@DimaBrt  I did not understand your comment. For me, everything works in Maple 2018.2.

@kambiz1199  You can use the  trace  option to show the intermediate positions of a ladder. See my updated answer above.

@Rouben Rostamian   I also do not know a short way. Here is a rather cumbersome way:

test := -2+exp(theta)+exp(-theta);
convert(test,trigh);
convert(%, tanh);
convert(%,exp);
factor(%);
numer(%)/simplify(denom(%));

@minhthien2016  Of course, this line is only needed to make these assignments  a:=...  and  b:=...  clear.

@acer  and  Carl. Thanks for replies! The trouble removed. 

I tried to test this procedure. I just copied the text of the procedure into a new worksheet. But during initialization, the error occurs:  Error, unexpected neutral operator

Can anyone confirm this situation? What could be the reason? I'm using Maple 2018.2

Error.mw

@hamideh 

restart;
plots:-implicitplot3d((x,y,z)->`if`(x<y and y<z,x^3*y*z-0.5,NULL), 0..5, 0..5, 0..5, style=surface, color=green, numpoints=500000, axes=normal, orientation=[55,75]);

@Al86  The idea of how to find these values is suggested by the graph of the implicit function 

(x-2)^2*(x-3)+epsilon^2=0

(see the code above). The straight line  epsilon=C  (C is a constant)  should cross this graph at exactly 2 points. One value is obvious  epsilon=0, and 2 others are easily found using the derivative:

epsilon1:=x->sqrt(-(x-2)^2*(x-3));
epsilon2:=x->-sqrt(-(x-2)^2*(x-3));
x1:=solve(D(epsilon1)(x)=0);
x2:=solve(D(epsilon2)(x)=0);
epsilon1(x1);
epsilon2(x2);

@David Sycamore

restart;
AlmostCarmichaelNumbers1:=proc(n::posint)
local L;
L:=ifactors(n);
if nops(L[2])>2 and L[2,1,1]<>2 and `and`(map(t->is(t[2]=1),L[2])[]) and irem(n-1,L[2,1,1]-1)=0 and irem(n-1,L[2,-1,1]-1)=0 and `and`(seq(irem(n-1,t[1]-1)<>0,t=L[2][2..-2])) then true else false fi;
end proc:

Example of use:

 select(AlmostCarmichaelNumbers1, [seq(i, i=1..20000, 2)]);
   [231, 1045, 1653, 4371, 4641, 5365, 6545, 8029, 9361, 10011, 10857, 13685, 13695, 15753, 16269, 18361, 18565]

Edit.

I first learned about this command from this post. It is not clear to me why it is necessary for all occasions to look for the appropriate command in Maple, if we can just write the simplest procedure. Example:

restart;
Dist:=(X,Y)->sqrt(add((X-~Y)^~2)):
X:=<1,2,3>: Y:=<2,3,4>:
Dist(X,Y);

To answer the question, we need to see your particular calculation. So submit here your code (as text and not a picture) using a bold green up arrow in the MaplePrimes editor, or simply upload your worksheet here using the same method.

@David Sycamore This combination of commands 

`and`(map(t->is(t[2]=1),L[2])[])

 makes the check for "squarefree".

Here is another example of how the procedure works up to some N (I took N = 1000):

seq([n,Check(n)], n=1..1000);
     

@David Sycamore 

a:= proc(n::posint)
option remember;
   if n=1 then return 1 fi;
   if type(a(n-1)+1,prime) then return a(n-1)+1 else
   if type(a(n-1),even) then return a(n-1)+2 else a(n-1)+3 fi; fi;
end proc:

seq(a(n), n=1..20);

@RaulGutsa  For 3d you can use in the procedure  P  the  plots:-pointplot3d  command:

Eq:=diff(y(x),x)=cos(x):
ics:=y(0)=0:
Sol:=dsolve({Eq, ics}, numeric):
P:=proc(s)
plots:-pointplot3d(eval([[x,y(x),x]], Sol(s)), style=point, symbol=solidcircle, symbolsize=15);
end proc:
plots:-animate(plots:-display,['P'(s)], s=0..Pi, scaling=constrained);

Present here in editable form this matrix so that I can copy it into my Maple. I have a program written many years ago and not published anywhere that allows you to step by step bring the matrix of an arbitrary size to the ReducedRowEchelonForm  (comments to the steps in Russian).