I Finding formula for nth term in sequence

AI Thread Summary
The discussion focuses on finding a formula for the nth term in a sequence defined by a recursive relationship. The initial term is given as f(1) = 0, and the formula involves a sum of previous terms with a modulo operation. Participants note inconsistencies in the pattern and suggest simplifying the approach by working with integers mod 10. The recurrence relations for even and odd indices are explored, revealing that odd values do not contribute significantly to the pattern. The conversation draws parallels between the recurrence and a shift register random number generator.
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.
 
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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 fn in the definition of fn.
You can ignore the "mod 1" and apply it afterwards in a suitable way, that won't change the sum.
 
mathman said:
Clarify your definition. You have f_n defined in terms of itself as well as previous values of the index.
mfb said:
Are you sure the sum ends with n? You use fn in the definition of fn.
You can ignore the "mod 1" and apply it afterwards in a suitable way, that won't change the sum.
Sorry I fixed it!
 
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
 
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