Sum of divisors

A divisor of an integer n, also called a factor of n, is an integer which divides n without leaving a remainder

http://en.wikipedia.org/wiki/Divisor

Here, I take the sum of the divisors of integers, then I'll be searching for squares and cubes such that the sum of its divisors is a square and a cube respectively.

Check my old post for 3-digit, 4-digit, 5-digit and 6-digit square numbers:

http://benvitale-funwithnum3ers.blogspot.com/2010/11/squares-looking-for-digits.html

The first square numbers are 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, ...

Here is a list of perfect cube numbers: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, ...

Now, I write their divisors and find their sums.

The factors of 4 (=2^2) are: 1 2 4

1 + 2 + 4 = 7

The factors of 8 (=2^3) are: 1 2 4 8

1 + 2 + 4 + 8 = 15

The factors of 9 (=3^2) are: 1 3 9

1 + 3 + 9 = 13

The factors of 16 (=4^2) are: 1 2 4 8 16

1 + 2 + 4 + 8 + 16 = 31

The factors of 25 (=5^2) are: 1 5 25

1 + 5 + 25 = 31

The factors of 27 (3^3) are: 1 3 9 27

1 + 3 + 9 + 27 = 40

The factors of 36 (=6^2) are: 1 2 3 4 6 9 12 18 36

1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 = 91

The factors of 49 (=7^2) are: 1 7 49

1 + 7 + 49 = 57

64 is an interesting number because it is both a square (8^2) and a cube (4^3)

The factors of 64 are: 1 2 4 8 16 32 64

1 + 2 + 4 + 8 + 16 + 32 + 64 = 127

The factors of 81 (=9^2) are: 1 3 9 27 81

1 + 3 + 9 + 27 + 81 = 121 = 11^2

So, 81 is the first square whose sum of its divisors is a square.

I need to find a cube.

The search continues ...