∏(m+1)

For instance 600 = 2^{3}×3×5^{2} has (3+1)(1+1)(2+1) = 24 divisors.

By expanding the product
∏_{p} (∑_{i} p^{i}),
for which in every sum ∑ :
0 ≤ i ≤ m for that prime p,
we get every term of expansion by choosing an exponent for each prime factor p.

Hence it is a divisor of N and the sum is the sum of divisors of N,
as we get all of them.

∑p^{i} can be written (p^{m+1}-1)/(p-1) as sum
of a geometric series, and finally :

S = ∏_{p} (∑_{i} p^{i}) =
∏(p^{m+1}-1)/(p-1)

For instance N = 600 : S = (1+2+4+8)(1+3)(1+5+25) = (16-1) × (9-1)/2 × (125-1)/4 = 1860

Script for calculation of sum of divisors

Let's speak now about perfect numbers.
The odd perfect numbers are unknown, and we think there are none.
However if there is one, it is more than 300 digits !
On the contrary, **even** perfect numbers are known since Euclid, and Euler proved
they are all :

2^{(p-1)}(2^{p}-1) with 2^{p}-1 being prime

The number 2^{p}-1 is not prime for every exponent p.

p must be prime, but it is not sufficient, for instance
2^{11}-1 = 2047 = 23x89.

Numbers M_{p} = 2^{p}-1 are primes (Mersenne primes) for
p = 2, 3, 5, 7, 13, 17, 19, 31...

That is Mersenne primes 3, 7, 31, 127, 8191, 131071, 524287, 2147483647...

and perfect numbers :

6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128...

- The "multiperfect numbers", equal to n times the sum of their divisors.
Mention the number given by Frenicle in 1643 :

n = 2^{36}× 3^{8}× 5^{5}× 7^{7}× 11 × 13^{2}× 19 × 31^{2}× 43 × 61 × 83 × 223 × 331 × 379 × 601 × 757 × 1201 × 7019 × 112303 × 898423 × 616318177

Could you do as well as Frénicle and just check that the sum of aliquot parts of n is 5n ? - We also know "amicable numbers" : The sum of proper divisors of each one is equal to the other

For instance 220 and 284. - We name "aliquot sequence" the sequence of the sum of divisors of the preceding number,
for instance, starting from 84 :
84, 140, 196, 203, 37, 1

Most of the time this sequence ends up with 1 (and stops as 1 has no proper divisor).

Sometimes this sequence seems to increase towards infinity : we don't know what happens with sequence starting from 276, we get numbers of more than 100 digits, whose sum of divisors are hard to compute !

In some cases, the sequence is periodic : returning to start point, for instance

12496, 14288, 15472, 14536, 14264, 12496 ...

This periodic sequences are named "sociable sequences".

This is solved by searching divisors of S in the form ∑p

For instance to search numbers whose sum of all divisors is 56 :

First of all get all divisors of 56 = 2³×7 : 1, 2, 4, 7, 8, 14, 28 and 56.

Next look for those of the form Q = 1+p+...p^{n}.

To check if a number q is of the form Q, we look for prime factors p of q-1.

For each one, check if q(p-1)+1 is a power of p.

For instance to check if q=31 is a Q number : 31-1 = 30 = 2x3x5

p = 2 : 31×1+1 = 32 = 2^{5} and 31 = 1+2+4+8+16

p = 3 : 31×2+1 = 63 not a power of 3

p = 5 : 31×4+1 = 125 = 5^{3} and 31 = 1+5+25

Hence 31 is of the form Q in two ways.

The only divisors of 56 of the form Q = 1+p+...p^{n} are :

4 = 1+3

7 = 1+2+4

8 = 1+7

14 = 1+13

Look for factorization of 56 with these divisors : 56 = 1×56 = 2×28 = 4×14 = 8×7 = 2×2×14 = 2×4×7 = 2×2×2×7.

Only factorizations 4×14 and 8×7 are made of Q numbers only.

4×14 = (1+3)(1+13) results into **N = 3×13 = 39**

8×7 = (1+7)(1+2+4) results into **N = 7×4 = 28**

All this doesn't work anymore if the sum of **proper** divisors is given
instead of **all** divisors.

Check first that all these numbers are primes (hard job with just paper and pencil !
The script giving the sum of divisors gives also the prime factorization)

The sum of all divisors of n is then :

S = (2^{37}-1) × (3^{9}-1)/2 × (5^{6}-1)/4 × (7^{8}-1)/6 ×
12 × (13^{3}-1)/12 × 20 × (31^{3}-1)/30 × 44 × 62 × 84 × 224 × 332 × 380 ×
602 × 758 × 1202 × 7020 × 112304 × 898424 × 616318178.

and we have to factorize all these numbers :

2^{37}-1 = 223×616318177 | (3^{9}-1)/2 = 9841 = 13×757 | |

(5^{6}-1)/4 = 3906 = 2×3^{2}×7×31 | (7^{8}-1)/6 = 960800 = 2^{5}×5^{2}×1201 | |

12 = 2^{2}×3 | (13^{3}-1)/12 = 183 = 3×61 | |

20 = 2^{2}×5 | (31^{3}-1)/30 = 993 = 3×331 | |

44 = 2^{2}×11 | 62 = 2×31 | |

84 = 2^{2}×3×7 | 224 = 2^{5}×7 | |

332 = 2^{2}×83 | 380 = 2^{2}×5×19 | |

602 = 2×7×43 | 758 = 2×379 | |

1202 = 2×601 | 7020 = 2^{2}×3^{3}×5×13 | |

112304 = 2^{4}×7019 | 898424 = 2^{3}×112303 | |

616318178 = 2×7^{3}×898423 |

S = 2

Sum of proper divisors is S - n = 5n QED.