@Thomas Richard 

For k=64/81 the polynomial ir reducible:

((3*w - 2)*(162*w^4 - 297*w^3 + 72*w^2 + 48*w + 32))/81

For solvable irreducibles, the values of k=m/n, |m|, |n| <= 100  are: -16/9, 1/81, 25/9, 80/81.

@Thomas Richard 

The value k=1/81 is interesting because the polynomial is irreducible and solve with option explicit works. 

@Carl Love 

Actually Im( ln(1+ z) ) does not have a limit at z=-1. The limit of ln(1+z) at z=-1 is "complex infinity" (denoted by infinity+I*infinity) in Maple.

Note that assuming does not make much difference between `<` and `<=`.

E.g.

limit(z^n, n=infinity) assuming abs(z)<=1; #   0

@David Sycamore 

Probably you have not noticed the new name for the procedure.

Download SP.mw

SP:= proc(N::posint, p1::prime:= 2)
local S:= Array([p1], datatype= integer[8]), Used:= table([p1= ()]), s, k, p:= p1, i,j,ok;
    for k from 2 to N do
        s:= `+`(convert(p, base, 10)[]);
        p:= s; ok:=false;
        for i do  for j by 2 to 10^i-1 do
            if p::prime and Used[p]<>() then ok:=true; break fi;
            p:= Scale10(s, i) + j;
        od; if ok then break fi;
        od;
        S(k):= p; Used[p]:= ()
    od;
    [seq(S)]
end proc:

@tomleslie 

The documentation also refers to the "angle between two given objects". What are the objects in this case? 
Anyway, it's a bug.

@Preben Alsholm 

Yes, it seems that you are right. 
I was able to show the uniquness of an analytic solution near 0 (|t| < 1/10).
It is not clear if X is C^2 only. [of course X will be C^oo for t<>0 but not necessarily analytic (for now)].

@mmcdara 

Why do you think that there is "the" exact solution. Maybe the ode has more than one (standard uniqueness theorems do not apply here).

@nrussoniello 

Each symbol is replaced with a random number, e.g. s1=77, s2=-665 ...

Suppose that the entries of a 2x2 matrix M(x,y) are polynomials in x,y.
The rank of M(x,y) is <=1 iff det(M(x,y)) is identical 0. E.g. if det(M(x,y)) = x^2 - y + x*y^7 then rank(M(x,y)) = 2.

If follows that if det(M(78,55)) <>0 then rank(M(x,y))=2. 
Your case is similar.

The sitation would have been more complicated if the rank for the matrix with symbols replaced were not maximal.Then we must repeat the procedure and take the maximal ranks, but accepting now that the true rank could be higher (but with a small probability).

If you don't want to read at least a tutorial, the shortest path would be to post here a simplified version of your problem.
Probably someone will upload a worksheet with the solution, and you will be able to adapt it for similar tasks.

@Carl Love 

This order is not total (linear) and has nothing to do with "lexicographic". Where is it used? How is x[2] computed using this order?

x[n] seems to be the least positive interger not in { x[1],...,x[n-1] } which satisfies ...

Do you mean the lexicographic order in N? But then 123 < 2 < 3, so why is x[2] = 3?

It is not only old code, but you have also replaced the old keyword array with the new one Array in a few places. Revert these changes.

@mmcdara The first solution is very elegant; of course I like it. But it was probably provided by the author/committee or maybe by a very smart student (who deserves congratulations & a special prize, if he/she exists).