@David Sycamore Here is a procedure to construct examples of the phenomenon being discussed. It works for any even number as the initial spacing, so not necessarily twin primes.

restart
:
(*>>>-------------------------------------------------------------------------------------
| Given primes p1 and p2 such that p2 > p1 and p2+(p2-p1) is also prime, this procedure  |
| attempts to return a set P of primes containing p1 and p2 such that there exists a     |
| *unique* e > 0 such that P +~ e is entirely prime. If successful, then necessarily     |
| e = p2-p1 and |P| = p2. If not, and execution is halted, then the global MAXP is the   |
| largest prime that was examined for possible inclusion in P.                           |
-------------------------------------------------------------------------------------<<<*)
FindFiniteExample:= proc(
   p1::prime, 
   p2::And(prime, satisfies(p-> p > p1 and isprime(2*p-p1)))
)
option `Author: Carl Love <carl.j.love@gmail.com> 3-Oct-2019`;
global MAXP:= p1;
local 
   e:= p2-p1, r2:= p1 mod p2, R2:= MutableSet(r2),
   p:= p1, P:= MutableSet(p1)
;
   while numelems(P) < p2 do
      p:= nextprime(p);
      if isprime(p+e) then
         r2:= p mod p2;
         if not r2 in R2 then MutableSet:-insert~([R2,P],[r2,p]) fi
      fi;
      :-MAXP:= p
   od;
   convert(P, set)
end proc
:
#Usage:
FindFiniteExample(3,5);
                       {3, 5, 11, 17, 29}
FindFiniteExample(3,7);
                   {3, 7, 13, 19, 37, 43, 67}
FindFiniteExample(5,11);
          {5, 11, 13, 17, 23, 31, 37, 41, 47, 73, 131}
FindFiniteExample(3,11);
        {3, 11, 23, 29, 53, 59, 71, 101, 131, 149, 173}
FindFiniteExample(3,17);
{3, 17, 23, 29, 47, 53, 59, 83, 89, 113, 137, 167, 179, 197, 449, 
  509, 617}
FindFiniteExample(5,17);
{5, 17, 19, 29, 31, 41, 47, 59, 61, 67, 71, 89, 137, 151, 179, 
  181, 227}
FindFiniteExample(7,13);
     {7, 13, 17, 23, 31, 37, 41, 47, 53, 61, 97, 103, 107}
FindFiniteExample(11,17);
{11, 17, 23, 31, 37, 41, 47, 53, 61, 67, 73, 83, 97, 103, 157, 
  263, 383}

So, it seems that examples are easy to find.

Edit: The code above has been simplified, although the original code was also correct. The output of the examples shown will in some cases be different.

@David Sycamore You asked about the notation P +~ e. This adds e to each element of P. Likewise both P mods~ p and irem~(P, p) compute the remainders mod p of all elements of P. See help page ?elementwise 

@David Sycamore I found a counterexample to my Conjecture 2 as stated. I suspect that these counterexamples are extremely rare, and perhaps there's only a finite number of them; and I'll restate the Conjecture in a form that I think is very likely true (as the Riemann hypothesis is very likely true).

I think that the smallest counterexample is 81345, whose prime factor set is P:= {3, 5, 11, 17, 29}. It's easy to prove that this has exactly one solution (so not an infinite number): P has a complete set of remaiders mod 5. In order, they are {3, 0, 1, 2, 4}. So, for any e > 0, P +~ e necessarily contains a multiple of 5, which may be 5 itself. Indeed, using e=2 yields {5, 7, 13, 19, 31}, which are all prime, so that's a solution. But using any other e, you'll get one member that's a multiple of 5 other than 5 itself, so it won't be prime.

Note that I carefully chose the primes in P to not have a complete set of remainders mod 3.

Here's the restated conjecture: Conjecture 2: Let P be a finite set of primes which does not have a complete set of remainders with respect to any of its members. Then there are an infinite number of positive integers e such that P +~ e contains only primes.

This modified conjecture is still a generalization of the famous twin-prime conjecture.

Here are some conjectures related to this:

Conjecture 1: Every case where there is no solution can be verified by the method that I showed.

Conjecture 2: In every case where there is a solution, there are an infinite number of solutions.

The second conjecture is a generalization of the famous twin-prime conjecture; so, it may be a consequence of the Riemann hypothesis. 

I applied my proof technique to all 465 no-solution cases in Tom's failure list. In all 465 cases, it is proven that there are indeed no solutions. This is intended to be run after generating Tom's list failure:

ProveNoSolution:= proc(n)
local P:= numtheory:-factorset(n), p:= min(P); 
   evalb(nops(irem~(P, p)) = p)
end proc
:  
(proven, unproven):= selectremove(ProveNoSolution, op~(1, failure));

MaplePrimes can't handle .maple workbook files. Can you convert your file to a .zip file and repost it?

What you've posted is not a differential equation.

@patrickfitzpatrick Please let me know whether the all-solutions code that I posted works adequately for you. In particular, Does it produce all solutions in a reasonable time for all the cases that you're interested in?

@acer Unfortunately, workbook .maple files don't work on MaplePrimes.

@Carl Love The algorithm for the all-solutions problem and its Maple implementation are in an Answer below, and I'm quite pleased with them.

@patrickfitzpatrick Both Kitonum's and my code give one solution--not all solutions--for a given n. I am working on an algorithm to efficiently find all solutions. It should take me a few hours.  

@Masooma The version released in 2018 is called "Maple 2018". There was also a version released several years earlier (2014?) called "Maple 18". Which of these two do you have? The command kernelopts(version) will tell you.

Why do you have size=[600,300] if you want another size?

@Masooma The only part of your code above that makes sense is the first line:

F := proc (x) options operator, arrow; x-2*f(x)/(D(f))(x) end procyou

It's more commonly written simply as

F:= x-> x - 2*f(x)/D(f)(x)

All the stuff with algsubs does nothing, and I don't know what you're trying to accomplish. The thing that you wrote with a second derivative could be meaningful, but it doesn't seem relevant to your goal of iterating the first function F.

By the way, what version of Maple are you using? I could retrofit my code to an earlier version if you need that.

@Masooma In order for the order of convergence to be 2, m must be the known multiplicity of the root (many thanks to Acer and VV for providing clarity on this). So, if we use m=2, and the root has multiplicity 2, then the order of convergence will be 2. If you apply my order-of-convergence approximator (code given above) to such a situation, you'll get an order very close to 2. The polynomial examples that you gave earlier do not have roots of multiplicity 2:

(x-1)^3                                   m=3
(x^2+2*x+1)^5 = (x+10)^10   m=10