[seqfan] Re: Gilbreath on Fibonacci
Jakob Schulz
bruderjakob17 at gmail.com
Tue Sep 19 10:25:09 CEST 2023
Dear Sean,
You can also just look back one row for a formal proof: Let d(n, 0) :=
F(n), and d(n, k+1) := |d(n, k) - d(n+1, k).
Claim: d(n, k) = ...
... F(n-k) if n >= k
... 0 if n <= k and n and k are equivalent mod 3
... 1 if n <= k and n and k are not equivalent mod 3
(Note that the patterns overlap if n=k, but they result in the same value.)
Now, do induction on k. The case k = 0 is trivial.
In the induction step, do a case distinction:
Case 1: n >= k + 1
Case 2: n < k + 1
...Case 2.1: n and k+1 are equivalent mod 3
...Case 2.2: n+1 and k+1 are equivalent mod 3
...Case 2.3: n+2 and k+1 are equivalent mod 3
Inserting the induction hypothesis and doing simple calculations (using
that F(n+1-k)-F(n-k) = F(n-(k+1)) in Case 1) finishes the proof of the
claim.
Best,
Jakob
El mar., 19 sept. 2023 4:01, Sean A. Irvine <sairvin at gmail.com> escribió:
> Hi,
>
> In terms of getting some traction on the sequences arising from the
> Gilbreath transform, it looks to me like the Gilbreath transform of the
> Fibonacci numbers which is apparently 0,1,1 repeated should be easy to
> prove. In this case, terms with index >= n in row n, are simply the
> Fibonacci numbers themselves and that after each set of three rows we have
> one additional copy of 0,1,1 at the start of the row. So I think you could
> make this more formal with some kind of induction looking back 3 rows?
>
> 1: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
> 2: 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
> 3: 1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
> 4: 0, 1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, ...
> 5: 1, 0, 1, 1, 0, 1, 1, 2, 3, 5, 8, 13, ...
> 6: 1, 1, 0, 1, 1, 0, 1, 1, 2, 3, 5, 8, ...
> 7: 0, 1, 1, 0, 1, 1, 0, 1, 1, 2, 3, 5, ...
> ...
>
> Sean.
>
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