# Distributions of Primes

#### Jarhyn

##### Wizard
Ok, another weird thread documenting my strange journey through number theory: Exactly How many primes are there in a region of numbers?

This is a strange one and requires a visualization of the patterns that exist in the Sieve of Eratosthenes to understand. Also, I might add, I wasn't expecting this but I started getting vibes from the movie Pi.

For a given N, lets ask "how many primes are there less than".

I've graphed out some iterations of sin(x*pi/P)/n(P) By counting all primes mapped (1 of them, not counting "1"), we get a number: 3-1.

Lets estimate the number of primes, then, under 12. For that, it looks like the number is 7-1. so, 6.

That's wrong though. It's off by 1.

It breaks down once the calculation reaches the first square of the first prime that is not exactly known: 9.

Now, as you can see, I've added another line. This represents another step: reducing the error.

To reduce the error, you need to find the COUNT of cross products and power products. To do this, I start squaring primes, and for every square and cross product less than 12, I can subtract 1 from my estimation. So, 6-1 is 5. Hooray, I have corrected my error.

There are 5 primes under 12, and 4 remaining lines. For regions above 12, it appears that there are "about" 4 primes per segment.

Between 0 and 60, there are "about" 5+4*(5-1) primes. so 21. This is wrong though. It's off by a few.

To correct the error, we need to figure out a couple more primes, though that's easy because the tool we use to get the estimate has already revealed them in it's pattern because this system will always locate all primes quickly under (highest solved)^2. To know how many primes worth of error we have exactly, we need to know the last prime less than square root of 12, which is 3. There is only one of them, so our error is 1.

I can then say "primes 5 and 7, but not 11, square less than 60." This is 2, with a single cross product, so 3; My estimate is wrong by 3. 21-3 is 18. There are 18 primes under 60. I don't really know what they are. For using this system to find out how many primes are OVER 60, however, I'm going to only subtract 5 and its powers under 60, which is 2. Why? Because I haven't gotten to 420 yet, and this describes my estimator for this epoch. so while I know there are exactly 18 primes under 60, I'm going to estimate that there are 16 primes for every region over 60.

Lets estimate using this count out to 420! 18+16*(7-1) gives us roughly 114. But there's an error again
Now our corrective primes do not include 5: We have accounted for this with our "exact count under 60." To resolve the error, we need to again subtract an exact number of primes' cross products and powers. To do this we are going to subtract from our inaccurate estimate "primes greater than 5 whose cross products are less than 420". This means calculating all primes again whose squares are less than 420, and then counting their cross products less than 420. I know the primes in question are 7, 11, 13.
7*7, 11*7, 13*7, 11*11, 11*13, 13*13.

In this part of the problem it is factorial. It will not exactly remain this way: This table becomes large, as if one of the primes in this epoch were to have a square whose product with another prime in the epoch is less than the target number, that also would be an error term. It isn't yet, but it will be. To fix this, we will have to add every "power" of a prime, in addition to the primes themselves, whose value is less than the greatest prime whose square is less than the doubled primorial being observed to the list of "error terms to cross-productize".

at any rate, we have identified our error as 6. There are exactly 108 primes less than 420. Again, because I haven't gotten to 4620 yet, I'm again only going to correct it for 7. This is again because this is where the pattern repeats exactly on the sieve. There are THREE powers of 7 less than 420, so 3. so while there are exactly 108 primes less than 420, for every region OVER 420, there are going to be "about" 108-3, so 105.

Lets do 4620. There are 108 primes less than 420, and for every region of 420 numbers greater than, there are "about" 105 primes. That's going to be about 108+105*(11-1) =1158...

#### steve_bank

##### Diabetic retinopathy and poor eyesight. Typos ...

When I looked at prime numbers what stood out wasthat it looked like a 1st order differential equation impulse response. In electrical terms a low pass filter.

I thought it might be possible to dervise a cmplex S doman function to derive a frequncy response. The scale factor being Nprimes/dx.

#### Jarhyn

##### Wizard

When I looked at prime numbers what stood out wasthat it looked like a 1st order differential equation impulse response. In electrical terms a low pass filter.

I thought it might be possible to dervise a cmplex S doman function to derive a frequncy response. The scale factor being Nprimes/dx.
Now here's the kicker: I'm pretty sure I see a model of operation here somewhere which would not just count them but to also locate the primes it has "counted" this way, since lower primes are tabled out by the repetition of 2*primorial. The sequence at 2*primorial that predicts inaccurately can be made accurate by operating a filter sieve on the higher order primes separate from the lower sieve. Then the intersection of those sieves can be used to create an even larger filter on their 2*primorial base.

In this way, reducing the previous field of solutions sequentially allows fast calculation of the lower region of the progression and (slightly less onerous) calculation of the higher region of the progression.

#### steve_bank

##### Diabetic retinopathy and poor eyesight. Typos ...
I don't see where sine comes in. Sin(N*x) represents harmonics which is what you plotted.

I assume difference tables have been calculated. Does that lead anywhere?

#### Jarhyn

##### Wizard
I don't see where sine comes in. Sin(N*x) represents harmonics which is what you plotted.

I assume difference tables have been calculated. Does that lead anywhere?
Ok, so, I haven't calculated much of anything? The sin(x*pi/n) was to visualize the Sieve of Eratosthenes as a mechanical interaction, so I could see where the patterns were: it let me count the primes without having to actually look at them.

Primes are, fundamentally, harmonic in their operation.

Look at all locations where the line on sin(x*pi) cuts on either graph uniquely: they are prime. Eventually there is a harmonization where patterns of primes will repeat (with growing error).

The error is only ever extant at (lowest unpatterned prime)^2, and can be calculated by finding the powers of all primes under (2*primorial) under (2*primorial) plus all primes under (2*primorial), and finding their cross products. The error calculation in "primes under" doesn't require finding the cross products just counting how many there are, which is an easier ask.

#### Jarhyn

##### Wizard
I don't see where sine comes in. Sin(N*x) represents harmonics which is what you plotted.

I assume difference tables have been calculated. Does that lead anywhere?
Ok, so, I haven't calculated much of anything? The sin(x*pi/n) was to visualize the Sieve of Eratosthenes as a mechanical interaction, so I could see where the patterns were: it let me count the primes without having to actually look at them.

Primes are, fundamentally, harmonic in their operation.

Look at all locations where the line on sin(x*pi) cuts on either graph uniquely: they are prime. Eventually there is a harmonization where patterns of primes will repeat (with growing error).

The error is only ever extant at (lowest unpatterned prime)^2, and can be calculated by finding the powers of all primes under (2*primorial) under (2*primorial) plus all primes under (2*primorial), and finding their cross products. The error calculation in "primes under" doesn't require finding the cross products just counting how many there are, which is an easier ask.
Observe, in the 12 pattern sequence including "3" as a line, look how many unique zeroes there are. For "three given primes", there are exactly four places where only sin(x*pi) zeroes. Hence this is what is used to create the estimate, and what defines the growth of the error term, namely the exclusion of "5" when we surpass 25, it's square. As you will note the error is 0 under 25, because the sequence of 12 doubles perfectly to 24.

#### Jarhyn

##### Wizard
so, @steve_bank @Swammerdami I'm taking a look at building the error/difference table for this, but I was playing around today to see if I could gin up a function that would essentially "be zero if prime", but I could not. The best I could do today was a function that would "be not zero if prime and not counted less than the square of the first uncounted prime": (1-(cos(2pi*x/z(1)))*(1-(cos(2pi*x/z(2)))... Notably this misses on "1"...

For example, If I were to graph (1-(cos(2pi*x/2))*(1-(cos(2pi*x/3)), the zeroes of this particular function exist at all real non-prime numbers less than 5^2, and at the points of all primes less than and equal to 3. The error rate builds as per OP.

So I can coerce this thing into spewing "isn't prime" but I can't coerce it to spitting out "is prime". Interestingly this does check each number for primeness exactly once, and it can calculate WHETHER a number is prime within a margin of error determined by n by just setting x.

Essentially if (1-(cos(2pi*x/2))*(1-(cos(2pi*x/3))*...(1-(cos(2pi*x/2))*(1-(cos(2pi*x/z)) for x = 0, where z >= sqrt(z), x is not prime, and for each prime less than, if (1-(cos(2pi*x/z))=0 that number is not a factor.

This is all a fact because 1-(cos(2pi*x/2) is really just (start a roller at 0, track the height of the peg, whose full period is 2), and so this is really just a clever Sieve of Eratosthenes that perfectly builds the "lookup table" using waveforms.

Ideally, I would coerce this thing so that at all original zeroes, the value is imaginary, and at all the original non-zero points, be zero.

#### Jarhyn

##### Wizard
$$x\prod_{n=1}^{2\cdot3}\left(1-\frac{\left(\frac{\pi x}{2}\right)^{2}}{n^{2}\pi^{2}}\right)$$
I did a bit of reading and figured out in some respects how to operate the infinite product operators and some googling to find out how to express product expansions properly so as to converge on sine.

This allowed me to express a finite product which will, without error, express for a given finite region the number of primes expressed by the number of iterations of this that are themselves productized. I'm not sure how to simplify such expressions yet, so anyone who can explain this, I'm all ears.

This yields results like as follows, which will predict all primes less than 7^2. Apparently by messing with the index, you can make the function not actually nuke out primes when they are counted. Hooray.
$$x\prod_{n=2}^{\frac{7^{2}}{2}}\left(1-\frac{\left(\frac{\pi x}{2}\right)^{2}}{n^{2}\pi^{2}}\right)\cdot x\prod_{g=2}^{\frac{7^{2}}{3}}\left(1-\frac{\left(\frac{\pi x}{3}\right)^{2}}{g^{2}\pi^{2}}\right)\cdot x\prod_{b=2}^{\frac{7^{2}}{5}}\left(1-\frac{\left(\frac{\pi x}{5}\right)^{2}}{b^{2}\pi^{2}}\right)$$
So this, with each product taken to the infinite product and index = 2 will zero at all non-prime numbers.

This will also be true of the infinite product of these NOT using primes, I think? Since all non-prime numbers will invoke zeroes since it's a product on those zeroes.

I'm pretty sure this is expressed by $$x\prod_{b=2}^{\infty}\left(x\prod_{n=2}^{\infty}\left(1-\frac{\left(\frac{\pi x}{b}\right)^{2}}{n^{2}\pi^{2}}\right)\right)$$

I'm fairly certain for all finite iterations of the product, this system is nonzero at all primes (though the range of these ends up having to be bounded differently for each piece of the product), but I'm not sure if it converges to a line at zero in the infinite product, and it only goes up to (next prime squared).

If I could figure out how to coerce my "always positive cosine function version" from the previous post into an infinite product like this, I'm pretty sure it would not converge to zero?

This should simplify too, since it's an infinite product of an infinite product which should itself be... merely an infinite product.?

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#### Jarhyn

##### Wizard
After some consideration, it seems that any such infinite product must converge to zero, as the slope of a wave converges to zero at infinite wavelengths, and the system is created though an infinite product.

So while it is possible for bounded products to indicate primes, it is impossible for unbounded products to indicate them.

You can't multiply it by infinity to get it away from zero.

You can't take it's square root to get it away from zero into I even if it were "negative cosine version": the Riemann sphere places 0 at a convergence from I all the same, so it still converges to zero.

You can't multiply it by any finite amount.

You can't divide by it.

You can't square it.

It is purely and eternally guaranteed that any thing built out of the sieve of Eratosthenes to the infinite end of that process must converge to zero.

So while there are an infinitude of primes they must become infinitely difficult to compute as prime{N} approaches infinity, at least in this way.

Edit: if you could find a cosine* function which mutates a sine wave into a more progressively square wave at the top and a narrowing dive to zero at the zeroes such that it converges to a discontinuity between 0 and 1... That would do it.

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#### Jarhyn

##### Wizard
So I wanted to play with using the Fourier Expansion to create square waves that start at zero and then pass through the second, third, fourth, fifth, etc indexes. of the sine wave rather than the first.

Unfortunately, the Fourier expansion doesn't seem to want to play nicely with the first index. I might have to disjoint it or run the Fourier Expansion differently across the first iteration?

But it still converges towards less.

I expect that I could make it converge towards infinity at non-zero points? If I could do that, it would mean that I would have difference between 0 and infinity in the function, but then, 0 times "convergence to infinity" appears not-well-defined? It may result in a nonzero value there. And infinite sums and products themselves may converge to natural numbers so again, I can let go that way.

This means that the only seemingly possible solution at infinite scales requires a function whose shape (for the 2-wave) is continuous between 0 and 4, and only discontinuous after that at every even integer, and 1 and -1 at all other points.

At that point you could adjust the period on your index from 2 to infinity as your variance of infinite product and you would have a "prime selection function" for all primes and their counts.

#### Jarhyn

##### Wizard
So I did some experimentations with infinite sums of a finite integration of Heaviside's step function, as h(0) = 0.

I was able to get it to bound down to 0 and progressively move to narrowing 0's and higher values of K while still well bounded between 0 and 1.

This supplies my "2 line".

I now need to work on getting the periodicity controlled variably so that I can infinitely productize it.

Ideally I'll also make it reflect on '0'.

Strangely, the Heaviside convergent square wave is going to be generated by an infinite sum rather than infinite product.

I dunno what's going to happen when I try to simplify this monstrosity... If it can be simplified.

#### Jarhyn

##### Wizard
T

So, I managed to coerce a binary square wave out of the Heaviside Step Function that converges to a square wave without the messiness of the sine convergence.

By squaring the wave function, I'll get "narrowing instantaneous zeroes" like I wanted as K approaches infinity, and peaks approaching 1 as k approaches infinity.

Now for hitting it with an infinite product function between whole number width expansions, and I will have an infinite function that only zeroes on all nonprimes.

#### Jarhyn

##### Wizard
$$\prod_{v=1}^{g}\left(\left(1-\frac{2}{1+e^{\left(2g\left(x\right)\right)}}\right)^{2}\ \cdot\prod_{n=1}^{g}\ \left(1-\frac{2}{1+e^{\left(\left(2g\right)\left(\frac{x}{\left(v+1\right)}-\left(n+1\right)\right)\right)}}\right)^{2}\right)$$

@lpetrich @Swammerdami, either of you folks know how I can simplify this monstrosity?

This seems to have done the trick. As g approaches a limit of infinity, this reveals all primes.

This is a combination of the Heaviside Step function, as an infinite product of an infinite product. I've adjusted it to index from 1, could adjust it to index from 0, and I could probably also make it symmetrical, but I'm not going to do any of those things.

I could probably also make it skip "1", but I'm also not going to do that. It's like one extra term.

edit: I lied. I added the terms. Also realized I could move the bit out of the product and lose nothing.
$$\left(\left(1-\frac{2}{1+e^{\left(2g\left(x\right)\right)}}\right)\left(1-\frac{2}{1+e^{\left(2g\left(x-1\right)\right)}}\right)\prod_{v=1}^{g}\left(\prod_{n=1}^{g}\ \left(1-\frac{2}{1+e^{\left(\left(2g\right)\left(\frac{x}{\left(v+1\right)}-\left(n+1\right)\right)\right)}}\right)\right)\right)^{2}$$

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#### Jarhyn

##### Wizard
$$\prod_{v=1}^{g}\left(\left(1-\frac{2}{1+e^{\left(2g\left(x\right)\right)}}\right)^{2}\ \cdot\prod_{n=1}^{g}\ \left(1-\frac{2}{1+e^{\left(\left(2g\right)\left(\frac{x}{\left(v+1\right)}-\left(n+1\right)\right)\right)}}\right)^{2}\right)$$

@lpetrich @Swammerdami, either of you folks know how I can simplify this monstrosity?

This seems to have done the trick. As g approaches a limit of infinity, this reveals all primes.

This is a combination of the Heaviside Step function, as an infinite product of an infinite product. I've adjusted it to index from 1, could adjust it to index from 0, and I could probably also make it symmetrical, but I'm not going to do any of those things.

I could probably also make it skip "1", but I'm also not going to do that. It's like one extra term.

edit: I lied. I added the terms. Also realized I could move the bit out of the product and lose nothing.
$$\left(\left(1-\frac{2}{1+e^{\left(2g\left(x\right)\right)}}\right)\left(1-\frac{2}{1+e^{\left(2g\left(x-1\right)\right)}}\right)\prod_{v=1}^{g}\left(\prod_{n=1}^{g}\ \left(1-\frac{2}{1+e^{\left(\left(2g\right)\left(\frac{x}{\left(v+1\right)}-\left(n+1\right)\right)\right)}}\right)\right)\right)^{2}$$
$$J\left(x\right)=\left(\left(1-\frac{2}{1+e^{\left(2g\left(x\right)\right)}}\right)\left(1-\frac{2}{1+e^{\left(2g\left(x-1\right)\right)}}\right)\prod_{v=1}^{g}\left(\prod_{n=1}^{g}\ \left(1-\frac{2}{1+e^{\left(\left(2g\right)\left(\frac{x}{\left(v+1\right)}-\left(n+1\right)\right)\right)}}\right)\right)\right)^{2}$$

I was going to say I'm bad at this, but I'm curious if that's just the imposter syndrome talking at this point.

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#### steve_bank

##### Diabetic retinopathy and poor eyesight. Typos ...
I have no idea what yiuare trying to get at but...

A step function by definition is a tertcal instantaneous changee from one state to another, and it does not exist in physical reality. A zero time step requires and infinite number of terms in a Fourier series, or infinite energy.

A uni step is not a square or rectangular wave

There is a proof that guarantees for any function there is one ad only one Fourier series and vice versa. So if you are trying to construct a square wave for specific zeros in a sine wave there is only one set of sine waves for a square wave.

A square wave is a special case of rectangular;ar waves. By varying the duty cycle of a rectangular wave you can vary the Fourier series frequency content. A square wave is a rectangular wave with a 50% duty cycle.

Something from my old archive you can play with creating waveform. The technique in synthesis is to create a function f(x) and take the Fourier Transform, or to create a record of the desired frequencies and phases and taking the inverse transform yielding the function f(x).. It is almost never done by direct integration due to complexity. It is done digitally by the FFT.

So if you try to fit a sine series to a square wave it is what is, there is only one series for a square wave.

Scilab script
clear
tmax = 1 // time seconds

_length = 10000 // time record legth
dt = tmax/_length
// harmonics
// n _even_odd
// 1 1 all harmonics ramp
// 1 2 odd harmonics square wave
// 2 2 e ven harmonics sawtooh
n = 2
even_odd = 2
_terms = 50
t = 0

for i = 1:_length y(i) = 0;end;
for i = 1:_terms
for j = 1:_length
x(j) = sin(2*%pi*n*t)/n
t = t + dt
end
for k = 1:_length y(k) = y(k) + x(k);end;
n = n + even_odd
end

a = scf(1)
clf(a)
plot

#### Jarhyn

##### Wizard
I have no idea what yiuare trying to get at but...

A step function by definition is a tertcal instantaneous changee from one state to another, and it does not exist in physical reality. A zero time step requires and infinite number of terms in a Fourier series, or infinite energy.

...
Read the wiki article on the Heaviside step function. https://en.wikipedia.org/wiki/Heaviside_step_function

The idea is to take a function that has a roughly "sigmoid" behavior but which approaches "squareness" with a single point at infinite slope at a "middle value". Which for Heaviside is at 1/2, so I multiply it by two and subtract that from 1, so that it wiggles between 1 and -1 and transitions in each wave at 0. As k, the "squareness" value increases, it goes towards narrowing width, but is still just 0 there.

I manipulated this in such a way to, as lim(k->infinity), productizes infinitely to hold this behavior, and then take its square so that the value is always positive. I realize I didn't need to set the index targets for the infinite sum at the limit of k, specifically; I could have just indicated infinity. But I wanted only a single dependent "infinity" to operate towards.

In some ways this does indicate "infinite energy", because one of the terms in the product (several actually) is a limit as k approaches infinity.

The whole point was to have something that gets insane in a very particular way at infinite values.

#### Jarhyn

##### Wizard
It's interesting that there's this behavior at 1/2 in the base function I pulled this out of. It is interesting because it creates a line that makes the zeroes of the function really hard to pull out. There's a symmetry of all points around that line in the infinite sum that generates it as K approaches infinity. Is this "the critical line"? It would explain why the zeroes have to be symmetrical up and down, left and right: Because the slope is infinite across that region.

#### Jarhyn

##### Wizard
$$lim_{k\to\infty}J\left(x\right)=\left(\prod_{v=1}^{k}\left(\prod_{n=1}^{k}\ \left(1-\frac{2}{1+e^{\left(\left(2k\right)\left(\frac{x}{\left(v+1\right)}-\left(n+1\right)\right)\right)}}\right)\right)\right)^{2}$$

This is the main equation that I really want to simplify, but I'm not really sure that's even possible. Note the 1-2/... which is modifying the natural centers off of 1/2.

#### Jarhyn

##### Wizard
Derp, forgot, -k, not like it matters for this, but some crazy shit happens in the nonconvergence of the unsquared result wave otherwise.

$$\left(\prod_{v=0}^{k}\left(\prod_{n=0}^{k}\ \left(1-\frac{2}{1+e^{\left(\left(-2k\right)\left(\frac{x}{\left(v+2\right)}-\left(n+2\right)\right)\right)}}\right)\right)\right)^{2}$$

#### Jarhyn

##### Wizard
@steve_bank So, long story short, it turns out this gave me a solution to Riemann's and P=NP.

I'm willing to discuss why I'm wrong but I'm also only willing to do so if you're willing to consider my discussion that argues that you are wrong that I am wrong.

Double. Fucking. Factors... Also, a new square wave function!

#### steve_bank

##### Diabetic retinopathy and poor eyesight. Typos ...
Eureka!!! WE have been eagerly awaiting a new square wave function.

I have no idea what you are doing. I can tell you the utility of the Heaviside step function, the unit step, has wide mathematical application in electric systems. The unit step is mathematically convoluted with a system transfer function to get the step response which is important for several reasons. An electrical step signal is widely used to test physical systems.

The other two major matmatical stimuli are the sine wave and the unit impulse.

A different approach for you would be to look at it as a digital simulation instead of closed direct mathematics.

A sqare wave becomes an array of ones and zeros. Point by point the digital square wave is input to a digital model of the distribution of primes.

Square waves or more generally rectangular waves usualy in reduce to a form of Fourier series.

Another thing to consider is the dc average of the square wave which may cause problems in your math. The average value of a square wave from 0 to 1 is 0.5. From -1 to 1 the average is 0.

You can try a search on La Place transform pair square wave. You might find a differential equation.

#### steve_bank

##### Diabetic retinopathy and poor eyesight. Typos ...
Exponentials can be used to create the rise and fall edges of a square wave if you wnat a continuous function.

Vary tau to adjust the rate of chnage of the edges. Vary k for amplitude.

n = 1000
tmax = 1.
dt = tmax/n
t = 0:dt:tmax
k = 1.
tau = .01
for i = 1:length(t)
yon(i) = 1. - k*%e^(-t(i)/tau)
yoff(i) = k*%e^(-t(i)/tau)
end

plot2d(t,yon)

##### Member
Off topic...

The square of every prime (p>4) is a multiple of 24, plus 1

p2 = 24k +1

#### Jarhyn

##### Wizard
Off topic...

The square of every prime (p>4) is a multiple of 24, plus 1

p2 = 24k +1
Actually not off topic.

24 is related to a primorial by a factor of 4.

Is there a point as well where this becomes true of other multiples of primorials, on some other range? I'm curious of the relationship. And whether you can link me to a proof.

And now I want to take a look at the relationships between lower primorial multiples and their neighbors at p<=4.

##### Member
Off topic...

The square of every prime (p>4) is a multiple of 24, plus 1

p2 = 24k +1
Actually not off topic.

24 is related to a primorial by a factor of 4.

Is there a point as well where this becomes true of other multiples of primorials, on some other range? I'm curious of the relationship. And whether you can link me to a proof.

And now I want to take a look at the relationships between lower primorial multiples and their neighbors at p<=4.

p2 = 24k + 1
p2 - 1 = 24k
(p+1)(p-1) = 24k

note: 24 = 2*3*4
since (p-1), p, (p+1) are consecutive numbers

p cannot be a multiple of 2,3 or 4

But (p-1) and (p+1) have to a factor of 2

And (p-1) and/or (p+1) have to have factors of 3 and 4

Therefor (p-1)(p+1) is a multiple of 24

(p-1)(p+1) = 24k

p^2 = 24k +1

#### Jarhyn

##### Wizard
Off topic...

The square of every prime (p>4) is a multiple of 24, plus 1

p2 = 24k +1
Actually not off topic.

24 is related to a primorial by a factor of 4.

Is there a point as well where this becomes true of other multiples of primorials, on some other range? I'm curious of the relationship. And whether you can link me to a proof.

And now I want to take a look at the relationships between lower primorial multiples and their neighbors at p<=4.

p2 = 24k + 1
p2 - 1 = 24k
(p+1)(p-1) = 24k

note: 24 = 2*3*4
since (p-1), p, (p+1) are consecutive numbers

p cannot be a multiple of 2,3 or 4

But (p-1) and (p+1) have to a factor of 2

And (p-1) and/or (p+1) have to have factors of 3 and 4

Therefor (p-1)(p+1) is a multiple of 24

(p-1)(p+1) = 24k

p^2 = 24k +1

I don't have the time or inclination to sort out your post for logic errors given the fact it is riddle with typos.

#### Jarhyn

##### Wizard
Off topic...

The square of every prime (p>4) is a multiple of 24, plus 1

p2 = 24k +1
Actually not off topic.

24 is related to a primorial by a factor of 4.

Is there a point as well where this becomes true of other multiples of primorials, on some other range? I'm curious of the relationship. And whether you can link me to a proof.

And now I want to take a look at the relationships between lower primorial multiples and their neighbors at p<=4.

p2 = 24k + 1
p2 - 1 = 24k
(p+1)(p-1) = 24k

note: 24 = 2*3*4
since (p-1), p, (p+1) are consecutive numbers

p cannot be a multiple of 2,3 or 4

But (p-1) and (p+1) have to a factor of 2

And (p-1) and/or (p+1) have to have factors of 3 and 4

Therefor (p-1)(p+1) is a multiple of 24

(p-1)(p+1) = 24k

p^2 = 24k +1

I don't have the time or inclination to sort out your post for logic errors given the fact it is riddle with typos.
...
It's not an unreasonable request for an established proof on a site or resource devoted to such, given this is a claim I have not seen. I would like to see more indepth discussion on it.

##### Member
Off topic...

The square of every prime (p>4) is a multiple of 24, plus 1

p2 = 24k +1
Actually not off topic.

24 is related to a primorial by a factor of 4.

Is there a point as well where this becomes true of other multiples of primorials, on some other range? I'm curious of the relationship. And whether you can link me to a proof.

And now I want to take a look at the relationships between lower primorial multiples and their neighbors at p<=4.

p2 = 24k + 1
p2 - 1 = 24k
(p+1)(p-1) = 24k

note: 24 = 2*3*4
since (p-1), p, (p+1) are consecutive numbers

p cannot be a multiple of 2,3 or 4

But (p-1) and (p+1) have to a factor of 2

And (p-1) and/or (p+1) have to have factors of 3 and 4

Therefor (p-1)(p+1) is a multiple of 24

(p-1)(p+1) = 24k

p^2 = 24k +1

I don't have the time or inclination to sort out your post for logic errors given the fact it is riddle with typos.
...
It's not an unreasonable request for an established proof on a site or resource devoted to such, given this is a claim I have not seen. I would like to see more indepth discussion on it.
This is elementary. Learn to talk to your peers. I have no interest in ever interacting with you again.

#### Jokodo

##### Veteran Member
Off topic...

The square of every prime (p>4) is a multiple of 24, plus 1

p2 = 24k +1
Actually not off topic.

24 is related to a primorial by a factor of 4.

Is there a point as well where this becomes true of other multiples of primorials, on some other range? I'm curious of the relationship. And whether you can link me to a proof.

And now I want to take a look at the relationships between lower primorial multiples and their neighbors at p<=4.

p2 = 24k + 1
p2 - 1 = 24k
(p+1)(p-1) = 24k

note: 24 = 2*3*4
since (p-1), p, (p+1) are consecutive numbers

p cannot be a multiple of 2,3 or 4

But (p-1) and (p+1) have to a factor of 2

And (p-1) and/or (p+1) have to have factors of 3 and 4

Therefor (p-1)(p+1) is a multiple of 24

(p-1)(p+1) = 24k

p^2 = 24k +1

I don't have the time or inclination to sort out your post for logic errors given the fact it is riddle with typos.
That was a proof.

Among any three consecutive integers, such as p-1, p, p+1, one must be a multiple of 3; for p>3, the choice is reduced to p-1 and p+1 since p, by definition is not a multiple of 3.

Among any two consecutive even integers, such as p-1 and p+1 for any p>2, one must be a multiple of 4, while both are, by definition, multiples of 2.

So (p-1) * (p+1), for p>3, will always be the multiplication of a multiple of 2 with a multiple of 4, with one of them also being a multiple of 3. The product is thus both a multiple of 8 and a multiple of 3. That means it has to be a multiple of 24.

Since (p-1)*(p+1) == p² - 1, or equivalently p² = (p-1)*(p+1) + 1, it follows that p² for p>3 is a multiple of 24, plus 1.

#### Jokodo

##### Veteran Member
Alternatively, you just base your proof on the fact that any prime>3 is representable as 6k+/-1, in addition to the simple fact that odd +/-/* odd = even, as does even +/-/* even

for odd k and p=6k-1: (6k - 1)² = 36k² - 12k + 1 = 36 times an odd number minus 12 times an odd number = 12 times an odd number minus 12 times (another) odd number = 12 times some even number = 24 times an integer, plus one

for even k and p=6k-1: (6k - 1)² = 36k² - 12k + 1 = 36 times an even number minus 12 times an (other) even number = 12 times an eveb number minus 12 times (another) even number = 12 times some even number = 24 times an integer, plus one

for odd k and p=6k+1: 36k² +12k + 1 = 36 times an odd number plus 12 times an odd number = 12 times an odd number plus 12 times (another) odd number = 12 times some even number = 24 times an integer, plus one

for even k and p=6k+1: (6k - 1)² = 36k² + 12k + 1 = 36 times an even number plus 12 times an (other) even number = 12 times an even number plus 12 times (another) even number = 12 times some even number = 24 times an integer, plus one

That exhausts the possibilities

#### Jokodo

##### Veteran Member
Of course, odd*odd=odd. The exposition makes clear that's what I was using.

#### Jarhyn

##### Wizard

Of course, odd*odd=odd. The exposition makes clear that's what I was using.
Well, it's not a link but it does provide the actual discussion I was looking for in clear terms, so thank you.

At some point I looked it up myself (the origin of the row being that I couldn't because I was on mobile and very busy and would be for much of the day), and asked the question "well, what's the deal with K, how does it grow?"

I interested to see that when I looked up the progression of K that they were in fact a progression of pentagonal numbers.

My real question is what is it about k=22,... That makes these pentagonal numbers lie about whether their result in k^2/24+1/24=**** that makes this number nonprime?

I'm going to look into the relationship of these "liars" and why they happen later.

I also found a paper from 2019 about a proposed relationship between pentagonal numbers and equal partitions? But it'll take me a few weeks to get enough understanding built up to see where it's going or if it ever actually makes it there.

****This is almost certainly written out wrong, since again I'm on mobile and can't reference any notes and my memory is shit.

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#### Jokodo

##### Veteran Member

Of course, odd*odd=odd. The exposition makes clear that's what I was using.
Well, it's not a link but it does provide the actual discussion I was looking for in clear terms, so thank you.

At some point I looked it up myself (the origin of the row being that I couldn't because I was on mobile and very busy and would be for much of the day), and asked the question "well, what's the deal with K, how does it grow?"

I interested to see that when I looked up the progression of K that they were in fact a progression of pentagonal numbers.

My real question is what is it about k=22,... That makes these pentagonal numbers lie about whether their result in k^2/24+1/24=**** that makes this number nonprime?

I'm going to look into the relationship of these "liars" and why they happen later.

I also found a paper from 2019 about a proposed relationship between pentagonal numbers and equal partitions? But it'll take me a few weeks to get enough understanding built up to see where it's going or if it ever actually makes it there.

****This is almost certainly written out wrong, since again I'm on mobile and can't reference any notes and my memory is shit.
My first proof is literally Adam's slightly reworded...

#### Jarhyn

##### Wizard

Of course, odd*odd=odd. The exposition makes clear that's what I was using.
Well, it's not a link but it does provide the actual discussion I was looking for in clear terms, so thank you.

At some point I looked it up myself (the origin of the row being that I couldn't because I was on mobile and very busy and would be for much of the day), and asked the question "well, what's the deal with K, how does it grow?"

I interested to see that when I looked up the progression of K that they were in fact a progression of pentagonal numbers.

My real question is what is it about k=22,... That makes these pentagonal numbers lie about whether their result in k^2/24+1/24=**** that makes this number nonprime?

I'm going to look into the relationship of these "liars" and why they happen later.

I also found a paper from 2019 about a proposed relationship between pentagonal numbers and equal partitions? But it'll take me a few weeks to get enough understanding built up to see where it's going or if it ever actually makes it there.

****This is almost certainly written out wrong, since again I'm on mobile and can't reference any notes and my memory is shit.
My first proof is literally Adam's slightly reworded...
With discussion, and some other posts besides, and enough attention to grammar to make me confident that what I was reading was not somehow incomplete.

The second post you posted was more what I was looking for as an extension from 6k+/-1.

As to the pentagonal nature of K among the primes, I don't suppose you could connect me with some information about results such as k=51 which result in a root of 25?

But then how do the pentagonal numbers interrelate with the first two primes, 2 and 3? It is this just the fact that the first pentagon is partitioned exactly to the primes 2 and 3?

#### lpetrich

##### Contributor
- 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.

#### Jarhyn

##### Wizard
- 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.

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#### lpetrich

##### Contributor
So 24*pntg(k)+1 is the square of a prime? Where pntg(k) = k*(3k-1)/2

In general, 24*pntg(k) + 1 = (6k-1)2

So the aforementioned condition is equivalent to 6k-1 being a prime.
• 6k-1: 5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95, 101, 107, 113, 119, 125, 131, 137, 143, 149, 155, 161, 167, 173, 179, 185, 191, 197, 203, 209, 215, 221, 227, 233, 239, 245, 251, 257, 263, 269, 275, 281, 287, 293, 299, 305, 311, 317, 323, 329, 335, 341, 347, 353, 359, 365, 371, 377, 383, 389, 395, 401, 407, 413, 419, 425, 431, 437, 443, 449, 455, 461, 467, 473, 479, 485, 491, 497, 503, 509, 515, 521, 527, 533, 539, 545, 551, 557, 563, 569, 575, 581, 587, 593, 599
• 6k+1: 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109, 115, 121, 127, 133, 139, 145, 151, 157, 163, 169, 175, 181, 187, 193, 199, 205, 211, 217, 223, 229, 235, 241, 247, 253, 259, 265, 271, 277, 283, 289, 295, 301, 307, 313, 319, 325, 331, 337, 343, 349, 355, 361, 367, 373, 379, 385, 391, 397, 403, 409, 415, 421, 427, 433, 439, 445, 451, 457, 463, 469, 475, 481, 487, 493, 499, 505, 511, 517, 523, 529, 535, 541, 547, 553, 559, 565, 571, 577, 583, 589, 595, 601
where I've bolded the prime numbers in the lists.

#### Jarhyn

##### Wizard
So 24*pntg(k)+1 is the square of a prime? Where pntg(k) = k*(3k-1)/2

In general, 24*pntg(k) + 1 = (6k-1)2

So the aforementioned condition is equivalent to 6k-1 being a prime.
• 6k-1: 5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95, 101, 107, 113, 119, 125, 131, 137, 143, 149, 155, 161, 167, 173, 179, 185, 191, 197, 203, 209, 215, 221, 227, 233, 239, 245, 251, 257, 263, 269, 275, 281, 287, 293, 299, 305, 311, 317, 323, 329, 335, 341, 347, 353, 359, 365, 371, 377, 383, 389, 395, 401, 407, 413, 419, 425, 431, 437, 443, 449, 455, 461, 467, 473, 479, 485, 491, 497, 503, 509, 515, 521, 527, 533, 539, 545, 551, 557, 563, 569, 575, 581, 587, 593, 599
• 6k+1: 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109, 115, 121, 127, 133, 139, 145, 151, 157, 163, 169, 175, 181, 187, 193, 199, 205, 211, 217, 223, 229, 235, 241, 247, 253, 259, 265, 271, 277, 283, 289, 295, 301, 307, 313, 319, 325, 331, 337, 343, 349, 355, 361, 367, 373, 379, 385, 391, 397, 403, 409, 415, 421, 427, 433, 439, 445, 451, 457, 463, 469, 475, 481, 487, 493, 499, 505, 511, 517, 523, 529, 535, 541, 547, 553, 559, 565, 571, 577, 583, 589, 595, 601
where I've bolded the prime numbers in the lists.
I can see the errors evolving as interactions od the primes not a factor of 24.

I wonder if there is a way to evolve the restriction of K to select out multiples of 5, as a specific subset of the pentagonals.

Edit: oh, I see it now: (6k-1)^2 isn't going to incorporate the factors outside of it's primorial component 6. You need to use a bigger primorial somehow, there, to weed out higher nonprimes at the expense of missing lower ones.

#### lpetrich

##### Contributor
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.

#### Jarhyn

##### Wizard
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.

#### Jokodo

##### Veteran Member
- 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

##### Wizard
- 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

##### Contributor
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}

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

##### Wizard
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}

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

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

#### Jarhyn

##### Wizard
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

##### Contributor
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

##### Wizard
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.