Michael Rao wrote a program producing a tree similar to the one in [Brent, Cohen, te Riele] that proves that N > 10

We use it to prove that:

(1) N > 10

(2) Ω(N) >= 101 (since then, we have pushed the computation to Ω(N) >= 115),

(3) the largest component (i.e. factor p

(4) Ω(N) >= 2ω(N)+51.

(5) If we consider the Euler form N=p

We obtain (1), (2), and (3) in this paper and (4) in this paper.

Using (2), (4), and the result in [Clayton, Hansen], we get Ω(N) >= max(115, 2ω(N)+51, (99ω(N)-187)/37).

The bound (5) is unpublished. It is tailor-made to boost Carl Pomerance's heuristic argument, since we can replace m > 10

The proof tree looks like this (26Mb for the bound N > 10

The slides of a talk in Montpellier (2014).

- We branch on the overall largest available prime factor instead of branching on the largest available prime factor "from the previous line".

- We do not use the arguments of the B25 and B30 bounds described in [BCR].

- We use instead a method to circumvent the roadblocks similar to the one used in [Hare] to show Ω(N) >= 75. It is described below.

- In the end, we have to conclude that an odd perfect number not divisible by any of the forbidden factors must be greater than the target bound. To do this, we use an improved version of the argument in [BCR] that is described below.

tXXXX contains composite numbers encountered when targetting the lower bound 10

The format is "p q c" where c is a composite factor of σ(p

There are all and only the composites that appear firstly on their respective branch:

If you factor one of them, then the proof tree will be reduced, and this factorization will not become useless because of the factorization of some other number.

The format for p

The list and the weights are not updated very often but every (partialy) factored composite is removed within a few days.

"p q+1 f" where f is a factor of σ(p

The compressed file checkfacts.txt.gz (12/2022) is 660 Mb.

Then we branch on all prime factors below this bound (except the forbidden or already considered ones).

Example: Consider the roadblock σ(11

To ensure that the unknown factors of C301 have a negligible contribution to the abundancy, we check that C301 has no factor less than 10

Thus, there are at most 301/9=33 such prime factors and each of them contributes at most 1+10

So the abundancy of the factors 11

Suppose our target bound is 10

Suppose p=853. We must have at least 511 other prime factors >= 853 because 1.10000004*(1+1/853)

But then 11

So we have to rule out every prime factor smaller than 853 except 19, 127, 7 (forbidden) and 11 (already considered).

When branching on a prime in order to circumvent a roadblock, we might encounter a new roadblock, so we have to apply this method recursively.

NB: I should eventually change this example since the factors of the C301 are known now.

Smaller primes p are prefered since their contribution to the abundancy (less than p/(p-1)) is higher and their contribution to the OPN (at least p

With these rules, the best thing to do is to take every allowed prime in order. Then, the OPN gets bigger than 10

As the target bound increases, we forbid more small primes so that this argument still works.

We also refine the argument. We consider the component (i.e., the prime together with its exponent) rather than the prime alone, and we use a bit more the fact that some factors are forbidden.

The efficiency of the component p

We deal with multiplicative increasings, which expains the logarithms, and then make the ratio between the contribution of the component p

Remarks:

a < b implies eff(p,a) > eff(p,b)

p < q implies eff(p,a) > eff(q,a)

Then:

- for each allowed prime p, we find the smallest a such that σ(p

Example: Consider p=23. σ(23

- we sort these components p

- we compute the product of such components as long as the abundancy of the product remains smaller than 2.

This product is greater than 10

E_q_k.c is a C/GMP code to compute the set E(q,k) in [BCR].