kth order Fibonacci numbers series [closed]

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I am reading an article on Fibonacci numbers at following link

http://xlinux.nist.gov/dads/HTML/kthOrderFibonacci.html

F(k)n = 0 for 0 ≤ n ≤ k-2

i am not getting what about above statement.

For example when k = 3 and n =2, 0 <= 2 < 1 which is not making sense? can any one please elaborate and pls give an example first 10 numbers 3rd order Fibonacci numbers

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The statement you quoted indicates that the first k-1 numbers in the sequence are zero.

if f(k,n) is zero for all n such that 0 <= n < k-2, then f(3, n) is zero for all n such that 0 <= n <= 1. So f(3,0) and f(3,1) are both zero.

Second Order:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34...

Third Order:

0, 0, 1, 1, 2, 4, 7, 13, 24, 44...

Fourth Order:

0, 0, 0, 1, 1, 2, 4, 8, 15, 29...

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For k=3 and n=2, you are looking at the wrong part of the definition. In your case, n = k-1, so you would you the second part of the definition or, F(k)k-1 = 1, so when k=3 and n=2, f(k) = 1.

For 3rd order, n=0 to n=10, you would have 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81

edit for not being able to add =)

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Basically you can't sum the k values preceding n if n < k - 1, simply because there aren't enough numbers. :) as for your example, since n = k - 1 then f(n = 2) = 1.

n    f    reason
--------------------------------------------------
0    0    by definition (because n <= k - 2 = 1)
1    0    see above
2    1    by definition (because n = k - 1 = 2)
3    1    1 + 0 + 0
4    2    1 + 1 + 0
5    4    2 + 1 + 1
6    7    4 + 2 + 1
7    13   7 + 4 + 2
8    24   14+ 7 + 4