Distributions of Primes

[Jarhyn]

Try k = 5k' + {0, 1, 2, 3, 4} -- one gets 30k' + {-1, 1, 7, 11, 13, 17, 19, 23}

One can normalize to get within range 0 to 30: {1, 7, 11, 13, 17, 19, 23, 29}

For 6k +-1, one gets {1, 5}

For 2k + 1, one gets {1}

If one tries this approach, one will get composite numbers divisible by 7, like 49. So one will have to repeat this construction with 7: 7*(30k+{}) ...

After that, 11, 13, 17, 19, ad infinitum.

I'm not sure I follow your language here.

 

[Jokodo]

 Pentagonal number - A000326 - OEIS - Pentagonal numbers: a(n) = n*(3*n-1)/2. - 0, 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, 247, 287, 330, 376, 425, 477, 532, 590, 651, 715, 782, 852, 925, 1001, 1080, 1162, 1247, 1335, 1426, 1520, 1617, 1717, 1820, 1926, 2035, 2147, 2262, 2380, 2501, 2625, 2752, 2882, 3015, 315

I don't see what they have to do with primes, because only one of them is a prime: 5.

Look at the factors K in p^2=24k+1 and compare the list to the series of pentagonals.

My first question, the very first on seeing Adam's post was "how does K evolve on p" and I got the first few numbers and googled them. It steps off at 25^2.

Edit: it's just that some numbers are liars about whether they lead to a prime square or not. Maybe the squares of prime squares and their squares? Again, I'll probably look closer when I get home.

It doesn't "step off". As the proof with 6k+/-1 shows, this is a property of any number that doesn't have 2 or 3 as a prime factor, which of course includes all prime (>3), but also 25, 35, 49, 55, 65, 77..., I.e numbers derived by multiplying any of those.

 

[Jarhyn]

 Pentagonal number - A000326 - OEIS - Pentagonal numbers: a(n) = n*(3*n-1)/2. - 0, 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, 247, 287, 330, 376, 425, 477, 532, 590, 651, 715, 782, 852, 925, 1001, 1080, 1162, 1247, 1335, 1426, 1520, 1617, 1717, 1820, 1926, 2035, 2147, 2262, 2380, 2501, 2625, 2752, 2882, 3015, 315

I don't see what they have to do with primes, because only one of them is a prime: 5.

Look at the factors K in p^2=24k+1 and compare the list to the series of pentagonals.

My first question, the very first on seeing Adam's post was "how does K evolve on p" and I got the first few numbers and googled them. It steps off at 25^2.

Edit: it's just that some numbers are liars about whether they lead to a prime square or not. Maybe the squares of prime squares and their squares? Again, I'll probably look closer when I get home.

It doesn't "step off". As the proof with 6k+/-1 shows, this is a property of any number that doesn't have 2 or 3 as a prime factor, which of course includes all prime (>3), but also 25, 35, 49, 55, 65, 77..., I.e numbers derived by multiplying any of those.

"Step off" is indicative of that. Sorry you don't like the way it's stated? @lpetrich understood me just fine, even if I'm having an issue understanding him back entirely.

 

[lpetrich]

I'll be more systematic.

We know that no prime after 2 is divisible by 2, so we should look in odd numbers:
2k + 1

We now try excluding everything divisible by 3:
6k + {1, 5}
Number of add-on numbers: (2)

Now everything divisible by 5:
30k + {1, 7, 11, 13, 17, 19, 23, 29}
(8)

Now everything divisible by 7:
210k + {1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 209}
(48)

Some of these add-on numbers are composite: 121=11², 143=11*13, 169=13², 187=11*17, 209=11*19

I recall an Isaac Asimov science essay from long ago where he proposed that numbers of form 6k+1 and 6k+5 would be good for studying the distribution of primes, because they filter out many composite numbers without being overly complicated.

 

[Jarhyn]

I'll be more systematic.

We know that no prime after 2 is divisible by 2, so we should look in odd numbers:
2k + 1

We now try excluding everything divisible by 3:
6k + {1, 5}
Number of add-on numbers: (2)

Now everything divisible by 5:
30k + {1, 7, 11, 13, 17, 19, 23, 29}
(8)

Now everything divisible by 7:
210k + {1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 209}
(48)

Some of these add-on numbers are composite: 121=11², 143=11*13, 169=13², 187=11*17, 209=11*19

I recall an Isaac Asimov science essay from long ago where he proposed that numbers of form 6k+1 and 6k+5 would be good for studying the distribution of primes, because they filter out many composite numbers without being overly complicated.

It's the set notation on addition that's throwing me, I guess? I don't see what's going to be done with the generated polynomials, exactly.

I expect the composite numbers you are seeing are exactly the numbers whose squares, powers, and cross products are less than (I'm not sure if it's going to be primorial or 2*primorial? Some comstant against primorial. Maybe next primorial?

It's a sieve of Eratosthenes.

 

[lpetrich]

I've written 6k + {1, 5} as a shorthand for
6k + 1
6k + 5

 

[Jarhyn]

I've written 6k + {1, 5} as a shorthand for
6k + 1
6k + 5

I figured out as much but I'm still thinking of the form "p^2=24k+1", so trying to see a single polynomial expansion in this and failing miserably. Thankfully I get the shape of the numbers this process produces, fundamentally a sieve of Eratosthenes losing accuracy at numbers whose squares and crosses are less than some value determined by the sieve's error table but I'm just not seeing "the full shape of the machine driving out primes".

Essentially, I can't quite see the polynomial expansion that "pentagonal" is getting filtered on so as to filter it's errors.

Ideally I'd like to understand this so as to express this as an infinite series or product.

 

[lpetrich]

I'll derive it. For 24*n+1 where n is a pentagonal number, then by definition, n = k*(3*k-1)/2

That gives us 24*k*(3*k-1)/2 + 1 = 12*k*(3*k-1) + 1 = 36*k² + 12*k + 1 = (6*k + 1)²

Then, p² = (6*k+1)² gives us p = 6*k+1

 

[Jarhyn]

I'll derive it. For 24*n+1 where n is a pentagonal number, then by definition, n = k*(3*k-1)/2

That gives us 24*k*(3*k-1)/2 + 1 = 12*k*(3*k-1) + 1 = 36*k² + 12*k + 1 = (6*k + 1)²

Then, p² = (6*k+1)² gives us p = 6*k+1

Oh, I got that part, I'm just trying to understand the evolution to constraining the subset of pentagonals.

Essentially, we've defined an n(k) such that n(k) is pentagonal(k) that constrains n.

I'm trying to ascertain what the form of the result is when n1(k) where n1(k) misses composites of only numbers 7 and greater as opposed to the original n(k) which misses composites of 5 and greater.