@one_man

Thank you for your interest! I was waiting for your question, because I know about your enthusiasm implicitly given curves. Setting the boundary of the figure is not provided in this way in my procedures, but still there is a simple solution to the problem. Here it is: 

P1 := plots[implicitplot]((x1+0.7e-1*sin(30*x2)^2)^2+(x2+0.7e-1*sin(30*x1)^2)^2-1 = 0, x1 = -2 .. 2, x2 = -2 .. 2, numpoints = 100000):

P2 := plottools[getdata](P1):

L := [convert(P2[3], listlist)]:

Perimeter(L);

Area(L);

plots[display](Picture(L, color = green, [color = black, thickness = 2]));

@Alejandro Jakubi 

Of course, my statement is not accurate from a mathematical point of view. Would be more accurate to write that there are examples which algsubs command decides better than applyrule command. Maybe you know the opposite examples?

Thanks for your comment!

@Alejandro Jakubi 

Of course, my statement is not accurate from a mathematical point of view. Would be more accurate to write that there are examples which algsubs command decides better than applyrule command. Maybe you know the opposite examples?

Thanks for your comment!

I didn't realize what  P[0] and  plotofmaxmin  stand for?

@Axel Vogt 

If we set

a:=cos(3*Pi/19)+cos(5*Pi/19)+cos(17*Pi/19):

b:=cos(Pi/19)+cos(7*Pi/19)+cos(11*Pi/19):

c:=cos(9*Pi/19)+cos(13*Pi/19)+cos(15*Pi/19):

then with Maple easily find

simplify([a+b+c, a*b*c, a*b+a*c+b*c]);

                    [1/2, -7/8, -3/2]

This result gave me the idea to associate in a single system known symmetric expressions  a+b+c, a*b*c, a*b+a*c+b*c  and unknown  expressions  a^(1/3)+b^(1/3)+c^(1/3),   a^(2/3)+b^(2/3)+c^(2/3)   and   a^(1/3)*b^(1/3)+a^(1/3)*c^(1/3)+b^(1/3)*c^(1/3) .  

To make up such a system, I used three identities, which are valid for any reals

(a^(1/3)+b^(1/3)+c^(1/3))^3=3*(a^(1/3)+b^(1/3)+c^(1/3))*(a^(2/3)+b^(2/3)+c^(2/3))-2*(a+b+c)+6*a^(1/3)*b^(1/3)*c^(1/3),

a^(2/3)+b^(2/3)+c^(2/3)=(a^(1/3)+b^(1/3)+c^(1/3))^2-2*(a^(1/3)*b^(1/3)+a^(1/3)*c^(1/3)+b^(1/3)*c^(1/3)),

(a^(1/3)*b^(1/3)+a^(1/3)*c^(1/3)+b^(1/3)*c^(1/3))^3=(a*b+a*c+b*c)+6*a^(2/3)*b^(2/3)*c^(2/3)+3*a^(1/3)*b^(1/3)*c^(1/3)*(a^(1/3)+b^(1/3)+c^(1/3))*(a^(2/3)+b^(2/3)+c^(2/3))-3*(a+b+c)*a^(1/3)*b^(1/3)*c^(1/3)

Firstly I got these identities by hand and then checked in Maple. They based on the theory of symmetric polynomials. It is known that any symmetric polynomial can be expressed as a polynomial in the elementary symmetric polynomials. My first two identities follow from the expressions

x^3 + y^3 + z^3 = 3*x*y*z + (x + y + z)^3 - 3 (x + y + z)*(x*y + x*z + y*z)  and

x^2 + y^2 + z^2 = (x + y + z)^2 - 2*(x*y + x*z + y*z) 

in which it was substituted

x = a^(1/3) ,  y = b^(1/3) ,  z = c^(1/3)   and

the third identity is obtained by hand through expansion of

 (a^(1/3)*b^(1/3)+a^(1/3)*c^(1/3)+b^(1/3)*c^(1/3))^3

Carl Love! 

Thanks for the elegant proof. I wonder whether can your approach  find the value of LHS in the real radicals if RHS is unknown?

I propose a different proof of this remarkable identity in which  directly constructed a polynomial, whose root is the value of LHS, and this is expressed in radicals. As this approach is completely unrelated to your method, I am creating a separate post.

@Carl Love

I mean only real radicals without complex numbers.

@Markiyan Hirnyk

Yes, you are right! With my method of counting, there are duplicates. The method will be true only if  nops(A intersect B)=3 . 

@Markiyan Hirnyk

Yes, you are right! With my method of counting, there are duplicates. The method will be true only if  nops(A intersect B)=3 . 

@Markiyan Hirnyk

Let  A union B ={a, b, c, d, e} is a subset of  S. Then  A  and  B  are subsets of  {a, b, c, d, e} .  From the second condition about the intersection follows that  nops(A intersect B)>=3 . Suppose  nops(A intersect B)=3 . The number of such pairs  A  and  B  is  m (see code). Set C, which intersects with (A intersect B) in these 3 elements can be chosen in 2^(30-3) ways  and so on. 

@Markiyan Hirnyk

Let  A union B ={a, b, c, d, e} is a subset of  S. Then  A  and  B  are subsets of  {a, b, c, d, e} .  From the second condition about the intersection follows that  nops(A intersect B)>=3 . Suppose  nops(A intersect B)=3 . The number of such pairs  A  and  B  is  m (see code). Set C, which intersects with (A intersect B) in these 3 elements can be chosen in 2^(30-3) ways  and so on. 

@Carl Love 

You are right. But this inaccuracy can be easily fixed.

@Carl Love 

Thank you for a shorter and more elegant version of the procedure.

@Axel Vogt

The original problem is not a problem of interpolation. It was necessary to find a pattern in the first eight terms of the sequence and find a general term of this sequence to this pattern persisted on the next terms.

The procedure  GeneralTerm  finds  n_nd term of your sequence:

GeneralTerm := proc (n)

local a, i;

a[1] := 2;  a[2] := 1;

for i from 3 to n do

if irem(i, 3) = 0 then a[i] := a[i-2]+a[i-1] elif

irem(i, 3) = 1 then a[i] := a[i-2]+a[i-1]+2 elif

irem(i, 3) = 2 then a[i] := a[i-1]-1;

end if; end do;

a[n];

end proc:

Example:  the first 20 members of your sequence:

seq(CommonTerm(k), k = 1 .. 20);

2, 1, 3, 6, 5, 11, 18, 17, 35, 54, 53, 107, 162, 161, 323, 486, 485, 971, 1458, 1457

@Markiyan Hirnyk 

I do not see any problems to write code this animation, but I'm not going to do it for you.