Under what conditions is the number 2
^{n}*p an
abundant number? Where n is a positive integer and p is prime.
How about 3^{n}*p?
How about 2*p^{n}?
Recall an abundant number is one whose proper factors sum to more than the number. For example 18 is abundant because 1+2+3+6+9 > 18.
2^n * p has 2*(n+1) proper factors when p is not equal to 2, and 2*(n+2) proper factors if p = 2.
case 1  p>2: The 2*(n+1) factors are 1, 2, ... , 2^n and the same numbers, except for the last, times p. The total is (p+1)*((2^(n+1))1)  p*2^n. This is more than p*2^n when n >= 2.
case 2  p=2: 2^n * p = 2^(n+1), which has proper factors 1, 2, 4, ... , 2^n, which add up to 2^(n+1)  1, so in this case the number is never abundant.
In sum: 2^n * p is abundant unless p=2 or n = 1.
3^n * p has 2*(n+1) proper factors when p is not equal to 3, and 2*(n+2) proper factors if p = 3.
case 1  p<>3: The 2*(n+1) factors are 1, 3, ... , 3^n and the same numbers, except for the last, times p. The total without the "times p" is (3^(n+1)1)/2 as a repunit in base 3, making the total (p+1)*(3^(n+1)1)/2  p*3^n.
I don't think these ever become abundant.
case 2  p=3: 3^n * p = 3^(n+1), which has proper factors 1, 3, 9, ... , 3^n, which add up to (3^(n+1)  1)/2, so in this case also the number is never abundant.
2*p^n , in the case where p is 2, has already been covered as p*2^n where p is 2, and found not to be abundant.
2*p^n where p is not equal to 2 has 2*(n+1) proper factors: 1, p, p^2, ... p^n and also twice each of those except for the last. The first part of the sum has the total (p^(n+1)1)/(p1), so the overall sum is 2*(p^(n+1)1)/(p1)  2*p^n. These are also never abundant.
Of the three cases given, only 2^n * p has abundant subcases: those where p>2 and n>1.

Posted by Charlie
on 20151028 10:38:43 