[seqfan] Re: Project: sequences obtained from Gaussian Integers via Penney's binary method of encoding?

Kevin Ryde user42_kevin at yahoo.com.au
Sun Aug 28 11:51:36 CEST 2016

antti.karttunen at gmail.com (Antti Karttunen) writes:
> (but please tell if it is much older and discovered in some other country!)

Another early reference but which I haven't actually seen

    Solomon I. Khmelnik, "Specialized Digital Computer for Operations
    with Complex Numbers" (in Russian), Questions of Radio Electronics,
    volume 12, number 2, 1964.

> Or are there better ones for Gaussian Integers?

A few sequences go by square spiral (like Ulam plotted primes on).

> A066321 "Binary representation of base i-1 expansion of n:

Existing sequences I know,

    A066321    N on X axis, being the base i-1 positive reals
    A066322    diffs (N at X=16k+4) - (N at X=16k+3)
    A066323    N on X axis, number of 1 bits
    A137426    dY at N=2^k-1 (step to next replication level)
    A003476    boundary length / 2
                 recurrence a(n) = a(n-1) + 2*a(n-3)
    A203175    boundary length, starting from 4
                 (believe its recurrence is true)
    A052537    boundary length part A, B or C, per Gilbert's paper

"Boundary length" is with each Gaussian integer as a unit square.
(I think the diffs question in A066322 would be answered from Penney,
with X axis as hex digits 0,1,C,D, something, something :-).

The correspondence of base i-1 to twindragon interior unit squares means
various dragon curve sequences (like A003476) probably have
interpretations in i-1, but maybe at a stretch.

> "base i-1" is not really a good term to search for.

I've been happy enough with i-1 in non-oeis things.  A few people call
it twindragon because the limit fractal is the same, but for integers I
find that unclear.

Base i-m with other integer m is also possible, digits 0 to norm(i-m)-1.
Gilbert's boundary calculation is for that general case.  I think the
number of digits you then need in i-m to represent points relatively
close to the origin becomes even bigger than say the 5 bits you noted
for z=-1.


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