The Fibonacci series arose from an investigation into the growth of rabbit populations. Fibonacci proposed a model in which each pair of rabbits produced just two pairs in their lifetime, but always waited for one season before starting to reproduce.
Without this delay, the total population would increase exponentially, doubling from year to year. Table 1 shows the expected populating, increasing by a power of 2 each year.
Table 2 shows the effect of allowing the season delay.
Table 1 forms a Pascal Triangle, the numbers being the successive terms of the expansion of (1+1)n
Table 2 is formed by moving the successive rows of the triangle by one space to the right compared with the row above, the total number of rabbit families being 0, 1, 1, 2, 3, 5, …. This series, in which each term is the sum of the two preceding values is the Fibonacci series.
This is conveniently done using a spread sheet. After several seasons these numbers become very large (though less so than when there is no delay). Table 2a shows the figures for the first 59 seasons.
Looking at Table 2 we see that the population in the 7th season is 1 + 5 + 6 + 1 = 13, or 3 C 3 + 4 C 2 + 5 C 1 + 6 C 0 , which is another way in which values of the successive terms may be found.
Introducing the operator P, which generates each successive term of the series from its predecessor, three successive terms might be a, Pa, P2 a where 1 + P = P2 and a is the first of terms considered.
The quadratic x2 – x – 1 = 0 has roots x = ½( 1 ± √5). Calling the positive root P and the negative one Q we thus have:
PQ = –1, P+Q = 1, P = 1 + √5 = 1.618.. Q = 1 – √5 = –0.618..
and the operator turns out to be simply a number.
Any linear combination of P & Q will also behave like the Fibonacci numbers, so we can choose two multipliers, a and b, so that a + b = 0 and aP + bQ = 1 to launch the series.
Solving for a and b gives a = 1/√5 and b = –1/√5.
Successive terms f0, f1, f2, f3, . become
(P0 – Q0 )/√5, (P1 – Q1 )/√5, (P2 – Q2 )/√5, (P3 – Q3 )/√5, ….,
and in general fn = (Pn – Qn )/√5
[ To justify the choice of suffixes, we might have solved for a and b starting further along the sequence e.g.
a + b = 8 aP + bQ = 13 ( 8 = f6, 13 = f7 )
Solving for a and b: a√5 = 9 + 4√5 = P6 b√5 = 4√5 – 9 = Q6
f6, = a+ b = ( P6 – Q6 )/√5 ]
fn = (Pn – Qn )/√5 = (( 1 + √5)n – ( 1 – √5)n) / (2 n √5) = ( n√5 + 5 n C 3 √5 + 25 n C 5√5
+ …) / 2
By definition fm+1 = fm-1 + fm
So that fm+2 = fm + fm+1 = fm-1 + 2fm = fm-1f2 + fm f3
fm+3 = fm+1 + fm+2 = 2fm-1 + 3fm = fm-1f3 + fm f4
fm+4 = fm+2 + fm+3 = 3fm-1 + 5fm = fm-1f4 + fm f5
Leading to fm+n = fm-1fn + fm fn+1
For example f7 = f3+4 = f2f4 + f3 f5 = 1 x 3 + 2 x 5 = 13
or f7 = f4+3 = f3f3 + f4 f4 = 2 x 2 + 3 x 3 = 13
fmn is a multiple of fm and of fn
Putting Pm = A and Qm = B,
fmn = ( An – Bn )/√5
= (A–B)( An–1 B + An–2 B2 + An–3 B3 + …)/√5
but (A–B)/√5 = fm
therefore fmn is a multiple of fm and by symmetry is also a multiple of fn
These may be defined as f-numbers which are indivisible by any other f-number.
If mn is not prime, then fmn is clearly not prime. I believe that the converse is true, but have been unable to prove it.
PS In fact the sieve of Eratosthenes deals with the point.