a step beyond Euclid and Fermat, 3

Let's make our first step beyond Euclid and Fermat. Their sequences are so similar that they beg to be placed under a roof of a common generalization. Let a b be two integers. Let's define:

• euf_{a b}(0)  :=  a+b;
• euf_{a b}(n+1)  :=  (euf_{a b}(n) - b) ⋅ euf_{a b}(n) + b
  

for every n=0 1...

Then

• euf_{a b}(n+1)  =  b + ∏_k=0...n euf_{a b}(k)
  

for every n=0 1...

Of course, if  B := euf_{a b}(k) - a  for certain non-negative integer k then:

• euf_{a B}(n)  =  euf_{a b}(k+n)

for every n=0 1...

For special values of integers a b, the sequence (euf_{a b}(n) : n=0 1...) becomes the Euclid sequence, when (a b) := (1 1), or the Fermat sequence, when (a b) := (2 3); when b=1 then we suppress  b  by writing

euf_a(n)  :=  euf_{a 1}(n)
Thus:

• e(n) = euf₁(n) — the Euclid sequence;
• f(n) = euf₂(n) — the Fermat sequence.

On the other hand, when a:=0, we obtain the constant sequence of values b.

The following properties of an Euclid-Fermat sequence euf_{a b} are equivalent:

• integers a b are relatively prime, i.e. gcd(a b) = 1;
• euf_{a b}(k) and euf_{a b}(n) are relatively prime whenever k and n are different.

We may rewrite the above simple recursive formula, which expresses euf_{a b}(n+1) in terms of euf_{a b}(n) as follows:

• euf_{a b}(n+1)  =  (euf_{a b}(n) - b/2)² + b - b²/4

This formula allows to study the rate of increase (or decrease) of an Euclid-Fermat sequence.