@Markiyan Hirnyk Here's a plot illustrating the corollary. Adjust n to any positive integer. The convergence of the maximum modulus to 2 is very rapid.

n:= 6:
R:= fsolve(z^n = add(z^k, k= 0..n-1), complex):
plots:-display([
     plot([Re,Im]~([R]), style= point, symbol= diagonalcross, symbolsize= 16),
     plot([1,2], t= 0..2*Pi, coords= polar, thickness= 0, color= blue)
     ], scaling= constrained, axes= boxed, view= [(-2.1..2.1)$2]
      , labels= [Re,Im]
);

I also modified the code in the Answer to show clearly what I meant by union~.

@Abdoulaye This plot command shows that the function is close enough to being 2*Pi periodic that there is no visible gap between the plots:

Fu:= eval(F, piecewise= piecewise[undefined]):
plot([Fu, eval(Fu, t= t-2*Pi)], t= op([1,1,1], F)..op([-2,-1,-1], F)+2*Pi);

On the other hand, clearly it is not exactly periodic because the horizontal piece 0.4977821578 isn't an exact match to the horizontal piece 0.4977820960.

But I wonder what is your true goal here. Are you actually trying to construct a periodic function by duplicating one fundamental period? If so, then that is easy to do, and I can show you how. Or are you truly trying to prove that a given function is periodic?

@shakuntala I already told you that you can plot using the command

plot([seq(X2||k, k= 1..1)], 0..blt);

Did you have trouble with that? Given that you actually only have one value of k in your original loop, you can simplify that to

plot(X21, 0..blt);

But it's X21, not X2. What gives you trouble in understanding that? Are you trying to follow some printed instructions that tell you to plot X2?

Please put Replies to an Answer below that Answer, not above.

@Abdoulaye Please either present the function in a plaintext form or attach a worksheet. I can't use it in the form that you present.

@shakuntala Well, like I said, X2 by itself is meaningless. There's

X21(0);

     .332057384255589

There's no need to use print.

@Markiyan Hirnyk I redefine A and S analogously for the three-consecutive problem. We have S[n] = S[n-1] union {members of S[n-1] that it's safe to add n to, unioned with {n}} union {members of S[n-1] that it's safe to add both n and n-1 to unioned with {n, n-1}}. So A[n] = A[n-1] + A[n-2] + A[n-3]. So, the asympototic growth factor is the maximum modulus of the roots of z^3 = z^2 + z + 1, which is 1.839....

Corollary: For any positive integer n, the maximum modulus of the roots of z^n = sum(z^k, k= 0..n-1) is strictly between 1 and 2.

@Markiyan Hirnyk I'm still thinking about the case of three consecutive numbers.

@Markiyan Hirnyk See the Maple code that I just added to the Answer. You can check that every member of the sequence S contains the empty set {}. Anyway, the empty set doesn't contain n+1 either!

P.S. I think that I see what you're saying. My union~ was a poor choice of symbols, and it cannot be literally translated into Maple code. However, the actual Maple code that I added works, and produces the same sets as does your code.

@Thomas Richard I was naively answering the Question as posed. I didn't realize that this was about solving an ODE. Of course, I see that now that you've pointed it out.

Hmm, I can only find one reference to this method: Susanne Raynor, "Cubic Spline Method for Solving Second-Order Differential Equations Theory and application to the Thomas-Fermi Model for Ions", Chemical Physics, Volume 66, Issue 3, 15 April 1982, Pages 409-415.

Unfortunately, it's behind an Elsevier pay wall.

@tomleslie 

Okay, I crashed it. Setting Digits:= 10 causes it to run very slowly and crash on the fourth iteration; so, same as Tom. Using Digits:= 15, it runs pretty fast, and without error.

@digerdiga No, you totally misunderstood me. I had no doubt that it crashed for you. I was just saying that it didn't crash for me to indicate that I'm rather limited in analyzing that situation further.

Oddly enough, I have the same version as Tom, 1133417.

@digerdiga Trying your original code in Maple 2016, I don't get a crash.

@digerdiga No, it has nothing to do with the log. If you use int, then every time that you call the procedure, it will attempt to do the integral symbolically first. That attempt will fail and will be in vane---a complete waste of time.

@digerdiga No, Int is not really needed. But compare the time difference between using Int and int.