GUY, FOCUS ON THIS π
I(B) is defined βΊ Ξ± Γ P(detect) > 1
The Math
Recurrence:
Closed form (take logs, solve Fibonacci, exponentiate back):
Where F_k is the k-th Fibonacci number: 0, 1, 1, 2, 3, 5, 8, 13...
With your seeds a_1 = 1,\ a_2 = 3,\ a_3 = 12:
Since a_1 = 1, the first term vanishes, so:
Verification:
n
F_{n-2}
12^{F_{n-2}}
Result
3
1
12^1
12 β
4
2
12^2
144 β
5
3
12^3
20,736 β
6
5
12^5
248,832
Wait β let me recheck term 6:
a_6 = a_5 \cdot a_4 = 20736 \times 144 = \mathbf{2,985,984} = 12^7... and F_4 = 3, F_5 = 5...
The Fibonacci indices just depend on how you seed F_0, F_1. The clean version:
The exponent is Fibonacci. The base is your branching constant (3Γ4 = 12).
