Perfect 2 draft page

Perfect Numbers: A Solution

Once the first few numbers are found, a pattern emerges:

1 6 1,2,3,6

2 28 1,2,4,7,14,28

3 496 1,2,4,8,16,31,62,124,248,496

4 8128 1,2,4,8,16,32,64,127,254,508,1016,2032,4064,8128

The middle pair of factors are very distinctive: 2 & 3, 4 & 7, 16 & 31 and 64 & 127.The first is a power of 2, the second, one less than twice as much. (2^k, and 2^(k+1)-1, where k is an integer). But some powers of 2 are missing; what about 8, & 32?

8x15=120, and 32x63=2016. Neither are perfect numbers.

What is it that 3,7,31,127 have, and 15 & 63 don't have??

Yes, the 'perfects' are primes, and the other two are not!

But before following this up, (and it's to do with Mersenne primes...) consider the factor sums of the perfect numbers....

1+2+3+6 = 1+2+3x1+3x2

= 1x(1+3) + 2x(1+3)

= 1x4 + 2x4

1+2+4+7+14+28 = 1+2+4+7x1+7x2+7x4

= 1x(1+7) + 1x(1+7) + 1x(1+7)

= 1x8 + 2x8 + 4x8

1+2+4+8+16+31+62+124+248+496 = 1+2+4+8+16+1x31+2x31+4x31+8x31+16x31

= 1x(1+31) + 2x(1+31) + 4x(1+31) + 8x(1+31) + 16x(1+31)

= 1x32 + 2x32 + 4x32 + 8x32 + 16x32

1+2+ ............. +4064+8128 = 1x128 + 2x128 + 4x128 + 8x128 + 16x128 + 32x128 + 64x128

In each case, the perfect number is half the sum of a geometric series.

Number first term common ratio number of terms

6 4=2^2 2 2

28 8=2^3 2 3

486 32=2^5 2 5

8128 128=2^7 2 7

The series has an initial term 'a' which is 2^n; the common ratio, 'r', is 2, and there are 'n' terms.

(to be continued & the layout considerablyimproved!!)