Hacker Newsnew | past | comments | ask | show | jobs | submitlogin

Yes. One has the identity:

1/(1 - x - x^2) = 1 + x + 2x^2 + 3x^3 + ...

where the Fibonacci numbers are the coefficients on the right. Try writing it out! The basic idea is that because F_n = F_{n - 1} + F_{n - 2}, everything will neatly cancel out.

This is an example of a "generating function". Anyway, plug in x = 0.1, and then divide by 100 to see the behavior described in the post.

One also gets

1/9899 = 0.00010102030508132134559046368320032326498...

1/998999 = 0.0000010010020030050080130210340550891442334...

and so on.



More to the point this function trivially statisfies f(x)-1-x = X (f(x)-1) + X^2 f(X). You should be able to convince yourself that this is equivalent to the fact that its power series satisfies the Fibonacci recurrence, and that this power series starts with 1 + x.

You can also use this to easily find generating functions that satisfy other starting conditions.




Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: