Theorem: All even perfect numbers are on the third diagonal of Pascal's Triangle.

Proof: The proof is detailed in the following paper.

The Perfect Numbers and Pascal's Triangle by Robert A. Antol Iowa State University original copyright (c) 1974

   (1)   n=(p1**a1)*(p2**a2)*...*(pr**ar)

A positive integer is called a perfect number if it is equal to the sum of all its positive divisors other than itself. The sum of divisors of a number n with the prime factorization (1) is

   (2)   s(n)=((p1**(a1+1)-1)/(p1-1))*((p2**(a2+1)-1)/(p2-1))*...
              ...*((pr**(ar+1)-1)/(pr-1))

The condition for a perfect number may then be given by n=s(n)-n or equivalently, s(n)=2n.

Euclid argued that if 2**p-1 is prime for p>1, then

   (3)   P=(2**(p-1))*(2**p-1)

For any positive integer m and any integer k satisfying 0<=k<=m, the binomial coefficient (m k) is defined by

   (4)   (m k)=m!/(k!*(m-k)!)

Use will now be made of the configuration known as Pascal's Triangle in which the binomial coefficient (m k) appears as the (k+1)st number in the (m+1)st row, as in [5].

                  k=0
                /   k=1
   m=0 -----  1   /   k=2                      (0 0)
   m=1 ---  1   1   /   k=3               (1 0)     (1 1)
   m=2 -  1   2   1   /              (2 0)     (2 1)     (2 2)
   m=3  1   3   3   1           (3 0)     (3 1 )    (3 2)     (3 3)

         Integrally                  Binomial Coefficient (m k)

                        Figure 1

All even perfect numbers can be shown to lie on the third diagonal of Pascal's Triangle (see Figure 1). The restriction for m is that it must be equal to a Mersenne prime plus one; that is, m=Mp+1. Setting k equal to 2 (since the third diagonal of Pascal's triangle is k=2),

   (m k)=(Mp+1)!/(2!)*(Mp+1-2)!=(2**p)!/((2!)*(2**p-2)!)=
         ((2**p)*(2**p-1)*(2**p-2)!)/(2!*(2**p-2)!)=((2**p)*(2**p-1))/2=(2**(p-1))*(2**p-1)=P,

As in [5], we now note that each number in Pascal's triangle is the sum of the numbers in the preceding diagonal (see Figure 2):

   (m k)=SUM (i=k-1 -> m-1) (i k-1)

                 1+2+3=6
           1   /
         1   1
       1   2   1
     1   3   3   1
   1   4   6   4   1
           -

       Figure 2

We have seen that all even perfect numbers are on the third diagonal of Pascal's triangle. Hence, the second diagonal would generate the perfect numbers. That is, every even perfect number is the sum of the first (2**p)-1=Mp (Mersenne prime) numbers:

   P=SUM (i=1 -> Mp) i

We now observe that the elements of the third diagonal are the triangular numbers and every even perfect number is triangular in shape (see [2]). (See Figure 3.)

                 1
               1   1
             1   1   1

   The perfect number 6 with base Mp=3

             Figure 3

According to Burton [1] there are 24 even perfect numbers known to date (1976). The first 5 and their associated Mersenne primes are given in Table 1.

   p     Mp      P=Mp*(2**(p-1))
   -----------------------------
   2     3       6
   3     7       28
   5     31      496
   7     127     8,128
   13    8,191   33,550,336

             Table 1

We now have several different ways of computing perfect numbers. We must first compute Mersenne primes Mp. Knowing the Mersenne primes, we can:

1. compute P=Mp*(2**(p-1)), using Euclid's formula,
2. compute P=SUM (i=1 -> Mp) i, summing up the first Mp positive integers, or
3. with m=Mp+1 and k=2, compute P=(m k).

References