@Ramakrishnan  I took frames = 121 so that one of the animation frames matched the value of the parameter a = -3, at which a qualitative change in the structure of the solution occurs. Therefore, the step for the parameter  a  can be taken from several variants: 0.5, 0.2, 0.1, 0.05, ... . I chose the latter option and then the number of frames will be  6 / 0.05 + 1 = 121

@Ramakrishnan  Substantially reduce the number of frames, for example to 60, or write so:

restart;
A := plots:-animate(plot, [x^3+a*x+2, x = -4 .. 4, -15 .. 15, color = blue, thickness = 2], a = -6 .. 0, frames = 120): 
A;

Also note that I slightly simplified the penultimate line of my code above (for C).
I advise you to think about why I took in the code  frames=121  instead of frames=120 .

@vv  Very interesting how did you get these results? My way for some reason fails for large n (even option remember does not help), and Markiyan's way takes a long time:

P(10^5);
   Error, (in P) too many levels of recursion

restart;
ts:=time():
x(1) := 1: N := 100000; for n to N do x(n+1) :=
sum(convert(x(n), base, 10)[j]*10^(nops(convert(x(n), base, 10))-j), 
j = 1 .. nops(convert(x(n), base, 10)))+n end do:

x(10^5);
time()-ts;

                          N := 100000
                           9966045232
                             84.281
 

@Markiyan Hirnyk  I often visit this site  http://www.mathforum.ru/forum/list/1/

Your expression can be greatly simplified if you reduce the fraction by  sqrt(s+thetac)  because

         A1*Dc*alpha1*s^2+A1*Dc*alpha1*s*thetac=A1*Dc*alpha1*s*(s+thetac)

@MrMarc  If you need a single solution, then you need to impose an additional condition. For example, the condition  x=y  leads to the unique solution:

solve({x+y = 373320, z = (x+y) / 0.44 - y -  y* (1 - 0.99), x=y});

         {x = 186660., y = 186660., z = 659927.9456}
 

 rlopez Many thanks for the detailed analysis of the code in  IntOverDomain  procedure and the great work you have done for this.

@acer  This is the perfection of skill!

@mohkam7 

restart;
MakeColored:=(s,c)->Typesetting:-mo(convert(s,string), mathcolor = c):

MakeColored(`ϵ`, "Green");
MakeColored(x, "Red")[MakeColored(i, "Red")];

@Markiyan Hirnyk  So what?

@Markiyan Hirnyk  My answer is just an alternative solution to the problem (without Statistics package). It remains to round 4.98 .. to the nearest integer. To be more pedantic, it is necessary to compare the values of the sums for n=4 and n=5 and choose the one that is closer to 0.95:

lambda:=2.6:
sum(lambda^k*exp(-lambda)/k!, k=0..4);
sum(lambda^k*exp(-lambda)/k!, k=0..5);

                          0.8774234888
                          0.9509628480
 

@idol050279  You have already received so many different tips and examples that it's time you made something yourself. Give your attempts to resolve and clearly indicate what exactly is causing you difficulties.

@rlopez  I think for the beginner your method is the simplest and most understandable solution to the problem. When I wrote about two ways, I just forgot about this method. So we already have 3 ways.

@Markiyan Hirnyk  You did not answer my question (...to display successive frames of animation). On the animation bar (if you use  plots:-animate  command) there is a special button that allows you to do this.

@Markiyan Hirnyk  How do you expect to display successive frames of animation with a specific step in this case?