If you use the above triangle, then you get the sequence b1=1,b2=1,b3=2+1=3,b4=2+2=4,b5=3+4+1=8,b6=3+6+3=12,b7=4+9+7+1=21,....
In this way we get the sequence {1, 1, 3, 4, 8, 12, 21, ...}.
Here we colored the numbers to show how numbers on diagonals add to a sequence.
The above sequence satisfies the following equation.
f(1)= 1, f(2)=1 and f(n)=f(n-1)+f(n-2)+
We can generalize this fact.
Let p be a natural number. Similarly by using {U(p,n,m,1), n = 1, 2, 3, ...,10 and m = 1, ...,n} we get the following sequence of Definition 1.
Definition 3.
When p = 4, the sequence of Definition 3 is {1,1,2,3,6,9,15,24,40,64,104,168,273,441,714,1155,1870,3025,4895,7920}.
There is a very simple relation between this sequence and the Fibonacci sequence F(n).
We omit the proof of theorems and lemmas here. If you want to read proofs, please click “Proof.”
Theorem 3. Suppose that f(n) is the sequence of Definition 3 for p = 4. Then
We omit the proof of this theorem, since this is a special case of Theorem 5.
From the above relation it is easy to get the following elegant relation.
Theorem 4. Under the same condition of Theorem 3 we have the following relations.
Remark. It took more than 1 month for us to discover the above relations, and these relations are going to be published in Reference [8].
It is well known that F(n+1)/F(n) converges to the golden ratio when we make n bigger, and this is one of the reason why Fibonacci sequence is very useful in many field of science and technology.
By the above formula our sequence has the same property, and we are looking for new application of our sequences.
After we submitted these relations to the journal, we could generalize the formula and got the following theorem.
Theorem 5. Suppose that p=4q for some natural number q, then the sequence of Definition 1 satisfies the following relations. Here n is non negative integer.