[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.

Before your reply, I seemed to have a good answer to only one item you
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
table for adding mantissas of logarithms, though the equivalent to adding
the characteristics is more complicated to define than for regular
(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,


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?)

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