Related to this question:
For any given sequence of digits, does a Fibonacci number exist ending with such sequence?
If not, it would be nice to find the smallest counterexample. (in other words, for example, showing that no Fibonacci number ends in, let's say, 12345)
I tried several approaches (like in answers to the linked question), but none of them worked.
No fibonacci number is congruent to 4 or 6 modulo 8. They cycle [0,1,1,2,3,5,0,5,5,2,7,1] . So modulo 1000 there are numbers you can't reach.
More info:
$$\begin{align} \text{modulus} && \text{Fibonacci period} && \text{# of missing residues} \\ 2 && 3 && 0\\ 2^2 && 6 && 0\\ 2^3 && 12 && 2\\ 2^5 && 48 && 11 = 2^5 \cdot \frac{11}{32}\\ 2^8 && 3 \cdot 2^7 && 88 = 2^8 \cdot \frac{11}{32} \\ 2^{15} && 3 \cdot 2^{14} && 5637 = 2^{15} \cdot \frac{11}{32} \\ 2^{16} && 3 \cdot 2^{15} && 11264 = 2^{16}\cdot \frac{11}{32} \\ 2^{20} && 3 \cdot 2^{19} && 360448 = 2^{20} \cdot \frac{11}{32} \\ 5 && 20 && 0 \\ 5^2 && 100 && 0 \\ 5^3 && 500 && 0 \\ 5^4 && 2500 && 0 \\ 10 && 60 && 0 \\ 10^2 && 300 && 0 \\ 10^3 && 1500 && 250 \\ 10^4 && 15000 && 3125 = 10^4 \cdot \frac{5}{16}\\ 10^5 && 150000 && 34375 = 10^5 \cdot \frac{11}{32}\\ 10^6 && 1500000 && 343750 = 10^6 \cdot \frac{11}{32}\\ \end{align}$$
Conjectures:
Modulo $2^n$, the period is $3 \cdot 2^{n-1}$. For $n \ge 5$, exactly $\frac{11}{32}$ of the residues will not be present in the cycle.
Modulo $5^n$, the period is $4 \cdot 5^n$. All residues are present.
Modulo $10^n$, for $n \ge 3$, the period is $15 \cdot 10^{n-1}$. For $n \ge 5$, exactly $\frac{11}{32}$ of the residues will be missing.