 mersenneforum.org Gaussian Aliquot Sequences? How to run in Pari/GP?
 Register FAQ Search Today's Posts Mark Forums Read  2016-03-06, 18:32 #1 Stargate38   "Daniel Jackson" May 2011 14285714285714285714 22·151 Posts Gaussian Aliquot Sequences? How to run in Pari/GP? Anyone think of Gaussian aliquot sequences? This is the only page I could find on that subject (3rd problem). For example: divisors(2)=[1,1+i,1-i,-1+i,-1-i] sum(positive divisors(2))=(1+i)+1=2+i (prime) That terminates in just 1 step. Anyone know how to run this in pari/gp? I tried sumdiv(), but it includes all the conjugates (positive and negative) and everything ends up summing to 0. Last fiddled with by Stargate38 on 2016-03-06 at 18:35   2016-03-06, 19:03   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts Quote:
 Originally Posted by Stargate38 Anyone think of Gaussian aliquot sequences? This is the only page I could find on that subject (3rd problem). For example: divisors(2)=[1,1+i,1-i,-1+i,-1-i] sum(positive divisors(2))=(1+i)+1=2+i (prime) That terminates in just 1 step. Anyone know how to run this in pari/gp? I tried sumdiv(), but it includes all the conjugates (positive and negative) and everything ends up summing to 0.
the reason it might get 0 is that (1+i)+(-1-i) = 1-1+i-i = 0+0 =0 so in theory without having some way so they don't cancel there's only one possible answer 0. so you would have to limit it to positive divisors ( and depending on how you define that you'll get different answers because maybe even the integer part of the gaussian divisors cancel out and leave only a sum of imaginary numbers depending on how you define positive ( positive a positive b both of those). I suck at PARI/gp so I don't know but just my first thought. same thing happens on the integers themselves. if you don't limit to positive divisors they will all cancel out as well as (-2)(-1) = 2 =(1)(2) and these would cancel otherwise. edit: okay but the sum of proper divisors of 2+i is 1 so in theory the end of both sequences ( through the whole numbers and through the Gaussian numbers would either have to line up or there could be no one answer to what the aliquot sequence of any whole number is etc.) edit: doh it could also be you not using capital I instead of lowercase like I did when thinking about it.

Last fiddled with by science_man_88 on 2016-03-06 at 19:41   2016-03-07, 14:07 #3 CRGreathouse   Aug 2006 10111001010102 Posts If you have the divisors you can just do Code: sum(i=1,#v, if(imag(v[i])<0, 0, v[i]))   2016-03-07, 14:50   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20B116 Posts Quote:
 Originally Posted by CRGreathouse If you have the divisors you can just do Code: sum(i=1,#v, if(imag(v[i])<0, 0, v[i]))
so divisors with real part that's negative are allowed through ?   2016-03-07, 18:46 #5 science_man_88   "Forget I exist" Jul 2009 Dumbassville 8,369 Posts I've realized that gaussian primes can also be everywhere on the plane so my thought about what positive is doesn't matter: still not sure if the lines mean absolute value or complex modulus ( they use the same symbol and I think one is called the complex absolute value but I haven't read up enough on these things to say anything useful though). I'll admit I did try to progrma for it. divisors(2+0*I) is the same as divisors(2) in my Pari/gp so certain values with b=0 might be a problem to get all the divisors. Last fiddled with by science_man_88 on 2016-03-07 at 18:51   2016-07-23, 06:13 #6 AndrewWalker   Mar 2015 Australia 8210 Posts Hi you can use the following procedure I wrote in Pari, this uses the factor form where the factors are in the first quadrant, which nicely is what Pari gives for complex numbers! More in my next message. CSigma(n)= { local(s,v,i); s=1; v=factor(n); for(i=1,length(v~), if(abs(v[i,1])>1, s=s* (v[i,1]^(v[i,2]+1)-1)/( v[i,1]-1 ) ) ); return(s) } Eg: ? CSigma(-1105 + 1020*I) %68 = -3744 - 208*I ? CSigma(-1105 + 1020*I)-(-1105 + 1020*I) %69 = -2639 - 1228*I ? CSigma(-2639 - 1228*I)-(-2639 - 1228*I) %70 = -1105 + 1020*I   2016-07-23, 06:52   #7
AndrewWalker

Mar 2015
Australia

2×41 Posts Okay I've attached a list of amicable pairs I've found in a search done since this thread started.

When I saw this I remembered I had seen them discussed before, a search found a few references but I'm sure there must be others out there.

The online encyclopedia of integer sequences lists a few pairs submitted by T D Noe in 2005, but whether he discovered these, found any others or published any others I don't know.

https://oeis.org/search?q=gaussian+a...lish&go=Search

Sequences are A102924 (real part) and A102925 (imaginary part)

The only other reference I could find is this Masters Thesis by Ranthony Ashley Clark
http://encompass.eku.edu/etd/158/

They found 111 pairs (Table 3) although there is at least one duplicated pair and a typo
(1118256 + 70544i should be 118256 + 70544i )

My search takes this to 242 pairs but there may be a few from this paper I didn't find, I haven't checked completely.

Reference 5 in this thesis
Spira, Robert, \The Complex Sum of Divisors",
The American Mathematical Monthly, v. 68, February 1961, pp. 120-124

doesn't discuss amicable pairs but does discuss why the factor form I mentioned is a good one to use. Apparently it makes the sum of divisor function multipicative and gives it other nice properties. It is also the form used for gaussian mersenne primes and such which is discussed by Spira and on Wikipedia and elsewhere.

Anyway my search was hopefully complete up to gaussian integers with both complex and real parts less than 120,000 (in magnitude). This found some missed by Ranthony, however here things were slowing down a lot as it is a 2D search!

Some larger pairs were found using a list of small gaussian primes and multiplying 2, 3 or 4 together.

Others were found by looping over complex integers as before, but multiplying by
1+i, 1+2i or 1+4i, or a small power of one of these. This was inspired by how Ranthony had searched ( "common factors") and looking at the forms of the pairs they list.

I also did a pretty brief search for aliquot cycles among small values and found nothing.

Some questions from the above which may help future searches:

1) Do all gaussian amicable pairs have -i as the unit factor of at least one of the numbers (and never +i or -1)?

2) Do all gaussian amicable pairs have both numbers divisible by at least one of 1+i, 1+2i or 1+4i (and why these gaussian primes?)

3) Can methods used to find the normal amicable pairs, such as breeders, Thabit rules and the BDE method be put to any use here?

It would also be appreciated if anyone can find a way to automatically pretty print the pairs I've attached, I'd prefer not to do it manually!

Andrew
Attached Files GaussListMaster.txt (9.2 KB, 113 views)

Last fiddled with by AndrewWalker on 2016-07-23 at 07:21   2016-07-23, 07:37 #8 AndrewWalker   Mar 2015 Australia 2×41 Posts I'll add for reference the wikipedia page on the divisor function https://en.wikipedia.org/wiki/Divisor_function and the Mathworld page http://mathworld.wolfram.com/DivisorFunction.html which mentions Gaussian integers near the bottom. Last fiddled with by AndrewWalker on 2016-07-23 at 07:38   2017-02-18, 08:59 #9 garambois   Oct 2011 5348 Posts Does anyone know if there is an instruction in Sage software that can compute sigma (z), where z is an integer complex number of Gauss : the equivalent of the DivisorSigma[1, z, GaussianInteger-> True] function of Mathematica and sigma being the sum of the divisors extended to complex numbers according to Spira's proposition in 1961? See here for a better understanding of my question : https://oeis.org/A102506 or http://mathworld.wolfram.com/DivisorFunction.html   2017-02-18, 18:28 #10 Dr Sardonicus   Feb 2017 Nowhere 357410 Posts Interesting. If p is congruent to 1 (mod 4) then p = a^2 + b^2 (a, b positive integers). The two (ideal) factors of (p) in Z[i] then have generators a + b*i and b + a*i in the first quadrant [note that the product of these two complex numbers is (a^2 + b^2)*i]. The values of a and b can be determined using for example Pari's command qfbsolve((1,0,1),p). It is not clear to me what to do if p is congruent to 3 (mod 4). In this case, (p) is a prime ideal in Z[i], and the generators p and i*p would appear to have equal claim to the first quadrant. A different generalization of the usual sum-of-divisors function is the sum sigma of the *norms* of the integral *ideal* divisors of a given integral ideal. This is well-defined for any integral ideal (whether it is principal or not) in the ring of integers of any given number field. This sigma assumes positive integer values, and is a multiplicative function on the integral ideals of the given number field. That is, it is the product of its values for the prime-power ideal factors. In practice, its computation would require knowing the norms of the prime ideals, and the prime ideal factorization of the given integral ideal. In the case of a non-principal ideal, determining the factorization might also depend on how the ideal is given. In the ring of Gaussian integers, R = Z[i], the prime ideal Q = (1 + i)R has norm 2; sigma(Q^k) = 2^(k+1) - 1. If p is a prime congruent to 3 (mod 4) then P = pR is a prime ideal of norm p^2; sigma(P^k) = (p^(2*k + 2) - 1)/(p^2 - 1). If p is a prime congruent to 1 (mod 4), then pR splits into two prime ideal factors P1 and P2, each having norm p; sigma(P1^k) = sigma(P2^k) = (p^(k+1) - 1)/(p-1) Last fiddled with by Dr Sardonicus on 2017-02-18 at 18:30   2017-02-21, 10:05 #11 garambois   Oct 2011 22×3×29 Posts Thank you for your answer Dr Sardonicus. I ask me another question. If I had this instruction for calculate sigma(z) with z a Gaussian Integer, should I study the iterative process sigma(z) or sigma(z)-z (like iterative process sigma(n)-n with integers in the case of aliquot sequences) ? What would be the most interesting ?   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post devarajkandadai Software 0 2017-10-05 04:54 devarajkandadai Software 0 2017-07-11 05:42 schickel FactorDB 18 2013-06-12 16:09 Andi47 FactorDB 21 2011-12-29 21:11 schickel mersennewiki 0 2008-12-30 07:07

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