# Sum of digits in Fibonacci sequence

This is my first question here so please go easy on me.

If you add the digits of each number on the Fibonacci sequence until your number is less than 10, it seems that you get a pattern of 24 numbers which is:

1 2 3 5 8 4 3 7 1 8 9 8 8 7 6 4 1 5 6 2 8 1 9 1

Of course, I didn't prove that, I just used Python to calculate the first 10000 numbers so I believe that the statement may be true.

My questions are many...

The procedure you describe, applied to $n$, finishes with the remainder of $n$ when divided by $9$. As the remainder behaves well with respect to the sum ( $r_9(r_9(a)+r_9(b))=r_9(a+b)$) you can work always with remainders. It is easy to see that two consecutive terms determine the whole sequence and there are a finite number of possible pairs, so that the sequence has to become periodic.
• More precisely, you don't obtain the remainder if $n>0$ is a multiple of $9$, inwhich case you obtain $9$ instead of $0$. But that doesn't destroy the (eventual) periodicity. – Hagen von Eitzen Feb 6 '14 at 18:49