Another from t400.txt:
sigma(14401^42):
r1=1158907471343077935619801566199517072672891 (pp43)
r2=32228122620202488291078007880274499627503222729476471533211188839746944678614929711496390382583051276078527 (pp107)

There are only 3 left from t400.txt.

Chris K

One more from t400.txt:
sigma(25646167^22)
r1=241732720013567165779062022737045283199963612458984850323587786403 (pp66)
r2=1079969357685198287960377081659067614592724015988821193538145931990646184302628795054527 (pp88)

Now only 2 left.Both being worked on.

Chris K

From Pascal's t1000.txt
[code]
sigma(4359977845598272757076518581118691539852489438400131^2)

P47: 47824108000342125391768788390332417666796760267
P54: 917981206567007339636067114141506928684975294752855863
[/code]

The last but 1 from t400.txt:
sigma(86950696619^16)
r1=1670948488976619312266854176443903725456927887481 (pp49)
r2=381887997728473574000809042338770035787863465357943751153760836840843679326648778219602471602002965463721230443509295032209 (pp123)

This took 29 days.I probably should have used larger large prime bounds, the yield was getting very low by the end.

The last number should finish on Sunday.

Chris K

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.

The method requires an upper bound < 2 on the abundancy.
In a usual factor chain, branching on p^a means that we assume [tex]$p^{(a+1)t-1}$[/tex] || N for t>=1, and we use the upper bound p/(p-1).
If the obtained bound is > 2, we split the cases p^a || N and [tex]$p^{(a+1)t-1}$[/tex] || N for t>=2, for some of the p^a,
and we use the abundancy σ(p^a)/p^a when p^a || N.
If a=1, i.e. p is the special prime, then we can always take (p+1)/p for the abundancy,
since p^(4k+1) || N for k>=1 will be considered with p^(2k+1).

Splitting 3^2, 3^4, and 7^2 is sufficient to handle every roadblock but three:
7^4 2801^82 / 3^2 13^1 / 11^18 6115909044841454629^16 / 661^2 145861^2 1013122723^16
7^4 2801^82 / 3^8 757^2 14713^42 13^2 61^1 31^2 331^2 5233^2 42073^58
7^4 2801^82 / 3^8 757^2 14713^42 13^2 61^1 31^2 331^2 5233^70
In these cases, even if we split 3^8, 7^4, and 13^2 to obtain an abundancy < 2,
it would still be very close to 2. That means the method would be painfully slow.

Of course, factoring σ(2801^82) would solve the problem,
but factoring σ(1013122723^16) and σ(14713^42) is sufficient.

I think σ(1013122723^16) could be done with an octic. Or without.

But σ(14713^42) seems easier.

...Nevermind, both are easy.

[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.

Splitting 3^2, 3^4, and 7^2 is sufficient to handle every roadblock but three:
7^4 2801^82 / 3^2 13^1 / 11^18 6115909044841454629^16 / 661^2 145861^2 1013122723^16
7^4 2801^82 / 3^8 757^2 14713^42 13^2 61^1 31^2 331^2 5233^2 42073^58
7^4 2801^82 / 3^8 757^2 14713^42 13^2 61^1 31^2 331^2 5233^70
In these cases, even if we split 3^8, 7^4, and 13^2 to obtain an abundancy < 2,
it would still be very close to 2. That means the method would be painfully slow.[/QUOTE]

Could you advise what 'painfully slow' means in this context - it took about six CPU-years spread over six months of real time, including a fair amount of human effort, to do 2801^79-1.

[QUOTE]Of course, factoring σ(2801^82) would solve the problem,
but factoring σ(1013122723^16) and σ(14713^42) is sufficient.[/QUOTE]

Does this mean that, instead of the slightly laborious four months of sieving and five weeks of linear algebra to deal with 2801^79-1, I could have done two, or four, or even sixteen 175-digit SNFS jobs? I think, had I known that, I probably wouldn't have done the 2801^79-1.

These SNFS jobs are small enough that I haven't jumped on them because I assume somebody else is jumping on them.

σ(1013122723^16) will be finished in an hour but /--yawn--/ I am too sleepy to see it. Barely ~1 cpu day.

σ(1013122723^16) =
[FONT=Arial Narrow]1231949479100411529897025851988085403028139763773928951784266154446032590355758084727532424504951397079565676334712527005962074886398989859190241 =>[/FONT]
[FONT=Arial Narrow]p54: 351752993625173410346880723949165323504208539390997381
p91: 3502314127888196426687123407041014993415804675525648157451709730356670228428469371425350061
[/FONT]

From Pascal's t800.txt

8793883298660072049758955866197625356353667659779232173073915908926942524282405943205461

prp63 = 288759920293341103246038563933509244288496151453106018946722089
prp26 = 30453960818823725899887949

[QUOTE=Batalov;235657]σ(1013122723^16) will be finished in an hour but /--yawn--/ I am too sleepy to see it. Barely ~1 cpu day.[/QUOTE]
What was the polynomial used? Was it 1013122723*x^4-1 or x^6-1013122723 or something else?