This is not what I need, though. There might not even exist a solution like that, I just want to simplify the sum a(1)^3 + a(2)^3 + ... + a(n)^3

The Generating Function for your coefficients is

g[x_] := (x - x^2 - x^4)/(1 - 2 x + x^2 - 2 x^3 - x^4 - x^5 + x^7 + x^8)

Your coefficients then are obtained by (which is sort of a symbolic expression)

a[n_] := 1/n! Derivative[n][g][0]

e.g.

In[25]:=aa= a /@ Range[30]

Out[25]= {1, 1, 1, 2, 6, 14, 28, 56, 118, 254, 541, 1140, 2401, 5074, \
10738, 22711, 48001, 101447, 214446, 453355, 958395, 2025963, \
4282685, 9053286, 19138115, 40456779, 85522862, 180789396, 382176531, \
807895636}

But this as well gives only numerical values, which might be calculated quicker by recursion. As check

With[{n = 15},
 {aa[[n]],2 aa[[n - 1]] - aa[[n - 2]] + 2 aa[[n - 3]] + aa[[n - 4]] +    aa[[n - 5]] - aa[[n - 7]] - aa[[n - 8]]}]

Unfortunately that doesn't give a reasonable acces to a[n]^3 and its sum.

What about the following:

a[n_] := a[n] = 
  2 a[n - 1] - a[n - 1] + 2 a[n - 3] + a[n - 4] + a[n - 5] - 
   a[n - 7] - a[n - 8]
a[1] = a1;
a[2] = a2;
a[3] = a3;
a[4] = a4;
a[5] = a5;
a[6] = a6;
a[7] = a7;
a[8] = a8;

Then you can write:

Sum[a[k]^3, {k, 1, 10}] // Expand

Then you will have your expression. After that you can substitute the numbers for the a1, a2, ...

Good luck

This gives a symbolic output:

recEq = {a[n] == 
    2 a[n - 1] - a[n - 2] + 2 a[n - 3] + a[n - 4] + a[n - 5] - 
     a[n - 7] - a[n - 8]};
sol = RSolveValue[recEq, a, n]
Map[Sum[#, {k, 0, n}] &, Expand[sol[k]^3]]

The initial conditions are harder to apply.

@Henrik

How did you get that?

First by paper and pencil following the procedures described in

https://www.amazon.de/s?k=herbert+wilf+generatingfunctionology&i=stripbooks&__mk_de_DE=%C3%85M%C3%85%C5%BD%C3%95%C3%91&ref=nb_sb_noss

Then I decided to make a procedure:

gF[k_] := ( a[1] x + Sum[   x^m (a[m] - Sum[c[j] a[m - j], {j, 1, m - 1}]), {m, 2, k}])/(
 1 - Sum[c[j] x^j, {j, 1, k}])

This is not quite the same as described in the book as it starts with a[1] x instead a[0].

Then with (the cc's are the coefficients in the recursion-formula)

cc = {c[1] -> 2, c[2] -> -1, c[3] -> 2, c[4] -> 1, c[5] -> 1,    c[6] -> 0, c[7] -> -1, c[8] -> -1};

and the initial values

r0 = {a[1] -> 1, a[2] -> 1, a[3] -> 1, a[4] -> 2, a[5] -> 6,    a[6] -> 14, a[7] -> 28, a[8] -> 56};

you get

In[4]:= gF[8] /. cc /. r0    
Out[4]= (x - x^2 - x^4)/(1 - 2 x + x^2 - 2 x^3 - x^4 - x^5 + x^7 + x^8 )

If you want more information or a notebook send me an email

h.dolhaine@gmx.de

@Henrik

Ich finde Ihre Beiträge oft sehr interessant. Würden Sie mir Ihre email geben? Gehen Sie auch manchmal in die dmug?

LG HD