20160306, 18:32  #1 
"Daniel Jackson"
May 2011
14285714285714285714
601_{10} 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,1i,1+i,1i] 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 20160306 at 18:35 
20160306, 19:03  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Last fiddled with by science_man_88 on 20160306 at 19:41 

20160307, 14:07  #3 
Aug 2006
3·5^{2}·79 Posts 
If you have the divisors you can just do
Code:
sum(i=1,#v, if(imag(v[i])<0, 0, v[i])) 
20160307, 14:50  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20160307, 18:46  #5 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 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 20160307 at 18:51 
20160723, 06:13  #6 
Mar 2015
Australia
1010010_{2} 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 
20160723, 06:52  #7 
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/ (google and google scholar have a few other links as well) 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. 120124 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 Last fiddled with by AndrewWalker on 20160723 at 07:21 
20160723, 07:37  #8 
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 20160723 at 07:38 
20170218, 08:59  #9 
Oct 2011
520_{8} 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 
20170218, 18:28  #10 
Feb 2017
Nowhere
3×13×89 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 sumofdivisors function is the sum sigma of the *norms* of the integral *ideal* divisors of a given integral ideal. This is welldefined 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 primepower 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 nonprincipal 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)/(p1) Last fiddled with by Dr Sardonicus on 20170218 at 18:30 
20170221, 10:05  #11 
Oct 2011
520_{8} 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 ? 
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