> If you want a challange, dislodge some of my gigantic PRP trinomials

> at Henri's site:

> http://www.primenumbers.net/prptop/prptop.php

Actually I did browse them for proofs. One or two could be proven with a large

ECM effort.

This one is a good very good one to try:

2^64695-2^15-1 with 19476 digits.

N+1 = 2^15*(2^64680-1)

(2^64680-1) has 96 cyclotomic divisors.

T 64680={ 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 14, 15, 20, 21, 22, 24, 28, 30,

33, 35, 40, 42, 44, 49, 55, 56, 60, 66, 70, 77, 84, 88, 98, 105, 110, 120, 132,

140, 147, 154, 165, 168, 196, 210, 220, 231, 245, 264, 280, 294, 308, 330, 385,

392, 420, 440, 462, 490, 539, 588, 616, 660, 735, 770, 840, 924, 980, 1078,

1155, 1176, 1320, 1470, 1540, 1617, 1848, 1960, 2156, 2310, 2695, 2940, 3080,

3234, 4312, 4620, 5390, 5880, 6468, 8085, 9240, 10780, 12936, 16170, 21560,

32340, 64680 } [96]

So: (2^64680-1) =

phi(1,2)*phi(2,2)*phi(3,2)*phi(4,2)*.....*phi(21560,2)*phi(32340,2)*phi(64680,2)

There are aurifeuillian factors (L and M) as well when for phi(n,2) n=4*k and

k=odd.

L=2^h-2^k+1, M=2^h+2^k+1, h=2k-1. Take gcd's with phi(4*k,2) and L or M.

E.g. phi(2156,2) can be divided in a L and M part as 2156 = 4*539

L: 10781.81929.90317512080398683509507180285854441.P83

M:

2136469147429.111206916097779728932051224808777.1297662995123479965752936319854262257.P46

There is great deal of work already done in the cunninghamproject. I would

expect factorization to be more than 25% done. If one reaches 30% David

Broadhurst can do his KP magic.

Bouk de Water.

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