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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