How to get players to be curious and ask questions?

I have recently noticed that my players don’t really try to find out much about the world and story beyond the direct “what do we see?” I want to have them interact more with their environment and actually explore the reasons behind things without me leading them by the nose. “Why are these monsters working together?” “What are the villains trying to accomplish?” How do I get my players to be curious and ask those questions that start the gears of imagination like Who, What, Where, When, Why, and How?

These kinds of questions can really bring a world to life and add SO much depth to a story but they just don’t ask and if I tell them without them asking the game drags on and they begin to feel like I’m monologuing.

A curious prime counting approximation or just data overfitting?

I am not sure, if this is a research problem. If not I will move this question to ME: Let $ \Omega(n) = \sum_{p|n} v_p(n)$ , which we might view as a random variable. Let $ E_n = \frac{1}{n} \sum_{k=1}^n\Omega(k)$ be the expected value and $ V_n=\frac{1}{n} \sum_{k=1}^n(E_n-\Omega(k))^2$ be the variance. Then $ $ \pi(n) \approx \frac{n\gamma(\frac{V_n}{E_n},1.4854177\cdot \frac{V_n}{E_n^2})}{\Gamma(\frac{V_n}{E_n})}$ $ where $ \Gamma=$ Gamma function, $ \gamma=$ lower incomplete gamma function.

Background: I was trying to fit the gamma distribution to the random variable $ \Omega(k)$ ,$ 1 \le k \le n$ . The value $ 1.4854177$ is fitted to some data. My question is, if there is any heuristic why this approximation should be good, if at all, or if this is just an overfitting problem?

Below you can find some sage code which implements this:

def Omega(n):     return sum([valuation(n,p) for p in prime_divisors(n)])  means = [] variances = [] xxs = [] omegas = [Omega(k) for k in range(1,10^4)] for nn in range(10^4,10^4+3*10^3+1):     n = nn     omegas.append(Omega(n))     print "---"     m = mean(omegas[1:-1])     v = variance(omegas[1:-1])     shape,scale = m^2/v,v/m     xx = find_root(lambda xx : n*(lower_gamma(shape,xx*1/scale)/gamma(shape) ).N()-prime_pi(n),1,2)     xx = 1.4854177706344873     approxPrimePi2 = n*(lower_gamma(shape,xx*1/scale)/gamma(shape) ).N()     primepi = prime_pi(n)     print primepi, approxPrimePi2,shape.N(),scale.N(),xx     print "---"     print "err2 = %s" % (abs(primepi-approxPrimePi2)/primepi)     xxs.append(xx)     means.append(m.N())     variances.append(v.N()) 

A curious conjecture: $\{\varphi(m^2)/\varphi(n^2):\ m,n=1,2,3,\ldots\}=\{r>0:\ r\in\mathbb Q\}$

Let $ \varphi$ denote Euler’s totient function. It is easy to see that all those numbers $ $ \varphi(n^2)=n\varphi(n)\ \ (n=1,2,3,\ldots)$ $ are pairwise distinct.

I have the following surprising conjecture.

Conjecture. Any positive rational number $ r$ has the form $ \varphi(m^2)/\varphi(n^2)$ with $ m$ and $ n$ positive integers.

I have verified this for $ r\in\{a/b:\ a,b=1,\ldots,50\}$ . My computation shows that \begin{align}&\left\{\frac{\varphi(m^2)}{\varphi(n^2)}:\ m,n=1,\ldots,15000\right\}\\supseteq&\left\{\frac ab:\ 1\le a,b\le 50\ \&\ \{a,b\}\not=\{19,47\},\{37,47\}\right\}.\end{align} In addition, I have found that $ $ \frac{\varphi(12765^2)}{\varphi(18612^2)}=\frac{80879040}{102738240} =\frac{37}{47}$ $ and $ $ \frac{\varphi(39330^2)}{\varphi(55836^2)}=\frac{373792320}{924644160} =\frac{19}{47}.$ $

I have no good explanation for the conjecture, but I’m confident that it should be true.

QUESTION: Is the above conjecture true? Are there any supporting heuristic arguments?

Your further check of the conjecture is also welcome!

A very curious wifi problem

I was having some trouble with my wifi reception on the 2.4g band so I started up airodump to look for any offending signals. I found a strong hidden essid on the same channel I was on, so I logged on to my router and switched channel. Curiously the mac transmitting the hidden essid did the same thing. (?) I thought maybe this is someone playing around with a simple script? But what for? So in order to alleviate my wifi issues, I simply engaged the 5g band and logged on to that instead. However, when I engaged the 5g, I then found the same mac address transmitting the same hidden essid on the 5g band! (and no longer on the 2.5g band) I contacted rogers tech support asking them if they were at all aware of how my router would have a third mac address sending out a hidden beacon, but they were as dumbfounded as I am. No other router in the neighborhood is doing the same thing, only mine.

I’m hoping someone might be able to shed some light on why my router is doing this. Why does it have some third mac sending out a beacon I have zero control over? Why is it only my router, when everyone around me has the same rogers router?

Curious about building a mining software

I am trying to learn and understand more about mining software, and so I am trying to build a simple 1 to try it out.

I have been reading resources from SlushPool and a bit from the google link they shared, and came up with a sample of the below code.

# The below code is in Python 3 import socket  host = '' port = 6667  with socket.socket(socket.AF_INET, socket.SOCK_STREAM) as s:     s.connect((host, port))     s.sendall(b'{"id": 1, "method": "mining.subscribe", "params": []}')     data = s.recv(1024)  print(repr(data)) 

However, the response that I receive from the pool is:


However, I noticed that this response is different from the response that i’m supposed to be receiving. I say this is because the response i should be receiving (according to SlushPool) should be:

{"id": 1, "result": [ [ ["mining.set_difficulty", "b4b6693b72a50c7116db18d6497cac52"], ["mining.notify", "ae6812eb4cd7735a302a8a9dd95cf71f"]], "08000002", 4], "error": null}\n 

I understand that the id will be different in the result.. But i’m not sure why is it that my response is not the mining.subscribe that I should be getting.

Any ideas? Thanks!

my over curious wife

i am having problems with my wifes untrusting nature, she was tracking me on google timeline and when i shut that off on her she started using ip addresss lookup services to do it. then i got a new phone and called her with it onceand before i got home she had already tracked me by ip address. so how did she do that with only my phone number? it wasnt on an account(google) she knew about< so i cant find a way she did it.

A curious $q$-identity

Let $ [x]_{q}=\frac{1-q^x}{1-q}$ and $ \binom{x}{n}_{q}$ denote a $ q$ -binomial coefficient.

Let $ A_n(x,q)$ be the $ n\times n $ matrix with entries $ $ q^\binom{i-j}{2}\binom{i+j+x}{i-j+1}_{q},$ $ $ 0 \le i,j \le n-1,$ and $ n+1\le m \le 2n-1.$

Computer experiments suggest that $ A_n(-m,q) v_{n,m}=0$ if $ v_{n,m}$ is the vector with entries $ $ q^{(\lfloor{m/2}\rfloor-j) (\lfloor{m/2}\rfloor+5-3j)/2}\frac{[m]_q}{[m-j]_q}\binom{m-j}{j}_{q}$ $ for even $ m$ and $ $ q^{(\lfloor{m/2}\rfloor-j) (\lfloor{m/2}\rfloor+7-3j)/2}\frac{[m]_q}{[m-j]_q}\binom{m-j}{j}_{q}$ $ for odd $ m.$

Any idea how to prove this?

For $ q=1$ one can use Rothe’s formula, but I could not find a $ q$ -analogue which applies to the general case.

A curious inequality concerning binomial coefficients

Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.

Let $ a_1,a_2,\ldots,a_k$ be non-negative integers such that $ \sum_i a_i = A$ . Then, for any non-negative integer $ B \le A$ : $ $ \sum_{(b_1,\ldots,b_k): \sum_i b_i = B} \prod_i \frac{\binom{a_i}{b_i}}{\binom{A-a_i}{B-b_i}} \ge {\binom{A}{B}}^{2-k}. $ $ The sum on the left is over all tuples $ (b_1,b_2,\ldots,b_k)$ of non-negative integers, with $ b_i \le a_i$ for all $ i$ , whose sum is equal to $ B$ .

I’m curious if the definition 2.2 found in this unpublished paper at has been defined elsewhere in the math literature?

I found a definition in an unpublished paper and I’m curious if the edge-sequence definition 2.2 found in this paper at has been defined elsewhere in the math literature?