@umar khan   I do not understand the reason for the error. I downloaded your file, ran it and it works correctly. I am attaching my worksheet (in Maple 2016.2).

plots.mw

@Carl Love   ListTools:-LengthSplit(L,n)  is a sequence not a list.

@Markiyan Hirnyk   I can not confirm your first calculation. I think that  value(Int(...))  is equivalent to int(...) .

In Maple 2016.2:

z := exp(I*t): 
f := z^(1/2)*(diff(z, t)): 
value(Int(f, t = 0 .. 2*Pi));
                                                          -4/3

I have not submitted any SCR.

@Markiyan Hirnyk 

Compare:

Matrix([seq(diff(y,[x$i]),i=0..nops(y)-1)]);
Matrix(nops(y),(i,j)->eval(diff(y[k],x$n),[k=i,n=j-1]));

@nm   Since  diff  command works with lists (as element-wise function), the same thing can be written shorter:

restart;
y:=[exp(2*x), exp(-x), x*exp(-x), x^2*exp(-x)]:
Matrix([seq(diff(y, [x$i]), i=0..nops(y)-1)]);
LinearAlgebra:-Determinant(%);

@st104290   If Maple can not find an explicit formula as in your example, then you can use tomleslie's method. It will work for a wider class of recurrence equations.

@leafgreen  In the syntax for a histogram, the frequencies are used instead of probabilities.  [2, 4$4, 6$5]  means  [2, 4, 4, 4, 4, 6, 6, 6, 6, 6]

@leafgreen 

Statistics:-Histogram([2, 4$4, 6$5], discrete = true, thickness = 25, tickmarks = [[2 = `ℏ`, 4 = `0`*`ℏ`, 6 = 2*`ℏ`], [0.1, 0.4, 0.5]], view = [0 .. 7, 0 .. 0.55], axes = normal, axesfont = [times, roman, 14], color = cyan, title = "Problem 7(b)");
 

@Derein  Just plot these data:

plot([seq([k, x[k]], k=1..30)]);
 

@Markiyan Hirnyk  In your document  double_points.mw

in the line  
for b in {seq(0..3*n)} do

should be  6  instead of  3

@Markiyan Hirnyk   I prefer to use eval than parameters option because it's just shorter for typing.

@mayazteca5  See the appendix to my answer below. You can also do this with other plotting methods using  plots:-display  command.

@vv  

J(5, 1);
   Error, (in J) invalid input: J expects its 1st argument, n, to be of type posint, but received 0

In fact, your link contains a solution that works perfectly. You do not have to add "1" yet. Perhaps you are confused by the example (in the link) for  n=5  and  k=3 . But in this example, the numbering of people starts with 1, not with 0. I did not understand the logic of your code and why do you consider it to be improved.

Josephus:=proc(n,k)
if n = 1 then return 0 else
Josephus(n-1, k)+k mod n fi;
end proc:

Example of use:

Josephus(5, 3);

                                            3
 

@Markiyan Hirnyk   If you give a result that depends on the parameters, then the parameter values should be specified when this result is correct. Otherwise, what is the value of this result?