AND coprimes approach zero density in the natural numbers up to the infinite primorial.

cheers, Jamie

Your really need to stop referring to an infinite promorial. There is no such thing. Your propositions are better expressed as

(primes less than n) / (coprimes of n less than n) --> 0, as n--> ?

and

(coprimes of n less than n) / n --> 0, as n--> ?

However, neither proposition has been proved here. I supported my intuition by means of a hand-waving argument, but that's all it was. It wasn't a proof.

Sylvia.

No that is a totally different statement and not true, since the number of coprimes less than n varies up and down depending on n, it isn't a non oscillating relationship unless n values are restricted to certain classes of numbers, ie primorials (or factorials too probably).

An Infinite primorial concept is no big deal it is just the product of all primes, I specifically stated it for primorials only and not for any n.

cheers, Jamie