- 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 - I do not know pari, using Mathematica instead.

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

>>

>>

>