Take

p = 9372688136871853

Then

p-1 = 9372688136871852

If I can factor p-1 into primes:

{{2, 2}, {3, 1}, {13, 1}, {2321923, 1}, {25875679, 1}}

I can get full list of divisors of p-1:

{1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156, 2321923, 4643846, 6965769,

> 9287692, 13931538, 25875679, 27863076, 30184999, 51751358, 60369998,

The rank(r) of a residue r mod p is the smallest of the above divisors

> 77627037, 90554997, 103502716, 120739996, 155254074, 181109994,

> 310508148, 336383827, 362219988, 672767654, 1009151481, 1345535308,

> 2018302962, 4036605924, 60081334210717, 120162668421434, 180244002632151,

> 240325336842868, 360488005264302, 720976010528604, 781057344739321,

> 1562114689478642, 2343172034217963, 3124229378957284, 4686344068435926,

> 9372688136871852}

of p-1 such that

r^rank(r) = 1 (mod p)

Using PowerMod routine I am arriving at:

rank(2) = 3124229378957284 = (p-1)/3

rank(3) = 781057344739321 = (p-1)/12

Since rank(2) IS divisible by rank(3), some power of 2 (mod p) is equal to 3.

Hence p is a divisor of a number of the form 2^n-3, but it is rather

hard to find a particular n (although it should be possible in this

case given that p-1 has no huge prime factors).

Since rank(3) is NOT divisible by rank(2), NO power of 3 (mod p) is equal to 2.

Hence p is NOT a divisor of any number of the form 3^n-2.

Jarek

2008/12/29 Devaraj Kandadai <dkandadai@...>:> A problem in programming: Let us take the first 9372688136871853; how do

> we test whether it is a Mangammal prime or not? To form the seq A123239

> (Mangammal primes)(refer OEIS) I used the following programme in pari

> : {p(k)=(3^k-2)/p}; for (k=1,p-1,print(p(k))). Now this is ok when the

> primes are small; WE CHECK WHETHER WE GET ATLEAST ONE RATIONAL INTEGER- in

> which case the p being tested is NOT a Mangammal prime. If the prime to be

> tested is as big as that indicated in the begining of this post what should

> we do? I will be glad if any one can suggest a short cut in pari.

> A.K. Devaraj

>

> On Mon, Dec 1, 2008 at 5:26 PM, jarek372000 <jaroslaw.wroblewski@...>

> wrote:

>>

>> For over a week I have been using 40-60 64-bit computers in an attempt

>> to find 2 AP22 with common difference. I have selected 43# as the

>> common difference. While the first AP22 was found early in the search,

>> the second one appeared today. It also improves the current AP22 record.

>>

>> Known AP20+ with difference 43#

>> First terms of progressions are given

>>

>> 1 9372688136871853

>> 2 11735227242889999

>> 3 76240762416222539 AP21

>> 4 93490858594661729 AP22 !!!!!

>> 5 114146343711853721

>> 6 135651360264848653

>> 7 195625258610971297 AP21

>> 8 202860934798777373

>> 9 223789213311833843

>> 10 224957853888083671 AP21

>> 11 251672116721153519

>> 12 325435306756257757

>> 13 333012166298058323

>> 14 338275337330536643

>> 15 381336957506808803

>> 16 485191591159166291

>> 17 493052729074838717 AP21

>> 18 630202728690264047

>> 19 1006137974832655813

>> 20 1296572696606684143

>> 21 1351906725737537399 AP22 !!!!!

>> 22 1376274082220856487

>>

>> Jarek

>>

>>

>