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Let $\omega(n)$ count the number of distinct prime factors of the integer $ n \geq 2$. This $\omega(n)$ is called the prime omega function.

Inspired by these ideas :

Improved sieve for primes and prime twins?

and the idea that the prime twin counting function is close to

$$ \pi_2(x) \sim x \prod_{n=2}^{\sqrt x+2} (1 - 2/p_n) $$

$$\pi_2(x)\sim 2\Pi_2 \frac{x}{\log^2(x)},$$ where $\pi_2$ denotes the number of twin primes smaller than $x\in[0,\infty)$

the following question naturally occurs :

$$\pi_2(t^2) =^? \sum_{2<j<t^2} (-2)^{\omega(j)} (1/2)(\lfloor{\frac{t^2}{j}}\rfloor +\lfloor{\frac{t^2-2}{j}}\rfloor)$$ where $j $ are squarefree odd integers.

Is this formula correct or correct upto some fixed constant addition $C$ ?

OR maybe this one :

$$\pi_2(t^2) - \pi_2(t) =^? \sum_{2<j<t^2} (-2)^{\omega(j)} (\tfrac{1}{2})(\lfloor{\tfrac{t^2}{j}}\rfloor +\lfloor{\tfrac{t^2-2}{j}}\rfloor)$$ where $j $ are squarefree odd integers.

calculatormathematical
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mick
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    How does the derivation for these go? Have you tried testing the formula with code? – EffingLoveMath May 06 '24 at 22:52
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    @DanielDonnelly no not yet – mick May 07 '24 at 00:00
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    @DanielDonnelly feel free to test it and post the results ! – mick May 08 '24 at 10:39
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    I will if I have time. I'm coding on non-mathematical solar heat system now for work. Very complex and daunting. However, when I get board, let's do it. Would you like me to teach you how to code this up very simply in SymPy? – EffingLoveMath May 08 '24 at 20:23
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    I'm also wondering how this is derived, because my formulas always needed modular arithmetic in them (they are exact formulas however for counting twin prime averages in a growing interval - see my post history), and / or CRT, which makes analyzing them difficult. If yours actually lower bound the counting function, and you could prove that; then that would be nice. How are you deriving these? Thanks. – EffingLoveMath May 08 '24 at 20:24
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    @DanielDonnelly notice " inspired by these ideas " in the OP. This is where the omega part comes from. then sieving t^2 and t^2 - 2 seems natural. The sieving is essential by all primes and there composites, so in a sense every sieve is a mod system. The power of 2 comes from the brun sieve , we sieve all primes twice. – mick May 08 '24 at 21:43
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    I believe if you sieve all primes twice, you end up over sieving if you will, and so inclusion-exclusion is irremovable, but you have that, so I'm confused. Please teach me and in exchange I will teach you two derivations of essentially the same thing. One a sum over $r^2 = 1 \pmod k$ for varying $k$ and the other a CRT splitting.

    I can also teach you how to use SymPy effectively for part of your question.

     – EffingLoveMath May 09 '24 at 08:56
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    see first page of my blog for the derivations of the formulas that I use. The blog is linked from my profile page. But that's just one side of things. I prefer the sum over the residues of $r^2 = 1 \pmod k$ because it forms a monoid and or group depending, and then the other cases of $4$-separted, $6$-separated prime averages and so on... are the same freakin formula except you have a varying parameter $l = 1$ for twin primes and $l \gt 1$ for $4, 6, 8, \dots$ and then in the higher separated cases, you sum over the cosets of the twin prime "group" $l=1$ case. That's why it's cool. – EffingLoveMath May 09 '24 at 08:58
  • No, I have not been able to relate the counts for the $l \gt 1$ cases to higher cases for some strange reason. It's almost there, but not quite no matter what I try. That's despite every case being essentially the same formula up to coset multiple. So the twin prime sums over $H_1={ x \in \Bbb{Z}/k : x^2 = 1 \pmod k}$ but the $4$-separated prime formula sums over well. We want similarly $x^2 = l^2 \pmod k$ so work with that and see if it's related to the $x^2 = 1 \pmod k$ case (twin primes). I can barely think right now. Some dude at the bar wanted to kick my ass. Let's discuss nchat – EffingLoveMath May 09 '24 at 09:00
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    @DanielDonnelly if we were to consider prime cousins, the $t^2 - 2$ would become $t^2 - 4$. Does this make sense to you ? – mick May 09 '24 at 15:25
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    yes, I'm 100% already aware of this succinctness – EffingLoveMath May 09 '24 at 17:36
  • What makes you think your formula is correct? I believe your formula is equivalent to $$f(t)=\sum\limits_{\substack{3\le j\le t^2-1 \ & \ 2, \nmid, j \ & \ \mu(j)\neq 0}} \frac{1}{2} (-2)^{\Omega(j)} \left(\left\lfloor\frac{t^2}{j}\right\rfloor+\left\lfloor\frac{t^2-2}{j}\right\rfloor\right)\tag{1}$$ where $\mu(j)$ is the Möbius function. Your formula doesn't seem to be anywhere near correct, in fact it seems to evaluate increasingly negative as $t$ increases. – Steven Clark May 31 '24 at 19:40
  • Here's a table of some sample evaluations: $$\begin{array}{cccc} t & \pi_2\left(t^2\right) & \pi_2\left(t^2\right)-\pi_2(t) & f(t) \ 1 & 0 & 0 & 0 \ 2 & 1 & 1 & -1 \ 3 & 3 & 2 & -9 \ 4 & 5 & 4 & -20 \ 5 & 7 & 5 & -34 \ 6 & 9 & 7 & -49 \ 7 & 11 & 8 & -68 \ 8 & 13 & 10 & -88 \ 9 & 15 & 12 & -106 \ 10 & 15 & 12 & -129 \ \end{array}$$ – Steven Clark May 31 '24 at 19:46
  • @StevenClark that would be strange considering how close the formula is to the idea of the brun sieve... It seems like it should be close to $t^2 (1/2) (1 - 2/3)(1 - 2/5)(1 - 2/7)...$

    Not saying you are wrong. But definitely surprising if you are correct.

     – mick Jun 22 '24 at 23:02

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