# Fibonacci Squares with Arbitrary Starting Numbers

Consider the Fibonacci Sequence

0,1,1,2,3,5,8,13…

Each number in the sequence is the sum of the two that precede it. The sequence is defined as have 0 and 1 as its first numbers. But what if we chose different numbers to start with?

We’re going to make something I call a Fibonacci Square. It is a little bit of Fibonacci, a little bit of Pascal’s Triangle, and a little bit of my own creativity.

Let’s choose two arbitrary numbers to begin with. Let’s say, 6 and 2. We’ll call 6 f (the first number), and 2 s (the second number). We will now write our “Fibonacci” sequence as normal, but using f and s as our starting numbers instead. For the purpose of our “square” we are constructing, we’ll only write the first four terms for now.

6, 2, 8, 10

Okay, so we’ve got that. Now to continue the pattern we will start a new line below the one we’ve just written, and replace s (the previous second number) with the fourth number generated in our sequence above. This just means that our starting numbers (f and s) are now 6 and 10, respectively. Let’s write a new 4 term line with them.

6, 10, 16, 26

So our “square” so far is:

6,   2,   8, 10
6, 10, 16, 26

We can make more lines as much as we want with this technique. Just take the fourth number and replace it as the second starting number in the next line. The third line in this sequence would be:

6, 26, 32, 58

What is interesting to note is that a formula can be derived to determine what the fourth number on any given line will be. In fact, the formula can be applied to any length lines greater than 4. That means that if we put 5 terms on each line instead of 4, our formula would still work.

Let’s check out the formula:

$(F(t)^n * (s+\frac{F(t-1)}{F(t)-1}f))-\frac{F(t-1)}{F(t)-1}f$

• F(t) is the t’th term of the Fibonacci Sequence (defined as starting at 0)
• t = the number of terms on each line (in our example there are 4)
• s = the number that we originally had as our “second number”
• n = the line number, starting at one. So for the three lines we calculated, they would have been n = 1, 2, and 3, respectively.

So, for example, we wondered what the number would be on the end of the 20th line of the sequence we started above, but we don’t want to write all that out. We can plug the values into our equation to find the answer.

$(F(4)^{20} * (2+\frac{F(4-1)}{F(4)-1}6))-\frac{F(4-1)}{F(4)-1}6$

If we condense that down (remember that F(4) means find the 4th term of the Fibonacci Sequence, which is 2), we get that the value on the end of the twentieth line in this sequence is 8,388,602. Wow! As you can see, when one creates a sequence in this manner, there is exponential growth.

Anyway, this is one of my pet projects and I hope you will find it interesting. I will probably make another post on other things I have found with this sequence, but this post is getting a bit long for now.