Thursday, 30 June 2016

Revisiting Fibonacci Constants

In a previous post (four years ago), I explained the Fibonacci Series - where it comes from, how it originates, and how it can appear in nature.  I also talked about the golden ratio, which is the ratio between subsequent terms in the Fibonacci Series.

I've been doing some further reading and research on the Fibonacci Series, and have been doing some of my own calculations and investigations.

Firstly:  what happens if we extend the series, so that instead of just summing the two previous terms, we sum the three previous terms, or the four or five previous?

This has been done before (I wasn't too surprised), and these are known as the following:
Fibonacci - 2 terms
Tribonacci - 3 terms
Tetranacci (or quadranacci) - 4 terms
Quintanacci (or pentanacci) - 5 terms
Hexanacci - 6 terms


I struggled to find names for the higher-number terms, so I'm going to submit my own.

Heptanacci - 7 terms
Octanacci - 8 terms
Nonancci - 9 terms
Decanacci - 10 terms

I stopped at 10, as I found that the data I'd accumulated was enough to draw some interesting conclusions from.  Here are the first few terms of each of the series:

Fib:  0,1,1,2,3,5,8,13,21,34,55,89,144,233,377
Trib:  0,0,1,1,2,4,7,13,24,44,81,149,274,504,927
Tetra: 0,0,0,1,1,2,4,8,15,29,56,108,208,401,773
Quint: 0,0,0,0,1,1,2,4,8,16,31,61,120,236,464
Hex: ...0,0,1,1,2,4,8,16,32,63,125,248
Hept:  ...0,0,1,1,2,4,8,16,32,64,127
Oct: ...0,1,1,2,4,8,16,32,64,128,255,509,1016,2028,4048,8080,16128
Non:  ...0,1,1,2,4,8,16,32,64,128,256,511,1021,2040,4076,8144
Dec: ..,0,1,1,2,4,8,16,32,64,128,256,512,1023,2045,4088

Taking this raw data, I then started to look at the ratios between subsequent terms.  We've seen previously that the Fibonacci series has the golden ratio 1.61803... or (1+ sqrt(5))/2 but what about the other series?

Fib 1.61803
Trib 1.83929
Tetra 1.92756
Quint 1.965948
Hex 1.983583
Hept 1.991964
Oct 1.996031
Non 1.998029
Dec 1.999019

I haven't identified the expressions for each of these but have found from research that the tetranacci constant satisfies x + x-4 = 2.  

Plotting the number of terms being summed (or the n-number for the series) against the ratio gives this graph.
The question is:  will the ratio ever reach 2?  It looks like the line will head towards 2 as an asymptote, but will it reach 2 if N increases?

The answer is no, and my proof is as follows:


Take the N=10 series as an example:
0,1,1,2,4,8,16,32,64,128,256,512,1023,2045,4088

We can see after the initial 0 and 1, that the first few terms are exactly double the previous one.  This is because each term is the sum of all the previous non-zero terms, including both of the 1s.  1+1 = 2, 1+1+2 = 4, 1+1+2+4=8 etc.  In this case, the ratio of each term to its previous term is 2, each term is exactly double the previous one.

However, this doubling of terms eventually ends:  when the sum no longer includes all the previous terms, that is to say, when the sum no longer includes both of the 1s and all subsequent terms, then the ratio falls below 2.  In the N=10 example above, the ratio falls below 2 when we reach 1023.


1+1+2+4+8+16+32+64+128+256 = 512
1+2+4+8+16+32+64+128+256+512 = 1023

At this point, the ratio falls below 2 (1023/512 = 1.998.  This fall will occur for any and all series which follow the "sum of previous terms" pattern; as N increases, it just takes more terms, and the final ratio will get closer to 2, but will remain below it.  As an aside, Wikipedia states that the ratio for an n-nacci series tends to the solution, x, of the equation  (no proof given, although my data confirms it).

Next:  I will look at what happens when we sum the previous term and
half of the term before that...e.g.  N = 1.5, N= 2.5, N=3.5 etc.