An Abundant number is a number n for which the   sum of divisors   σ(n) > 2n,

or,   equivalently,   the   sum of proper divisors   (or aliquot sum)       s(n) > n.



E.G.

12   is abundant, it has the proper divisors     1,2,3,4 & 6     which sum to   16   ( > 12 or n);

       or alternately,   has the sigma sum of   1,2,3,4,6 & 12   which sum to   28   ( > 24 or 2n).



Abundant numbers are common, though even abundant numbers seem to be much more common than odd abundant numbers.

To make things more interesting, this task is specifically about finding   odd abundant numbers.



Task
Find and display here: at least the first 25 abundant odd numbers and either their proper divisor sum or sigma sum.
Find and display here: the one thousandth abundant odd number and either its proper divisor sum or sigma sum.
Find and display here: the first abundant odd number greater than one billion (109) and either its proper divisor sum or sigma sum.


References

   OEIS:A005231: Odd abundant numbers (odd numbers n whose sum of divisors exceeds 2n)
   American Journal of Mathematics, Vol. 35, No. 4 (Oct., 1913), pp. 413-422 - Finiteness of the Odd Perfect and Primitive Abundant Numbers with n Distinct Prime Factors (LE Dickson)


#!/usr/bin/python
# Abundant odd numbers - Python
 
oddNumber  = 1
aCount  = 0
dSum  = 0
 
from math import sqrt
 
def divisorSum(n):
    sum = 1
    i = int(sqrt(n)+1)
 
    for d in range (2, i):
        if n % d == 0:
            sum += d
            otherD = n // d
            if otherD != d:
                sum += otherD
    return sum
 
print ("The first 25 abundant odd numbers:")
while aCount  < 25:
    dSum  = divisorSum(oddNumber )
    if dSum  > oddNumber :
        aCount  += 1
        print("{0:5} proper divisor sum: {1}". format(oddNumber ,dSum ))
    oddNumber  += 2
 
while aCount  < 1000:
    dSum  = divisorSum(oddNumber )
    if dSum  > oddNumber :
        aCount  += 1
    oddNumber  += 2
print ("\n1000th abundant odd number:")
print ("    ",(oddNumber - 2)," proper divisor sum: ",dSum)
 
oddNumber  = 1000000001
found  = False
while not found :
    dSum  = divisorSum(oddNumber )
    if dSum  > oddNumber :
        found  = True
        print ("\nFirst abundant odd number > 1 000 000 000:")
        print ("    ",oddNumber," proper divisor sum: ",dSum)
    oddNumber  += 2