Finding formula for nth term in sequence

[JimbleJambler]

$\displaystyle f_{n} = ((\sum_{k= \lfloor{\frac{n}{2}}\rfloor}^{n-1}f_{k}) mod 1) + 0.1$
$\displaystyle f_{1} = 0$

I really would like to know where to begin for finding a formula for the nth term, I wrote out a bunch of the terms and couldn't really eyeball a pattern of any sort. I noticed sometimes it seems to be that f(n) = (f(n-1) * 2) mod 1, but that's not very consistent.

 

[mathman]

Clarify your definition. You have $$f_n$$ defined in terms of itself as well as previous values of the index.

 

[mfb]

Are you sure the sum ends with n? You use f_n in the definition of f_n.
You can ignore the "mod 1" and apply it afterwards in a suitable way, that won't change the sum.

 

[JimbleJambler]

Clarify your definition. You have $$f_n$$ defined in terms of itself as well as previous values of the index.

Are you sure the sum ends with n? You use f_n in the definition of f_n.
You can ignore the "mod 1" and apply it afterwards in a suitable way, that won't change the sum.

Sorry I fixed it!

 

[willem2]

It's easier to work with integers mod 10. I can only simplify it a bit.
Since f(2k) = f(k)+ ... + f(2k-1) + 1,
f(2k+1) = f(k)+ ... + f(2k-1) +f(2k) +1 = 2f(2k) so the odd values aren't interesting
f(2k+2) = 4f(2k) - f(k)
The recurrence is rather like a shift register random number generator, only with a growing shift register.
https://en.wikipedia.org/wiki/Linear-feedback_shift_register