Yes, quartic was faster by miles.
Overnight I started the other one, σ(14713^42); it will be done by the afternoon.

[QUOTE=Batalov;236129]Yes, quartic was faster by miles.
Overnight I started the other one, σ(14713^42); it will be done by the afternoon.[/QUOTE]
Thanks! And for σ(14713^42), what was the polynomial used? Was it x^4-14713 or 14713*x^6-1 or something else?

[URL="http://www.factordb.com/index.php?id=1100000000228976623"]σ(14713^42)[/URL] is done now
an easy sextic, sieving on alg. side (~4 CPU-days)

And here's the last result from t400.txt:
sigma(17707^42)
r1=3737617806240266760055033119426898876709357150159282314998557 (pp61)
r2=4541233905046125016491283138503334949108107587278687386026675752340534798119092376135171296366148994116878876867 (pp112)

Slightly later than I though it would be.

Chris K

[QUOTE=Pascal Ochem;235625]Many thanks to everyone who participated in the factorizations of 2801^73-1 and 3^607-1, and to Chris for the numbers in t400.

We have now a way to get around roadblocks and we obtain that an odd perfect number is greater than 10^1350.
The goal is to reach 10^1500.
...
Of course, factoring σ(2801^82) would solve the problem,
but factoring σ(1013122723^16) and σ(14713^42) is sufficient.[/QUOTE]

If you don't mind an uninformed question too much, where would an
odd perfect number between 10^1350 and 10^1500 show up among
these roadblocks? Do you check something for each prime factor found?

Not that I believe that there are any; just a question of how a factorization
pushes the limit up. -Bruce

[QUOTE=bdodson;236285]If you don't mind an uninformed question too much, where would an
odd perfect number between 10^1350 and 10^1500 show up among
these roadblocks? Do you check something for each prime factor found?

Not that I believe that there are any; just a question of how a factorization
pushes the limit up. -Bruce[/QUOTE]

See

[url]http://oai.cwi.nl/oai/asset/1613/1613A.pdf[/url]

[QUOTE=bdodson;236285]If you don't mind an uninformed question too much, where would an
odd perfect number between 10^1350 and 10^1500 show up among
these roadblocks? Do you check something for each prime factor found?

Not that I believe that there are any; just a question of how a factorization
pushes the limit up. -Bruce[/QUOTE]

Let me start by saying I know little of odd perfect numbers, but know about calculating to see if a number is perfect.
I don't know who, if anyone, is actually doing the calculations, but I know that if you know the prime factorization of a number you can [URL="http://mersennewiki.org/index.php/Aliquot_Sequences"]calculate the sum of its divisors[/URL], and thus show whether it is a perfect number. (I don't know if this is the same thing used here, but it certainly could do it)
e.g. an even perfect number:
[TEX]8128=2^6*127[/TEX]

[TEX]{{2^{6+1}-1}\over{2-1}}*{{127^{1+1}-1}\over{127-1}}-2^6*127=[/TEX]
[TEX](2^7-1)*{{127^2-1}\over126}-2^6*127 =[/TEX]
[TEX](2^7-1)*{{(127+1)(127-1)}\over126}-2^6*127 =[/TEX]
[TEX](2^7-1)*128-2^6*127 =[/TEX]
[TEX]127*2^7-2^6*127 =[/TEX]
[TEX]2^7*127-2^6*127 = 2^6*127 = 8128[/TEX]
Since the sum of the proper divisors (i.e. the divisors besides the number itself, hence the -2^6*127 at the end) equals the number, it is a perfect number. If you can find an odd number that this works out for, then you've found an odd perfect number.

[QUOTE=Mini-Geek;236289]Let me start by saying I know little of odd perfect numbers, but know about calculating to see if a number is perfect.
I don't know who, if anyone, is actually doing the calculations, but I know that if you know the prime factorization of a number you can [URL="http://mersennewiki.org/index.php/Aliquot_Sequences"]calculate the sum of its divisors[/URL], ...[/QUOTE]

Yes, but that's not the way to handle all of the (odd) numbers
between 10^1350 to 10^1500, we're certainly not finding all of
the divisors for each of them --- if I understand correctly, the issue
is the use of factorizations of "roadblocks". -bd

@Bob -- Not a single thing new since Brent's paper? Sigh!

bdodson,

There are new ideas in this search. They are explained on Pascal's site.

[QUOTE=Zeta-Flux;236350]bdodson,

There are new ideas in this search. They are explained on Pascal's site.[/QUOTE]

That would be [url]http://www.lri.fr/~ochem/opn/[/url]

a link that I found on oddperfect.org. Indeed, the very first
heading is "Differences with BCR". Richard's link for pub116 includes
"comments" on the lower bound of 10^300: "main part of the proof
is ... a tree ... [with] 12644 [leaves]", where each leaf is either a
contradiction or a verification that the smallest odd perfect number
is above 10^300.

From Pascal's site, there are composites t400 (just 3), t600 (lots, already),
t800, t1000 and t1200 (_really_ lots) --- with each composite first on
one of the branches. Then there are the roadblocks.

So, back to my question; an odd perfect number between 10^1350
and 10^1500 would show up as a leaf that doesn't produce a contradiction?
And a factorization cuts out all of the leaves on the branch for that
composite? If that's the case, a prime factor never could reveal an
odd perfect number; just trim the number of branches that could possibly
have a leaf that doesn't give a contradiction.

Uhm, that would be for factorizations of the "composites". But we're
being asked to factor "roadblocks". So now that I've looked at the
recommended reading, can someone tell me what a prime factor of
one of the roadblocks does? -Bruce

[QUOTE=bdodson;236478] ...
So, back to my question; an odd perfect number between 10^1350
and 10^1500 would show up as a leaf that doesn't produce a contradiction?
And a factorization cuts out all of the leaves on the branch for that
composite? If that's the case, a prime factor never could reveal an
odd perfect number; just trim the number of branches that could possibly
have a leaf that doesn't give a contradiction.

Uhm, that would be for factorizations of the "composites". But we're
being asked to factor "roadblocks". So now that I've looked at the
recommended reading, can someone tell me what a prime factor of
one of the roadblocks does? -Bruce[/QUOTE]

Uhm. No, that's not right. Eventually all of the composites factor
(and that's the case for 10^300, with only three needed for 10^400),
so there's still some reduction to the place where the branches would
have been if there were still any unfactored composites. -bd