@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.

@Markiyan Hirnyk

It is not necessary to build the animations having explicit parametric equations. Maple can plot this curve, for example, by plots:-intersectplot . Therefore it is possible to extract the data by, for example,  plottools:-getdata  and use them in the future for animation.

@Markiyan Hirnyk 

Why do not you ask your question to the author of this animation?

There specified an equation of a space curve, on which green hinge moves.

{(x1-2)^2+x2^2+(x3-2)^2-9=0, x1^6+x2^6+x3^6-4=0}

Therefore, there is no problem to write code for the animation.

Markiyan Hirnyk  

If you  in your post use the result obtained by not you, but another person, you should give the link. Since you did not, then I will have to do

http://forum.exponenta.ru/viewtopic.php?t=11284&postdays=0&postorder=asc&start=30

@Markiyan Hirnyk 

First, I build list  L . It is a list of frequencies for the sum of the digits in a sequence of six digits,  ie how many times this or that sum of of digits occurs.

L = [[0, 1], [1, 6], [2, 21], [3, 56], [4, 126], [5, 252], [6, 462], [7, 792], [8, 1287], [9, 2002], [10, 2997], [11, 4332], [12, 6062], [13, 8232], [14, 10872], [15, 13992], [16, 17577], [17, 21582], [18, 25927], [19, 30492], [20, 35127], [21, 39662], [22, 43917], [23, 47712], [24, 50877], [25, 53262], [26, 54747], [27, 55252], [28, 54747], [29, 53262], [30, 50877], [31, 47712], [32, 43917], [33, 39662], [34, 35127], [35, 30492], [36, 25927], [37, 21582], [38, 17577], [39, 13992], [40, 10872], [41, 8232], [42, 6062], [43, 4332], [44, 2997], [45, 2002], [46, 1287], [47, 792], [48, 462], [49, 252], [50, 126], [51, 56], [52, 21], [53, 6], [54, 1]]

For example, the sub-list  [30, 50877]  means that the sum of digits equal to 30 in the 6-digit sequence occurs 50877 times. The same means the list  M , only for 5 digits.

Note that 23 = 6 +6 +6 +5. Therefore  I find all representations of  N , which is given as argument of the procedure, as a sum of 4 numbers (the first number from 0 to 54, and so on). Get the list  S .

Then I go through the list  S in the cycle , and, for example for N = 100, element of the list  [20, 35, 30, 15]  in accordance with the well-known rule of combinatorics (rule of product) gives me  35127*30492*50877*13992  variants.

 .