Comments on A094865 from Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Aug 07 2008

Three methods of obtaining the sequence whose polynomials is characteristic of
the Cartan  A_4:5 - 20 x + 21 x^2 - 8 x^3 + x^4;

1) Matrix vector Markov: ( Bezout matrix form)
m = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {-5, 20, -21, 8}};
v(n) = m.v(n - 1).

2) recursion:
a(n) = 8*a(n - 1) - 21*a(n - 2) + 20*a(n - 3) - 5*a(n - 4).

3) Binet:
f(n)=(1/5)* 2^(-2 -n)* ((3 - Sqrt[5])^n(-5 + Sqrt[5]) - (-5 + Sqrt[5]) (5 + Sqrt[5])^n + (5 + Sqrt[5]) ((5 - Sqrt[5])^n - (3 + Sqrt[5])^n))

The importance of A_4 as a Cartan group is that it is associated with SU(5) which is the unbroken symmetry of the standard model of physics.

The root sum is:
a0 = Table[x /. Solve[5 - 20 x + 21 x^2 - 8 x^3 + x^4 == 0, x][[n]], {n, 1, 4}];
Apply[Plus, N[a0]]=8.

So the fifth balancing root is -8 in the polynomial:

Expand[(5 - 20 x + 21 x^2 - 8 x^3 + x^4)*(x + 8)];
40 - 155 x + 148 x^2 - 43 x^3 + x^5

1) Matrix vector Markov:
m = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {-5, 20, -21, 8}};
v(n) = m.v(n - 1).

2) Recursion:
a(n) = 8*a(n - 1) - 21*a(n - 2) + 20*a(n - 3) - 5*a(n - 4).

3) Binet:
f(n)=(1/5)* 2^(-2 -n)* ((3 - Sqrt[5])^n(-5 + Sqrt[5]) - (-5 + Sqrt[5]) (5 + Sqrt[5])^n + (5 + Sqrt[5]) ((5 - Sqrt[5])^n - (3 + Sqrt[5])^n))

Mathematica: (* 1*)
v[0] = {0, 0, 0, 1};
m = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {-5, 20, -21, 8}};
v[n_] := v[n] = m.v[n - 1];
Table[v[n][[1]], {n, 0, 30}]
(*2*)
a[0] = 0; a[1] = 0; a[2] = 0; a[3] = 1;
a[n_] := a[n] = 8*a[n - 1] - 21*a[n - 2] + 20*a[n - 3] - 5*a[n - 4];
Table[a[n], {n, 0, 30}]
(*3*)
Clear[f, g, n, a]
 f[n_Integer] = Module[{a}, a[n] /. RSolve[{a[n] == 8*a[n - 1] - 21*a[n -  2] + 20*a[n - 3] - 5*a[n - 4],a[0] == 0, a[1] == 0, a[2] == 0, a[3] == 1}, a[n], n][[1]] // FullSimplify] ;
Rationalize[N[Table[f[n], {n, 0, 25}], 100], 0]