c(1)=5, c(2)=29, c(n) = 6*c(n-1) - c(n-2)

( see https://www.research.att.com/~njas/sequences/A001653 )

What I find interesting is that apparently all Pythagorean triples can be generated with recurrence relations. I am far from being a mathematician, and so I cannot prove any of this.

It is sufficient to look at primitive triples where gcd(a,b)=1.
The possible values of the difference |b - a| in primitive triples are 1,7,17,23,31,... see https://www.research.att.com/~njas/sequences/A058529 .

When |b-a| = 7, we have :

c(1)=13, c(2)=17, c(3)=73, c(4)=97, c(n) = 6*c(n-2) - c(n-4)

The values of a will be generated by:

a(1)=5, a(2)=8, a(3)=48, a(4)=65, a(n) = 6*a(n-2) - a(n-4) + 14

When |b-a| = 17, we have :

c(1)=25, c(2)=53, c(3)=137, c(4)=305, c(n) = 6*c(n-2) - c(n-4)
a(1)=7, a(2)=28, a(3)=88, a(4)=207, a(n) = 6*a(n-2) - a(n-4) + 34

When |b-a| = 23, we have :

c(1)=37, c(2)=65, c(3)=205, c(4)=373, c(n) = 6*c(n-2) - c(n-4)
a(1)=12, a(2)=33, a(3)=133, a(4)=252, a(n) = 6*a(n-2) - a(n-4) + 46

and so on...

Does anyone know of an explanation please?

Regards
Andras Erszegi