Prime twin counting by $\pi_2(t^2) =^? \sum_{2<j<t^2} (-2)^{\omega(j)} (1/2)(\lfloor{\frac{t^2}{j}}\rfloor +\lfloor{\frac{t^2-2}{j}}\rfloor) +C$?

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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 :

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