What is the sum of the aliquot (proper) divisors of 3+i? a) The divisors are {1, 1+i, 1+2i, 3+i}. The sum of the first three numbers is 3+3i. b) Robert Spira (1961) defined a "complex sum of divisors" that is incorporated into Mathematica's DivisorSigma function: In[1]:= DivisorSigma[1,3+I] Out[1]= 2+6 I Of course that answer includes the divisor 3+i, so we subtract that from 2+6i to get -1+5i. c) Neither of the above. Spira's sum-of-divisors function is (I think) a product. So subtracting the 3+i after calculating the "sum" is a leap of faith.
Many folks are unaware that 5 is a perfect number. But 1 + 2+i + 2-i = 5. --Rich ---- Quoting Hans Havermann <pxp@rogers.com>:
What is the sum of the aliquot (proper) divisors of 3+i?
a) The divisors are {1, 1+i, 1+2i, 3+i}. The sum of the first three numbers is 3+3i.
b) Robert Spira (1961) defined a "complex sum of divisors" that is incorporated into Mathematica's DivisorSigma function:
In[1]:= DivisorSigma[1,3+I] Out[1]= 2+6 I
Of course that answer includes the divisor 3+i, so we subtract that from 2+6i to get -1+5i.
c) Neither of the above. Spira's sum-of-divisors function is (I think) a product. So subtracting the 3+i after calculating the "sum" is a leap of faith.
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In the ring Z[i], the units (divisors of 1) are +1, -1, +i, -i. Thus if d is a divisor of 3+i, so are -d, id, -id. In taking the sum of divisors, how do you make a canonical choice among the associates of each divisor? If you take them all, the sum is zero. In the rational integers, one chooses the positive divisor. -- Gene
________________________________ From: Hans Havermann <pxp@rogers.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Wednesday, June 8, 2011 11:50 AM Subject: [math-fun] Some Sum
What is the sum of the aliquot (proper) divisors of 3+i?
a) The divisors are {1, 1+i, 1+2i, 3+i}. The sum of the first three numbers is 3+3i.
b) Robert Spira (1961) defined a "complex sum of divisors" that is incorporated into Mathematica's DivisorSigma function:
In[1]:= DivisorSigma[1,3+I] Out[1]= 2+6 I
Of course that answer includes the divisor 3+i, so we subtract that from 2+6i to get -1+5i.
c) Neither of the above. Spira's sum-of-divisors function is (I think) a product. So subtracting the 3+i after calculating the "sum" is a leap of faith.
On Wed, Jun 8, 2011 at 12:07 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
In the ring Z[i], the units (divisors of 1) are +1, -1, +i, -i. Thus if d is a divisor of 3+i, so are -d, id, -id. In taking the sum of divisors, how do you make a canonical choice among the associates of each divisor?
First quadrant. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
On Wed, Jun 8, 2011 at 12:45 PM, Marc LeBrun <mlb@well.com> wrote:
="Mike Stay" <metaweta@gmail.com> First quadrant.
Leaving only rational perfects?
Then 5 is imperfect after all?
Perfection can only be up to a unit factor. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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