In my work, I've proved that the period of the pattern is 20, and so you can state that $[F_{[n]_{20}}]_5$ is equal to $[F_n]_5$. But this is still a recursive description of the pattern right? Is there any way that I can describe the pattern without using recursion?
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Why not $F_n\equiv F_{n+20}\mod 5$, or is that still to recursive? If so, a messy way of doing that might involve Bernoulli's formula. – Graviton Nov 08 '20 at 08:14
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Unfortunately, I think it is till too recursive. Okay, thanks, let me look into that! – VerilyThou Nov 08 '20 at 08:29
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One could use the methods explained in https://en.m.wikipedia.org/wiki/Linear_difference_equation , but work in the field of residues modulo $5$, which I will call $\mathbb Z_5$. In this field, the characteristic polynomial of the Fibonacci sequence:
$$x^2-x-1$$
has a double zero $x=3$, so the general solution of the Fibonacci recurrence relation in $\mathbb Z_5$ is:
$$F_n=(a+bn)\cdot 3^n$$
Tailoring $a,b$ to the initial conditions $F_0=0, F_1=1$ yields $a=0,b=2$, so: $F_n=2n\cdot 3^n$, or, back in $\mathbb Z$ we have $F_n\equiv 2n\cdot 3^n\pmod 5$.
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Could you explain a bit more how you got $$F_n=(a+bn)\cdot 3^n$$ from the characteristic polynomial of the Fibonacci sequence? – VerilyThou Nov 08 '20 at 08:55
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https://en.wikipedia.org/wiki/Linear_difference_equation#Solution_with_duplicate_characteristic_roots . $x=3$ is a double zero because $(x-3)^2=x^2-6x+9=x^2-x-1$ if all calculations are done $\pmod 5$. – Nov 08 '20 at 09:18
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