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

Peter Munn techsubs at pearceneptune.co.uk
Sun Sep 11 17:28:46 CEST 2022

```Many thanks for your characterisation, Don. You have speeded me forward
greatly. Especially, I didn't know where to start to define a norm.

covered: the subject of your puzzle... the units over integer pairs are
(F_{i+1}, F_i), i in Z, together with their negations. [That is ...,
(-1,2), (1,-1), (0,1), (1,0), (1,1), (2,1), (3,2), (5,3), ... ; also
negated. Extract either component from each pair to see the bi-infinite
Fibonacci series.]

I plan a multipication table based on the extended Wythoff array. I make
use of extending the array infinitely to the left as eW(.,.), because
(eW(n,k+1), eW(n,k)) and (-eW(n,k+1), -eW(n,k)) give all possible integer
pairs (excepting 0,0) exactly once.

The "positive" units are exactly the pairs (eW(0,k+1), eW(0,k)) that come
from the first row of the array (for which a row index of 0 seems to work
most neatly). The pairs from any other row are related to each other by a
positive unit multiple. So we can multiply two rows of the array to give
another row, for example:
- (3,1) is seen (reversed) on row 1 (indexed as above)
- (3,1) o (3,1) = (10,5) by my definition of "o"
- (10,5) appears reversed on a row that includes ..., 5, 10, 15, 25, 40,
65 ...
- this is row 15, which enters the unextended array (A035513) at 40, 65, ...
- so row 1 "multiplied by" row 1 gives row 15.

It's possible to show we will not get an answer that is in a negated row.

My estimation that this is valuable comes from a few anticipations:
(1) demonstration that this multiplication table is akin in usage to a
the characteristics is more complicated to define than for regular
logarithms;
(2) the full multiplication will be shown to be strongly related to
Knuth's Fibonacci multiplication;
(3) the multiplication will be shown to be related to the asymptotic
relationship of rows of the arrays as the column goes to infinity.

If anyone wishes to contribute, I've made a start at A357097.

Best regards,

Peter

On Sun, September 11, 2022 3:33 am, Don Reble via SeqFan wrote:
>> 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.
>
>     It's easy to map this ring into the real numbers.
>     Suppose that (a,b) = a+bx (for some real or complex x),
>     and that (a,b)o(c,d) = (a+bx)(c+dx) (ordinary multiplication).
>     Then (a,b)o(c,d) = ac + bdx^2 + (ad+bc)x.
>     If that equals (ac+bd) + (ad+bc-bd)x, then x^2 = 1 - x,
>     and x = (-1 +- sqrt(5)) / 2. Either mapping is good.
>
>> This seems to check out (with integer components) as a commutative
>> ring, possibly becoming a field if the components are rationals.
>
>     Those (any) subrings of the real numbers are commutative.
>
>     Let x,X be the two roots of that quadratic: xX = -1, x+X = -1.
>     The norm of (a+bx) is (a+bx)(a+bX)
> 	= a^2 + ab(x+X) + b^2(xX)
> 	= a^2 - ab - b^2
>     And 1/(a,b)
> 	= (a+bX) / [(a+bx)(a+bX)]
> 	= [a+b(-1-x)] / (a^2 - ab - b^2)
> 	= (a-b,-b) / (a^2 - ab - b^2)
>     For nonzero a,b, if the denominator is zero, then
>     (a/b)^2 - (a/b) - 1 = 0, and a/b is irrational.
>     So with rational coefficients, it's a field.
>
>
>     Today's puzzle: what are the units of the integer-coefficient ring?
>     (For which integral (a,b) is a^2-ab-b^2 = +1 or -1?)
>

```