Finding formula for nth term in sequence

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  • Thread starter JimbleJambler
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  • #1
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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Answers and Replies

  • #2
mathman
Science Advisor
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Clarify your definition. You have [tex]f_n[/tex] defined in terms of itself as well as previous values of the index.
 
  • #3
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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.
 
  • #4
JimbleJambler
Clarify your definition. You have [tex]f_n[/tex] 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.
Sorry I fixed it!
 
  • #5
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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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