## Re: 2 AP22 with the same difference

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• ... primer. Patience would have been a virtue. If I had gone just a little further, to 1371, I would have found my first solution. A couple of days ago I
Message 1 of 7 , Dec 16, 2008
--- In primenumbers@yahoogroups.com, "Mark Underwood" <mark.underwood@...>
wrote:
>(snip) Meanwhile, I'm about 90 percent done my run with finding seven relatively teenie
> weenie numbers up to 1365. :) No luck getting even another 63 primer, let alone a 64
primer.

Patience would have been a virtue. If I had gone just a little further, to 1371,
I would have found my first solution. A couple of days ago I started
on the problem again from another angle, which resulted in a simpler
program with what seems like twice the efficiency.
Within a day it found the seven numbers

1371, 1170, 825, 756, 700, 615, 414

whose 64 possible additive combinations are all prime.
The largest prime produced is 5851.
My search was exhaustive in terms of the largest prime produced,
so all other solutions must have the largest prime greater than 5851.

Alas there are only forty eight distinct primes out of a possible 64,
but beggars can't be choosers. :)

Mark
• Easy to test by programming inpari as follows: {p(k)=(%1)+k*(%3)} where (%1) represents the first term and (%3)=43#. Hope this is useful to all Devaraj On
Message 2 of 7 , Dec 17, 2008
Easy to test by programming inpari as follows: {p(k)=(%1)+k*(%3)} where
(%1) represents the first term and (%3)=43#. Hope this is useful to all
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
>
>
>

[Non-text portions of this message have been removed]
• 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
Message 3 of 7 , Dec 28, 2008
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
>
>
>

[Non-text portions of this message have been removed]
• 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},
Message 4 of 7 , Dec 29, 2008
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,

> 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}

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

> 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
>>
>>
>
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