The Proofs of Theorems and Lemmas in this Article Part 3.
Here we present proofs that are omitted throughout this article. The following proofs may look difficult, but we only use the definition of Fibonacci sequence and a well known simple formula.
The proof of Theorem 5
We are going to prove that the right side of (2) satisfy (1) of Definition 3 for p = 4q.
f(1)=1=F(1)F(2q)/F(2q).
f(2)=1=F(2)F(2q)/F(2q).
For k = 0,1,...,4q-3
f(4qn+k)+f(4qn+k+1) =(F(2qn+k)F(2qn+2q)+F(2qn+k+1)F(2qn+2q))/F(2q)
=(F(2qn+k+2)F(2qn+2q))/F(2q)=f(4qn+k+2) ...(3).
f(4qn+4q-2)+f(4qn+4q-1)= (F(2qn+4q-2)F(2qn+2q)+F(2qn+4q-1)F(2qn+2q))/F(2q)
=F(2qn+4q)F(2qn+2q)/F(2q)=f(4qn+4q) ...(4).
f(4qn+4q-1)+f(4qn+4q)=( F(2qn+4q-1)F(2qn+2q)+F(2qn+2q)F(2qn+4q))/F(2q)
=F(2qn+2q)F(2qn+4q+1)/F(2q) ...(5).
f(4qn+4q+1)=F(2qn+2q+1)F(2qn+4q)/F(2q) ...(6).
By using the well known formula
F(n-m)= F(m+1)F(n)-F(m)F(n+1)
we can get
F(2qn+2q)F(2qn+4q+1)+F(2q)=F(2qn+4q)F(2qn+2q+1) ...(7).
Therefore by (5), (6) and (7) f(4qn+4q-1)+f(4qn+4q)+1=f(4qn+4q+1) ...(8).
Therefore by (3), (4) and (8) the right side of (2) satisfy the condition of (1) of Definition 3,
and we can finish the proof.
The proof of Theorem 7.
We are going to prove that the right side of (13) satisfy (1) of Definition 3.
By the property of Fibonacci sequence it is easy to get
f(1)=1
and
f(2)=1=
For k = 0,1,...,p-3
f(pn+k)+f(pn+k+1)
=
=
=f(pn+k+2) ...(14).
f(pn+p-2)+f(pn+p-1)
=
=
=f(p(n+1)) ...(15).
f(pn+p-1)+f(p(n+1))+1
= f(p(n + 1) + 1) ...(16).
Therefore by (14), (15) and (16) the right side of (13) satisfy (1) of Definition 3,
and we can finish the proof.
=
=
=
=
=
=f(p(n+1)) ...(11),
where we used the fact that F(0) = 0.
f(pn+p-1)+f(p(n+1))+1
f(1)=1=
= f(p(n + 1) + 1) ...(12)
Therefore by (10), (11) and (12) the right side of (9) satisfy the condition of (1),
and we can finish the proof.
=f(pn+k+2) ...(10).
f(pn+p-2)+f(pn+p-1)
The proof of Theorem 6
We are going to prove that the right side of (9) satisfy (1) of Definition 3.
By the property of Fibonacci sequence it is easy to get
and
f(2)=1= .
For k = 0,1,...,p-3
f(pn+k)+f(pn+k+1)
