@Carl Love 

That's interesting, but I find the resulting code unintuitive.
Probably most users would prefer:

J:=Iterator:-Combination(6,4):
seq(ArrayTools:-Alias(x[]+~1, [2,2]), x in J);

@emendes 

The denominator of R[n] is  4*2^(2^n).

Actually, using rsolve ==>  R[n] = -1/2*cos(arccos(3/4)*2^n)+1/2

Using this, R can of course be formed even for n >> 1000.

@JanBSDenmark 

If you want an explanation. In mathematics sqrt and arcsinh are multivalued functions (defined on C).
All CASes have problems in choosing the wanted branches for such functions.
E.g. with your notation, sol[2] is a solution for another branch:  - x*sqrt(4*x^2 + 1) + arcsinh(2*x) - 5 = 0.

@David Sycamore 

P calls PP (from my first answer), so, PP must be present and executed in order to use P.

Just execute (use the (!!!) icon) the attached:

PP.mw

@David Sycamore 

You do not have the procudure PP in the worksheet (and executed, of course).

@David Sycamore 

The procedure PP can be easily used for any such questions:

PPrime:=proc(p::prime,n::posint)  
evalb(nops([PP(n,p,[p],1)])=1)
end:

P := proc(n)
local p, P:=select(isprime, [seq(1..n)]);
select(PPrime, P, n)
end:

P(50)
     [2, 3, 5, 7, 11, 13, 19]

@David Sycamore 

Just replace

Sol[++nSol] := XXX

with

nSol:=nSol+1; Sol[nSol] := XXX

(in 2 places)

@Carl Love 

It works in 2018 (even if ++x etc are not documented).

@Rouben Rostamian  

There are 2 solutions: take the symmetry  (x,a) -> (-x, -a).
(For external tangency there are also 2 solutions).

@Kitonum 

A continuous function e.g. on R has a continuous antiderivative but a CAS does not necessarily find one.

E.g.  int(1/(2+cos(x)), x) .  Even for a human it's not trivial.

@Kitonum 

The antiderivative is computed correctly, but it is not continuous, which techically is not a bug:

restart;
F:=int(sqrt(1+((k*Pi)/l*cos((Pi*x)/l))^2), x) assuming k>0,l>0:
F22:=eval(F,[l = 2, k = 2]):
limit(F22,x=2,left) - limit(F22,x=0, right); 
evalf(%);

      4*EllipticE(Pi*I)/Pi
      4.609785320 - 0.*I

@mehdi jafari 

I tried DirectSearch instead of fsolve.
It seems that the system has no solutions, see the attached worksheet.

Inflation-Inverse-1-vv.mw

@mehdi jafari 

Your code is not usable. You have merged procedures and expressions: e.g. if F and G are procedures with parameter x, F(x) + G(x) makes sense but F + G(x) or F + x^2  are usually nonsenses.

In your first post, eq2 contains  int( T[rr]*r, r=s..t)
but T[rr]  does not depend on r (r appears as an integration variable).

So, before turning Eq1,...,Eq4 into procedures, you must clarify the equations mathematically.
Note that if you are going to have double integrals (which is not clear for now), fsolve will probably be inacceptably slow or even fail.

@Kitonum 

I think the OP wants to insert rational numbers with minimal denominators between the elements of a sorted  list(realcons).
I don'd know why he does not try to state the question more clearly.

@acer 

Strange indeed. A "normal" answer should have been

X:= n -> piecewise(n<10, [1,22,-308,-3003,21021,105105,-360360,-765765, 765765][n], 0);

obtained easily using rsolve({req, x(1)=1}, x, makeproc);

How did you think at assuming integer? Vote-up for this!