# [seqfan] Embedding Knuth's Fibonacci (circle) product in a ring defined over pairs.

Peter Munn techsubs at pearceneptune.co.uk
Sat Sep 10 14:31:34 CEST 2022

```Knuth uses a small circle for his associative product (A101330), so let's
use "o" for the product of pairs of my intended embedding.

Define (j1,k1) o (j2,k2) = (j1*j2 + k1*k2, j1*k2 + k1*j2 - k1*k2), with
(j1,k1) + (j2,k2) = (j1+j2, k1+k2) for addition as you might expect. This
seems to check out (with integer components) as a commutative ring,
possibly becoming a field if the components are rationals. [Definitely not
a field if the components are reals, as (1, -(sqrt(5)+1)/2) is a divisor
of zero.]

I rapidly run up against the limits of my knowledge here, and I'd be very
surprised if I am the first to define this ring, so a more general
characterisation of the ring and relevant pointers/references would be
much appreciated.

Anyway, it is easy to see that (1,0) is the multiplicative identity and
the integer ring Z embeds, j -> (j,0). Next, notice that multiplication by
(1,1) effects a generalised Fibonacci shift: (1,1) o (j,k) = (j+k,j).

The way I think Knuth's product embeds is k -> (f(k),k), where f(k) =
A022342(k+1), the Fibonacci successor to k (with f(0) = 0). If correct,
this gives (f(k1),k1) o (f(k2),k2) = (f(k1 o k2), k1 o k2). For example,
(7,4) o (7,4) = (65, 40) and 4 o 4 = 40, and using the relevant b-files I
find the identity correct for 1 <= k1,k2 <= 50.

If this is already well established by published research, perhaps we can
highlight this with a comment in A101330.

Best regards,

Peter

```