How can I convert a Classic Slideshow to an ordinary notebook?

I have several Classic Slideshow presentations originally created 3 or 4 years ago. I would like to convert these to ordinary Mathematica notebooks, removing all the slideshow-related paraphernalia including page breaks, etc.

If I simply select Format > Stylesheet > Default, the separate slides are still there, labelled “Slide 1 of 34” etc.. Is it then just a matter of manually removing all the page breaks, or is there more to it?

Related to this, (though it’s not what I want to do at present), what is the best way to convert from a classic slideshow to the newer Presentation style? Will Format > Stylesheet > PresenterTools do the job correctly?

Can I use Mathematica to find the values of $n$ for which the ordinary hypergeometric function is defined?

For some $ n$ , Mathematica cannot compute the solution to a problem I am working on, because the ordinary hypergeometric function is not defined for all values of $ n$ .

Can I use Mathematica to find the values of $ n$ for which the ordinary hypergeometric function is defined?

What is the relation between Tabaxi and ordinary cats?

I’m currently playing a Tabaxi Swashbuckler / Battlemaster (and I love it). I’m thinking of taking the Magic Initiate feat and thus gaining access to the Find Famliar spell. I thought it would be funny to have a cat familiar and have my character think of it as a kind of miniature avatar of himself. But even if a familiar is not a beast, but (in my case) a fey, I had to ask myself whether that would even make any sense. I am hence looking for answers from official source books to the following question:

How closely are Tabaxi and ordinary cats related? Do they maybe have a special bond, or would they rather find each other a bit odd?

Even though Tabaxi clearly have feline characteristics, the rules notably feature no mention of any special relation to cats (such as e.g. proficiency in animal handling when dealing with cats), so maybe Tabaxi are no closer to cats than humans are to orang-utans?

Can an ordinary whip be poisoned?

I’m running a whip-wielding ranger/rogue, and I’m wondering if it’s possible to poison a whip, both from a rules standpoint and as a practical matter.

Rules-wise, there’s this from the DMG:

Injury. A creature that takes slashing or piercing damage from a weapon or piece of ammunition coated with injury poison is exposed to its effects.

since a whip deals slashing damage, it suggests that it’s at least legal. The PHB seems to back this up:

Poison, Basic. You can use the poison in this vial to coat one slashing or piercing weapon or up to three pieces o f ammunition. Applying the poison takes an action…

Am I missing anything that specifies bladed or metal weapons? Then, assuming it’s legal, the other question is does it make sense? I’m no expert on whips, although at our local Renaissance Faire there’s a guy who dips his chainmail bullwhip in gas and lights it on fire, which suggests a whip might soak up a liquid, like a poison.

Is there anyone out there who can make a practical case for or against poisoning a whip?

How to compute a system of ordinary differential equations with initial condictions over a continuous range

I have some questions about Mathematica programming and would appreciate if you could help me.

I want to solve a system of ordinary differential equations \ [Mu] ‘[t] and \ [Lambda]’ [t] and each equation contains a large number of terms so it is impractical to write them explicitly. I express these terms as two functions F1 and F2 that depend on two parameters P1 and P2 and \ [Lambda] [t] and \ [Mu] [t].

I have been able to solve this system for a couple of initial conditions \ [Lambda] [0] = ic1 and \ [Mu] [0] = ic2, but I would like to solve my system of equations for a continuum of values ​​\ [Lambda] [0] = {0, …., Pi / 2} and \ [Mu] [0] = {0, …., Infinity} and then get \ [Lambda] [t] and \ [Mu] [t] and use them to perform an integral on \ [Lambda]= \ [Lambda] [0] ={0, …., Pi / 2} and \ [Mu] =\ [Mu] [0] ={0, …., Infinity} that are precisely our initial conditions.

I integrate the product of a function G in the time t (where \ [Lambda] [t] and \ [Mu] [t] are taken into account for a certain initial condition defined by the continuous ranges of the integral) with the same function, but in t = 0 (where the initial conditions are taken into account with the continuous ranges of the integral).

The structure of the program is:

ode = {\[Mu]'[t] ==  F1[p1, p2, \[Lambda][t], \[Mu][t]], \[Lambda]'[t] ==  F2[p1, p2, \[Lambda][t], \[Mu][t]], \[Mu][0] == {0, ....,   Pi/2}, \[Lambda][0] == {0, ...., Infinity}};   Sol = NDSolve[ode, {\[Mu], \[Lambda]}, {t, 0, 1},`Method -> "Some method to choose"]     \[Mu]1[t_] := Evaluate[\[Mu][t] /. Sol] // First \[Lambda]1[t_] := Evaluate[\[Lambda][t] /. Sol] // First   data = ParallelTable[{t,  NIntegrate[   G[p1, p2, \[Mu]1[      t] "for the initial condition \[Mu]=\[Mu][0]", \[Lambda]1[      t] "for the initial condition \[Lambda]=\[Lambda][0]"] G[p1,     p2, \[Mu] "=\[Mu][0]", \[Lambda] "=\[Lambda][0]"] , {\[Mu] "=`\[Mu][0](initial condition)", 0,    Pi/2}, {\[Lambda] "=\[Lambda][0](initial condition)", 0,    Infinity}, Method -> {"Some method to choose"}]}, {t, 0, 1}];` 

Root vs ordinary share permissions on Windows Server 2016

I can access \server\share using my current login.

To access \server\c$ I need to authenticate using an administrative account on server.

Once authenticated using the administrative account, my permissions to \server\share mirrors those of the administrative account when server is running Windows Server 2016, but mirrors my current login if it’s Windows Server 2008 R2.

Is there any way to disable the former?

Collapsed partitions and ordinary partitions

Adopt the standard notation for integer partitions, writing $ \lambda_1^{a_1} \cdots \lambda_k^{a_k}$ as shorthand for the partition $ a_1 \lambda_1 + \cdots + a_k \lambda_k$ with parts $ \lambda_1 > \cdots > \lambda_k \geq 1$ and multiplicities $ a_i \geq 1$ .

If $ \lambda = \lambda_1^{a_1} \cdots \lambda_k^{a_k} \vdash n$ , its collapsed image is the partition with multiple occurrences of parts removed, and is denoted by $ \underline{\lambda} := \lambda_1 \cdots \lambda_k$ . Introduce the set $ \underline{\mathcal{P}_n}=\{\,\underline{\lambda}: \,\lambda\vdash n\}$ as well as the all-familiar $ \mathcal{P}_n=\{\,\lambda: \,\lambda\vdash n\}$

QUESTION. Is this true? $ $ \#\underline{\mathcal{P}_n}=\sum_{j=0}^{n-1}\#\mathcal{P}_j$ $

Remark. For a related problem, see here:

how to decrypt an ordinary folder (not home, nor partition, nor .Private)

Following the procedure described here I could create an encrypted folder in an external drive.

mount -t ecryptfs /media/foo/bar /media/foo/bar 

Then I put some stuff inside, restarted my computer and I now get a few folders named

 ECRYPTFS_FNEK_ENCRYPTED.FWbqvPj9ejI71-QEsY3Zgt7-TRG7S-ek830T453HKzNQPfJjZVpbWWaxlE-- 

Which I don’t know how to mount again. All tutorials I look for speak about encrypted partitions, encrypted .Private folders or encrypted /home/<user>

Probability boxes (precious vs ordinary stones)

We play in a television game. We have 8 identical and indistinguishable boxes, each of which has 2 pebbles. Any pebble can be precious or not. We pick a box and the host, without looking at it, pulls out one of the pebbles. It turns out to be precious. The host then states that we have exactly 50% chance and the second pebble in this box is precious.

We assume that the host knows the distribution of the pebbles in the boxes – how many boxes the precious stones are 2, how many 1 and how many are 0. However, the host does not know what the pebbles are in a particular box because he does not distinguish the boxes himself.

If we know that initially the number of precious stones is not less than the number of precious, which of the following statements should be true?

A) If we ask for a change of box, we will have a better chance of the next drawn stone being precious.

B) If we ask for a change of box, we will not change our chances that the next downloaded stone will be precious.

C) If we ask for a change of box, we will have less chance of the next drawn stone being precious.

D) Initially, there were just 2 boxes with two precious stones in the game.

E) At first half the boxes had one precious and one precious stone.

F) Initially, there were an equal number of precious and precious stones in the game.

G) Initially the number of boxes with 2 precious stones is equal to the number of boxes with 2 precious stones.

H) The host is confused. There is no chance for a second precious stone in our box to be exactly 50% because there are an odd number of pebbles in the game.