0 like 0 dislike
0 like 0 dislike
Fibonacci numbers less than 1?
by

6 Answers

0 like 0 dislike
0 like 0 dislike
similar to how one can make an upside down Pascal's triangle
0 like 0 dislike
0 like 0 dislike
I've never really looked in to this so I might be wrong, but my first thought is that there probably isn't much to be learned about this extension beyond what can be learned about the normal Fibonacci numbers. The reason is that there doesn't seem to be any new behavior except for the alternating sign. These new numbers are just the same as the normal Fibonacci numbers aside from that. As a result, it seems like there's nothing really new to talk about.

Others have mentioned extending the fibonacci numbers to the complex plane. I've never studied that, but there's a good chance that by doing that you actually get new behavior and new things to study via complex analysis.
0 like 0 dislike
0 like 0 dislike
ye and you can do the same with the factorial :

3!=6

/3  → 2!=2

/2  → 1!=1

/1  → 0!=1

/0  → -1!=yes
0 like 0 dislike
0 like 0 dislike
From the pair (nth Fibonacci number,(n-1)th Fibonacci number) you can get the next pair by multiplying by the matrix

1 1

1 0

This matrix has inverse

0 1

1 -1

which you can use to find the "previous pair" (so for the one of interest, you take the difference of the current pair, as you say).

The first matrix is helpful in analysing the behaviour of the recursion rule on any starting pairs. There are two Perron-Frobenius eigenvalues. There's a leading or "expanding" direction, with eigenvalue value the golden mean, and a contracting direction with eigenvalue value -1/golden mean. That means almost all pairs converge towards the expanding eigenline, where the ratio converges to the golden mean. However, you can also take pairs on the contracting eigenline, where the ratio of successive terms is always exactly -1/golden ratio and they converge to 0.
0 like 0 dislike
0 like 0 dislike
As noted, the Fibonacci recursion is related to Pascal's triangle by diagonal sums. The next bump up in difficulty is the Catalan numbers, which are the analogue of the central binomial coefficients. That is, use Pascal's recursion with 1, -1 as the top of the triangle (now trapezoid) and the Catalan numbers go down the middle columns near the zeros..
0 like 0 dislike
0 like 0 dislike
Let me add one interesting thing about the reverse fibonacci numbers.
Start with any two numbers and apply the fibonacci algorithm. The ratio of adjacent numbers will approach the golden ratio (1+sqrt(5))/2 pretty fast. But the other direction?
It will actually approach the OTHER golden ratio, (1-sqrt(5))/2 . And yes, this works for any real numbers, not just 1 1.

Related questions

0 like 0 dislike
0 like 0 dislike
79 answers
coL_Punisher asked Jun 21
Regretting majoring in math
coL_Punisher asked Jun 21
0 like 0 dislike
0 like 0 dislike
61 answers
_spunkki asked Jun 21
Just ordered a Klein Bottle from Cliff Stoll. He sent me about 2 dozen pictures of him packing it up. Why is he so cute :)
_spunkki asked Jun 21
0 like 0 dislike
0 like 0 dislike
21 answers
Brands_Hatch asked Jun 21
Is set theory dying?
Brands_Hatch asked Jun 21
0 like 0 dislike
0 like 0 dislike
2 answers
a_dalgleish asked Jun 21
Contributing to the right math area, If all areas are equally curious
a_dalgleish asked Jun 21
0 like 0 dislike
0 like 0 dislike
5 answers
BrianDenver7 asked Jun 21
Is there a nice way to recast riemannian geometry in terms of principal bundles?
BrianDenver7 asked Jun 21

33.4k questions

135k answers

0 comments

33.7k users

OhhAskMe is a math solving hub where high school and university students ask and answer loads of math questions, discuss the latest in math, and share their knowledge. It’s 100% free!