Two runs completed.
1083 numbers proved prime on the first run.
43 numbers proved prime on the second run.

I can't see any more such PRPs in factordb now. But I'll keep the script handy in case another batch appear.

Chris

Proving this [URL="http://www.factordb.com/index.php?id=1100000000991811881"]PRP463[/URL] will allow N-1 proof of [URL="http://www.factordb.com/index.php?id=1100000000708879674"](168^1741+1)/169[/URL]

Primitive part of 77^447-1 ([URL="http://www.factordb.com/index.php?id=1100000001030669252"](77^447-1)/(77^149-1)/6007[/URL]) proved using N-1

Ditto [URL="http://www.factordb.com/index.php?id=1100000001029732958"](425^621-1)*(425^9-1)/(425^207-1)/(425^27-1)[/URL]

And [URL="http://www.factordb.com/index.php?id=1100000001003713424"](217^892+1)/(217^4+1)[/URL]

And [URL="http://www.factordb.com/index.php?id=1100000001014993209"](287^914+1)/(287^2+1)[/URL]

[URL="http://www.factordb.com/index.php?id=1100000001019480113"](583^824+1)/(583^8+1)[/URL]

[URL="http://www.factordb.com/index.php?id=1100000001019902227"](873^776+1)/(873^8+1)[/URL]

[URL="http://www.factordb.com/index.php?id=1100000001018848053"](765^788+1)/(765^4+1)[/URL]

[URL="http://www.factordb.com/index.php?id=1100000001051370888"](807^423-1)*(807^3-1)/(807^141-1)/(807^9-1)[/URL]

I moved ~200 numbers from combined proof to either + OR - the other day.
Most didn't even need very much help.

[URL="http://factordb.com/index.php?id=1100000001051369355"](1243^423-1)*(1243^3-1)/(1243^141-1)/(1243^9-1)[/URL]

What are the algebraic factors of a number like (1243^423-1)*(1243^3-1)/(1243^141-1)/(1243^9-1)-1 ? I've seen a few like that but could not work out what algebraic factors it would have.

Chris

[QUOTE=chris2be8;469582]What are the algebraic factors of a number like (1243^423-1)*(1243^3-1)/(1243^141-1)/(1243^9-1)-1 ? I've seen a few like that but could not work out what algebraic factors it would have.

Chris[/QUOTE]
The number shown by axn doesn't have any algebraic factors because *it* is the primitive factor of 1243^423-1. Is your question a typographical error? Or should we expect the additional "-1" to be of a form that provides algebraic factors??

Let [TEX]k=1243,[/TEX] and [TEX]p^2q = 3^2\times47 = 423[/TEX]

Then, [TEX]k^{p^2q}-1 = \frac{(k^p-1)(k^{p^2q}-1)}{(k^{pq}-1)(k^{p^2}-1)}\quad\times\quad\frac{(k-1)(k^{pq}-1)}{(k^p-1)(k^q-1)}\quad\times\quad\frac{k^q-1}{k-1}\quad\times\quad \frac{k^{p^2}-1}{k^p-1}\quad\times\quad\frac{k^p-1}{k-1}\quad\times\quad(k-1)[/TEX]

[QUOTE=chris2be8;469582]What are the algebraic factors of a number like (1243^423-1)*(1243^3-1)/(1243^141-1)/(1243^9-1)-1 ? I've seen a few like that but could not work out what algebraic factors it would have.

Chris[/QUOTE]

[CODE]f=factor( polcyclo(423)-1 )[,1]
vector(#f, i, poliscyclo(f[i]))
[/CODE]
[CODE][x - 1,
x,
x + 1,
x^2 - x + 1,
x^2 + x + 1,
x^22 - x^21 + x^20 - x^19 + x^18 - x^17 + x^16 - x^15 + x^14 - x^13 + x^12 - x^11 + x^10 - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1,
x^22 + x^21 + x^20 + x^19 + x^18 + x^17 + x^16 + x^15 + x^14 + x^13 + x^12 + x^11 + x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1,
x^44 - x^43 + x^41 - x^40 + x^38 - x^37 + x^35 - x^34 + x^32 - x^31 + x^29 - x^28 + x^26 - x^25 + x^23 - x^22 + x^21 - x^19 + x^18 - x^16 + x^15 - x^13 + x^12 - x^10 + x^9 - x^7 + x^6 - x^4 + x^3 - x + 1,
x^44 + x^43 - x^41 - x^40 + x^38 + x^37 - x^35 - x^34 + x^32 + x^31 - x^29 - x^28 + x^26 + x^25 - x^23 - x^22 - x^21 + x^19 + x^18 - x^16 - x^15 + x^13 + x^12 - x^10 - x^9 + x^7 + x^6 - x^4 - x^3 + x + 1,
x^135 - x^132 + x^126 - x^123 + x^117 - x^114 + x^108 - x^105 + x^99 - x^96 + x^90 - x^87 + x^81 - x^78 + x^72 - x^69 + x^63 - x^60 + x^54 - x^51 + x^45 - x^42 + x^36 - x^33 + x^27 - x^24 + x^18 - x^15 + x^9 - x^6 + 1]

[1, 0, 2, 6, 3, 46, 23, 69, 138, 0]
[/CODE]

which means, x^138-1, since all the given cyclotomic polys [1, 2, 6, 3, 46, 23, 69, 138] together constitute x^138-1

EDIT:- Or simpler, x^1-1, x^2-1, x^3-1, x^6-1, ... x^138-1 are all algebraic factors.