Magento 2: Use a model with a dynamic name

In the case you are interested in there is a simple generating (Dirichlet) series: $$ \sum_{n=1}^\infty \frac{d(n^2)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}.$$ From this you can either use a convolution argument or a Perron formula type argument to get an asymptotic formula. In particular, I believe it follows that $$\sum_{n\leq x} d(n^2) = \frac{3}{\pi^2}x \log^2 x +O(x \log x). $$ With more work, you can get lower-order terms of size $\asymp x \log x$ and $\asymp x.$

Edit: There seems to some disagreement on whether the coefficient of the leading order term is $\frac{3}{\pi^2}$ or $\frac{3}{2\pi^2}$. I believe that $\frac{3}{\pi^2}$ is correct. Here are three bits of reasoning: From the generating series, we have $$ \sum_{n\leq x}d(n^2) = \sum_{n\leq x} \sum_{k\ell^2=n} d_3(k)\mu(\ell) = \sum_{\ell\leq \sqrt{x}} \mu(\ell) \sum_{k\leq x/\ell^2} d_3(k) $$ where $d_3(n)$ denotes the number of ways to write $n$ as a product of three positive divisors and $\mu(\ell)$ is the Moebius function. By a standard estimate $$ \sum_{k\leq x/\ell^2} d_3(k) = \frac{x}{2\ell^2}\log^2(x/\ell^2) + O\left(\frac{x\log x}{\ell^2} \right)$$ from which it follows that $$ \sum_{n\leq x}d(n^2) = \frac{x \log^2 x}{2} \sum_{\ell \leq \sqrt{x}} \frac{\mu(\ell)}{\ell^2} +O(x\log x) = \frac{3 x}{\pi^2}\log^2 x +O(x\log x).$$ Alternatively, a Perron formula (e.g. Prime Number Theorem) type argument can be used to show that $$ \sum_{n\leq x}d(n^2) = \text{Res}_{s=1} \frac{\zeta^3(s)}{\zeta(2s)} \frac{x^s}{s} +o(x) = \frac{3x}{\pi^2}\log^2 x + O(x\log x).$$ Moreover, in Mathematica, you can use DivisorSigma[0, n^2] to calculate $d(n^2)$. For $x=1,000,000$ I get that $$ \frac{\pi^2}{3x\log^2x}\sum_{n\leq x} d(n^2) \approx 1.27305392....$$ The slow convergence to 1 is from the influence of lower-order terms. Notice, however, that the value it is not anywhere near $1/2$. However, if I define $F(x)$ to be the residue of$\frac{\zeta^3(s)}{\zeta(2s)}\frac{x^s}{s}$ at $s=1$, I get that $$ \frac{1}{F(x)}\sum_{n\leq x} d(n^2) \approx 1.0000073....$$

Unless someone has done some quantitative or qualitative research, I think this will be very hard to verify either way. There are certainly professors that are famous and have very vocal political opinions, such as Noam Chomsky, and many philosophers. However, it is hard to verify whether: their opinions help their career, their career helps people accept their opinions more readily, or whether their career could have gone even further without their opinions.

One thing to note, however, is that it used to be many philosophers seemed to take it as a duty to give public talks and so on, on politics (e.g., Bertrand Russell), and perhaps other academics did too. So although we probably cannot give an empirical answer right now, it does seem that in many cases social media is merely an extension of professor's work, even if they are giving thoughts on areas they do not research, as thinkers they may have a duty to contribute informed opinions based on their training.

Killing or subduing an enemy will result in a loss of points. A loss of points means it takes longer for you to unlock the improved abilities that 47 can gain and making it harder to gain the coveted "Silent Assassin" rating. However, depending on how you perform the kill and take care of the situation, you will gain points back. If executed correctly, you can end up with a net zero point change even after killing someone. For example, if you perform a silent kill (eg. fiber wire with nobody else looking), you will lose a set amount of points for killing someone, but then gain half of it back for doing the silent kill. Then, if you hide the body, you will gain the remaining points that were lost, resulting in no change. This also works for subduing people, then hiding them.

All in all, this method allows you to still earn a "Silent Assassin" rating while still being able to deal with certain people that are directly in your path.

Also, killing or subduing someone will gain you a set amount of instinct, so if you're running low and you depend on instinct, this is a viable way to get more. I'm not sure how instinct works on Purist difficulty.

Edit: It seems that there is no instinct on Purist difficulty, at least in the sense that you can't gain more. On Purist mode, trying to use instinct will only show you where your current target is and slow time down a bit, nothing more. This ability also appears to be unlimited.