Super Golden Ratio

[Albert Chan]

“Rabbit Math” – A Formula for the Magical Fibonacci Sequence

We use the same trick, for "Super" Fibonacci numbers generation function

x/(1-x-x^3) = x + x^2 + x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 6*x^7 + 9*x^8 + ...

Binet-like direct formula for "Super" Fibonacci numbers, has this form:

S(n) = [k1, k2, k3] * [r1, r2, r3] .^ n

[r1, r2, r3] are reciprocal roots of 1-x-x^3 = 0 (or, roots of x^3 = x^2+1)
[k1, k2, k3] are constants to be solved, so that S(1) = S(2) = S(3) = 1

XCAS> R := 1 ./ proot([-1,0,-1,1]);
→ [-0.232786+0.792552*i, -0.232786-0.792552*i, 1.46557]

XCAS> K := proot([-1,0,3/31,1/31]);
→ [-0.208619-0.183825*i, -0.208619+0.183825*i, 0.417238]

XCAS> [K*R.^29, K*R.^30, K*R.^31]
→ [27201.0, 39865.0, 58425.0]

Both cubics are in depressed form, easy to get (K,R) exact form

w = exp(i*2*pi/3) :;
kb := (1/62 + sqrt(93)*3/1922)^(1/3) :;
ka := +1/(31*kb) :;
rb := (1/2 + sqrt(93)/18)^(1/3) :;
ra := -1/(3*rb) :;

K := [ka*w+kb/w, ka/w+kb*w, ka+kb] :;
R := 1 ./ [ra*w+rb/w, ra/w+rb*w, ra+rb] :;

approx(K,R) matched proot values.
We just need to confirm K coefficients produce S(1) = S(2) = S(3) = 1, exactly.

XCAS> S(n) := K * R.^n
XCAS> simplify([S(1), S(2), S(3)])

After 31 seconds, we get back [1, 1, 1]