Michael Rao wrote a program producing a tree similar to the one in [Brent, Cohen, te Riele] that proves that an odd perfect number (OPN for short) is > 10^300.

We use it to prove that:

- an OPN is greater than 10^1500 (since then, we have pushed the computation to 10^2000),

- the total number of prime factors (counting multiplicities) of an OPN is at least 101,

- the largest component (i.e. factor p^a with p prime) of an OPN is > 10^62.

A draft of our paper is here.

The proof tree looks like this (26Mb for a bound of 10^540).

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 by Kevin Hare to show that an OPN has at least 75 not necessarily distinct prime factors. 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^XXXX.

The format is p q c where c is a composite factor of sigma(p^q).

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^q-1 is "p q-1 composite weight", where "weight" is the number of roadblocks involving the composite.

p q+1 f where f is a factor of sigma(p^q). p, q+1, and f are prime.

The compressed file checkfacts.txt.gz (10/2017) is 660 Mb.

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

Example: Consider the roadblock sigma(11^18) = 6115909044841454629, sigma(6115909044841454629^16) = C301.

To ensure that the unknown factors of C301 have a negligible contribution to the abundancy, we check that C301 has no factor less than 10^9.

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

So the abundancy of the factors 11^18, 6115909044841454629^16, and C301 is at most (1+1/10)*(1+1/6115909044841454628)*(1+1/10^9)^33 < 1.10000004.

Suppose our target bound is 10^2100 and p is the smallest prime factor of the OPN distinct from 11.

Suppose p=853. We must have at least 511 other prime factors >= 853 because 1.10000004*(1+1/853)^510 < 2.

But then 11^18*6115909044841454629^16*C301*853^511 > 10^2117.

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.

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^2, except for the special prime) is lower.

With these rules, the best thing to do is to take every allowed prime in order. Then, the OPN gets bigger than 10^300 before that the abundancy reaches 2.

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^a is defined as eff(p,a) = ln(sigma(p^a)/p^a)/ln(p^a).

We deal with multiplicative increasings, which expains the logarithms, and then make the ratio between the contribution of the component p^a to the abundancy and to the OPN.

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 sigma(p^a) is not divisible by 4 or some forbidden factor.

Example: Consider p=23. sigma(23^1), sigma(23^2), sigma(23^3), are respectively divisible by 4, 7, 4. So the exponent of 23 is at least 4.

- we sort these components p^a by decreasing efficiency.

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

This product is greater than 10^1735, as shown by this maple code.

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