Spread Periodicity

The nthn^{\text{th}} spread polynomial, denoted by Sn(s)S_n(s), is recursively defined by S0(s)0S_0(s) \equiv 0, S1(s)sS_1(s) \equiv s, and

Sn(s)2(12s)Sn1(s)Sn2(s)+2s, S_n(s) \equiv 2(1-2s) S_{n-1}(s) - S_{n-2}(s) + 2s,

for n2n \geq 2. We will, for the context of this note, restrict our discussion to the finite field Fp\mathbb{F}_p, where pp is an odd prime. As an example, we see that in F3\mathbb{F}_3,

S0(0)=0s0(1)=0s0(2)=0 S_{0}(0) = 0 \quad s_{0}(1) = 0 \quad s_{0}(2) = 0

S1(0)=0s1(1)=1s1(2)=2 S_{1}(0) = 0 \quad s_{1}(1) = 1 \quad s_{1}(2) = 2

S2(0)=0s2(1)=0s2(2)=1 S_{2}(0) = 0 \quad s_{2}(1) = 0 \quad s_{2}(2) = 1

S3(0)=0s3(1)=1s3(2)=0 S_{3}(0) = 0 \quad s_{3}(1) = 1 \quad s_{3}(2) = 0

S4(0)=0s4(1)=0s4(2)=2 S_{4}(0) = 0 \quad s_{4}(1) = 0 \quad s_{4}(2) = 2

S5(0)=0s5(1)=1s5(2)=1 S_{5}(0) = 0 \quad s_{5}(1) = 1 \quad s_{5}(2) = 1

S6(0)=0s6(1)=0s6(2)=0 S_{6}(0) = 0 \quad s_{6}(1) = 0 \quad s_{6}(2) = 0

In fact, for any positive integer nn we will have that Sn(s)=Sn+6(s)S_n(s) = S_{n+6}(s). Thus, a natural question to be asked is whether this phenomenon is also seen for arbitrary finite fields; as it turns out, it does and in fact we have Sn(s)=Sn+k(s)S_n(s) = S_{n+k}(s) in Fp\mathbb{F}_p, where k12(p1)(p+1) k \equiv \frac{1}{2}(p-1)(p+1) . How do we prove that this is true? Are we able to discern other patterns within particular values of ss? If so, how do we ascertain this mathematically?

Note by A Former Brilliant Member
11 months, 1 week ago

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