## Re: Primes with the representation of the year in them

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• Off Topic:  (Small is not uglier than big). (n^625+2) /2011 is prime for n= n = 2513759 , n = 8401967,... Puzzle:  Find another prime p such that there is n
Message 1 of 17 , Nov 24, 2011
Off Topic:  (Small is not uglier than big).

(n^625+2) /2011 is prime for n= n = 2513759 , n = 8401967,...

Puzzle:  Find another prime p such that there is n and k  and ( n^(p^k)+2)/2011 is prime. If you want it still harder, choose a fixed k = p-1 or k = p.

PS: I do not know if  "2011" could be found in the decimal representation of these primes at least once.

[Non-text portions of this message have been removed]
• ... The second one ( n = 8401967 ), ends in 2011 followed by 1121 other digits : ? 8401967^5^4 2011 10^1121%10^4 2011 Maximilian
Message 2 of 17 , Nov 24, 2011
On Thu, Nov 24, 2011 at 3:28 PM, Robin Garcia <sopadeajo2001@...> wrote:

> (n^625+2) /2011 is prime for n= n = 2513759 , n = 8401967,...
>
> PS: I do not know if  "2011" could be found in the decimal representation of these primes at least once.

The second one ( n = 8401967 ), ends in "2011" followed by 1121 other digits :

? 8401967^5^4\2011\10^1121%10^4
2011

Maximilian
• ... and also a PRP for n = 9128*2011 + 9 = 18356417. It is easy enough to obtain those as PRPs; proving primality might take you rather longer. Explanation:
Message 3 of 17 , Nov 24, 2011

> (n^625+2)/2011 is prime for n = 2513759, n = 8401967

and also a PRP for n = 9128*2011 + 9 = 18356417.

It is easy enough to obtain those as PRPs;
proving primality might take you rather longer.

Explanation: Using znprimroot and znlog we easily determine
that 3 is a primitive root modulo 2011 and that
znlog(-2,Mod(3,2011)) = 1250.
Thus the sequence in question is:
n = 9 mod 2011 and (n^625+2)/2011 is prime.

Then we easily obtain the first three PRPs:

\$ more tmp

ABC2 ((\$a*2011+9)^625+2)/2011
a: from 0 to 10000

\$ nohup pfgw -f -l -N tmp > & log &

[wait a short while]

\$ grep PRP pfgw.out

((1250*2011+9)^625+2)/2011 is 3-PRP! (1.2318s+0.0412s)
((4178*2011+9)^625+2)/2011 is 3-PRP! (1.6907s+0.0518s)
((9128*2011+9)^625+2)/2011 is 3-PRP! (1.7720s+0.0710s)

So all you have to do now, Robin, is to prove that
those are /prime/. Using Primo, you should be able to
complete those 3 task in this year of grace 2011.

Thanks for another timely observation,

David
• Yes, David . For 2011 to be a prime factor of  n^625+2  n must be 9 mod 2011 (or 9 mod 4022 if we wnat to avoid even n´s which are 0 mod 2). Though i had
Message 4 of 17 , Nov 24, 2011
Yes, David .

For 2011 to be a prime factor of  n^625+2  n must be 9 mod 2011 (or 9 mod 4022 if we wnat to avoid even n´s which are 0 mod 2). Though i had (sadly) to discover it experimentally.
Thanks also to Maximilian. I would like to know how to find this "2011" in the all-the-digits representation of the prime.  With Pari gp. Just don´t know how to do it.

And yet, David. What is the next prime  p such that  there is n such that : (n^(p^p)+2)/2011 is prime ??????

[Non-text portions of this message have been removed]
• I just forgot to remark the third PRP is : 18356417^625 +2  and  18356417 is prime.  Fourth is 26979585^625+2 . There must be somewhere some theory to tell
Message 5 of 17 , Nov 24, 2011
I just forgot to remark the third PRP is : 18356417^625 +2  and  18356417 is prime.  Fourth is 26979585^625+2 . There must be somewhere some theory to tell us how is it prime bases b generate primes b^n+c. I must say i do not know this theory.

[Non-text portions of this message have been removed]
• ... was very easy, using the simple identity znlog(-2,Mod(3,2011)) = 2*5^4. ... You say next , but have you already solved the puzzle for some at least one
Message 6 of 17 , Nov 24, 2011

> Yes, David.

> ((1250*2011+9)^625+2)/2011 is 3-PRP! (1.2318s+0.0412s)
> ((4178*2011+9)^625+2)/2011 is 3-PRP! (1.6907s+0.0518s)
> ((9128*2011+9)^625+2)/2011 is 3-PRP! (1.7720s+0.0710s)
was very easy, using the simple identity
znlog(-2,Mod(3,2011)) = 2*5^4.

> What is the next prime p such that there is n such that
> (n^(p^p)+2)/2011 is prime ?

You say "next", but have you already solved the "puzzle" for
some at least one prime? If so, what is that prime, after
which we are supposed to find the "next" prime? Moreover, do
you hold, privately, such a "next" prime, thereby ensuring
that your "next prime" puzzle might be solved by others?

In the puzzles that set here, I know in advance that there
is a solution that may be found by others in a reasonable
time. Do you have a private solution or are you just
fishing, hoping that others will succeed where you have
failed? A candid answer: "solution known to exist", or
"merely fishing", would be helpful, in this case.

Finally you have yet to produce certificates for the
probable primes that I list above, or to apologize for
claiming primality in situations where you have not yet
proven primality.

Best wishes

David (per proxy znlog and Primo)
• ... It is straightforward to extend the list of Robin s 2011 numbers that are probably prime: ((1250*2011+9)^625+2)/2011 is Fermat and Lucas PRP!
Message 7 of 17 , Nov 24, 2011

> you have yet to produce certificates for the probable primes

It is straightforward to extend the list of
Robin's "2011" numbers that are probably prime:

((1250*2011+9)^625+2)/2011 is Fermat and Lucas PRP! (3.5581s+0.2420s)
((4178*2011+9)^625+2)/2011 is Fermat and Lucas PRP! (3.0076s+0.1882s)
((9128*2011+9)^625+2)/2011 is Fermat and Lucas PRP! (3.2012s+0.2176s)
((13416*2011+9)^625+2)/2011 is Fermat and Lucas PRP! (4.1322s+1.0847s)
((43172*2011+9)^625+2)/2011 is Fermat and Lucas PRP! (4.3995s+0.3097s)
((45030*2011+9)^625+2)/2011 is Fermat and Lucas PRP! (4.2177s+0.1581s)
((52926*2011+9)^625+2)/2011 is Fermat and Lucas PRP! (5.4420s+0.1791s)
((77270*2011+9)^625+2)/2011 is Fermat and Lucas PRP! (5.6263s+0.2931s)
((84708*2011+9)^625+2)/2011 is Fermat and Lucas PRP! (4.5988s+0.2018s)

The largest of these 9 has 5142 decimal digits,
well within the comfort-zone of Marcel's Primo.

David
• Well; David. I know they are not hard to find with Primeform. And not too hard either with Pari gp and its ispseudoprime function. The important matter, if
Message 8 of 17 , Nov 24, 2011
Well; David. I know they are not hard to find with Primeform. And not too hard either with Pari gp and its "ispseudoprime" function. The important matter, if newcomers to the world of prmes read us, is to let them understand that you can find infinitely many primes with almost any form.

Yet i have been censored. I wrote this today at 2:21 and  was not  published. Please publish it. No more censorship! :

"I  asked you to try to find next prime p and a  n such that (n^(p^p)+2)/2011 is  very-very-very-very-probably prime because i found a prime p and a very-very-very-very probably prime n^(p^(p-1))/2011. You are allowed with both forms.

I do not need to find myself anything . Cause i am not a mathematician. Just an amateur. You are not a mathematician either, but  you are a  physicist.

¡ And the rest is just literature ..!

PS:  i would have continued philosophing but english is not my language and i am just an amateur philosopher. You´re the profesional here."

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• I post again the message since it seems to be lost and not published. n^(5^5)+2 is divisible by  2011 for: n=132 mod 2011 n=339 mod 2011 n=764 mod 2011 n=
Message 9 of 17 , Nov 25, 2011
I post again the message since it seems to be lost and not published.

n^(5^5)+2 is divisible by  2011 for:
n=132 mod 2011
n=339 mod 2011
n=764 mod 2011
n= 1048 mod 2011
n=1739 mod 2011

So the puzzle is much easier than what i thought. Though i found it experimentally, sadly. With excellent factordb program.

(n^(7^7)+2) /2011 must be prime somewhere for a value  n = 1646 mod 2011.These are huge PRP to be found with more than 2.6 million digits each.

I do not know if there is time to find one of these PRP before 2012.

[Non-text portions of this message have been removed]
• ... Since there are 365 days in 2011, you should search for a PRP of the form ((2*k*2011-365)^(7^7)+2)/2011. It is easy to show that there is none with less
Message 10 of 17 , Nov 25, 2011

> (n^(7^7)+2) /2011 must be prime somewhere

Since there are 365 days in 2011, you should search
for a PRP of the form ((2*k*2011-365)^(7^7)+2)/2011.
It is easy to show that there is none with less
than 4 million decimal digits.

> you can find infinitely many primes with almost any form

Puzzle 67: For p = 67, state the precise number of
primes of the form (n^(p^p)+2)/2011. Explain why 67
is the largest prime p for which such a puzzle is fair.

David (per proxy "znorder")
• () ASCII ribbon campaign () Hopeless ribbon campaign / against HTML mail / against gratuitous bloodshed [stolen with permission from
Message 11 of 17 , Nov 25, 2011
() ASCII ribbon campaign () Hopeless ribbon campaign
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[stolen with permission from Daniel B. Cristofani]

> Subject: [PrimeNumbers] Re: Primes with the representation of the year in them
> Date: Friday, November 25, 2011, 4:25 PM
>
>
>
>
> > (n^(7^7)+2) /2011 must be prime somewhere
>
> Since there are 365 days in 2011, you should search
> for a PRP of the form ((2*k*2011-365)^(7^7)+2)/2011.
> It is easy to show that there is none with less
> than 4 million decimal digits.
>
> > you can find infinitely many primes with almost any
> form
>
> Puzzle 67: For p = 67, state the precise number of
> primes of the form (n^(p^p)+2)/2011.

I don't see where that /2011 can be applied:

? polrootsmod(x^(67^67%eulerphi(2011))+2,2011)
[]~

> Explain why 67
> is the largest prime p for which such a puzzle is fair.

I see no reason why larger p wouldn't permit us to cover all everything divisible by 2011 with other primes (smaller, or larger). In fact, quite the opposite, I would expect it to be quite easy to arrive at a 0 answer.

Phil
• ... p=2: (n^4+2)/2011 is prime for n = 61119, 119871, 121449, 181779, 205911, 276729, 314505, ... p=3: n^27+2 is never divisible by 2011. p=5:
Message 12 of 17 , Nov 25, 2011
Robin Garcia wrote:
> What is the next prime p such that there is n such that :
> (n^(p^p)+2)/2011 is prime ??????

p=2: (n^4+2)/2011 is prime for
n = 61119, 119871, 121449, 181779, 205911, 276729, 314505, ...

p=3: n^27+2 is never divisible by 2011.

p=5: (8693685^3125+2)/2011 is Fermat and Lucas PRP! (151.9740s+0.0022s)
At 21682 digits I don't expect anybody to attempt ECPP.

p=7: n^823543+2 is only divisible by 2011 for n = 1646 (mod 2011).
The odd candidates are ((3657+4022*k)^823543+2)/2011.
The first without a factor below 10^9 is for k=33 with 4228693 digits.
I don't expect anybody to search prp's.

--
Jens Kruse Andersen
• ... Quite correct Phil. It is easy to prove that there are precisely 0 primes of the form (n^(67^67)+2)/2011 for the very good reason that there are precisely
Message 13 of 17 , Nov 25, 2011
Phil Carmody <thefatphil@...> wrote:

> > Puzzle 67: For p = 67, state the precise number of
> > primes of the form (n^(p^p)+2)/2011.
>
> I don't see where that /2011 can be applied

Quite correct Phil. It is easy to prove that
there are precisely 0 primes of the form
(n^(67^67)+2)/2011 for the very good reason that
there are precisely 0 integers of that form.

> > Explain why 67
> > is the largest prime p for which such a puzzle is fair.
>
> I would expect it to be quite easy to arrive at a 0 answer.

So what is the prime p > 67 for which Phil might
prove that (n^(p^p)+2)/2011 is never prime?

For myself, I was quite content to take p = 67
as largest prime divisor of the previous year :-)

David
• Thanks to everybody for the always positive feedback, with the exception of  David  (¡ oh, these Welsh !..). Thanks to Jens Kruse for his  always very
Message 14 of 17 , Nov 25, 2011
Thanks to everybody for the always positive feedback, with the exception of  David  (¡ oh, these Welsh !..).

Thanks to Jens Kruse for his  always very effective (and fast) finding of the computational bounds.

[Non-text portions of this message have been removed]
• ... For the good reason that 3 is a divisor of the order of -2 modulo 2011. The only other prime divisor is 67, which is why Puzzle 67 was so easy:
Message 15 of 17 , Nov 25, 2011
"Jens Kruse Andersen" <jens.k.a@...> wrote:

> p=3: n^27+2 is never divisible by 2011.

For the good reason that 3 is a divisor of the order
of -2 modulo 2011. The only other prime divisor is 67,
which is why Puzzle 67 was so easy:

print(factor(znorder(Mod(-2,2011))))

[3, 1; 67, 1]

David (per proxy "znorder")
• ... The concept of a covering set is familiar in the context of (n^q + c)/d, with (n, c, d) held fixed, but not (to me at least) in the present context, with
Message 16 of 17 , Nov 25, 2011
Phil Carmody <thefatphil@...> wrote:

> I see no reason why larger p wouldn't permit us to cover
> everything divisible by 2011 with other primes

The concept of a covering set is familiar in the context of
(n^q + c)/d, with (n, c, d) held fixed, but not (to me at least)
in the present context, with (q = p^p, c = 2, d = 2011) held fixed
and only the base, n, allowed to vary.

> I would expect it to be quite easy to arrive at a 0 answer.

I am unable to do so except in the trivial cases p = 3 and p = 67.
On the contrary, I conjecture that for every prime p that does not
divide znorder(Mod(-2,2011)) = 3*67 there is an infinitude of primes
of the form (n^(p^p)+2)/2011, with fixed p, even though we
presently know of no probable prime in any case with p > 5.

David (leaving himself uncovered?)
• ... Yeah, I m clearly confused. This happens occasionally. Phil
Message 17 of 17 , Nov 26, 2011
> Phil Carmody <thefatphil@...> wrote:
> > I see no reason why larger p wouldn't permit us to cover
> > everything divisible by 2011 with other primes
>
> The concept of a covering set is familiar in the context of
> (n^q + c)/d, with (n, c, d) held fixed, but not (to me at least)
> in the present context, with  (q = p^p, c = 2, d = 2011) held fixed
> and only the base, n, allowed to vary.
>
> >  I would expect it to be quite easy to arrive at a 0 answer.
>
> I am unable to do so except in the trivial cases p = 3 and
> p = 67.
> On the contrary, I conjecture that for every prime p that does not
> divide znorder(Mod(-2,2011)) = 3*67 there is an infinitude
> of primes
> of the form (n^(p^p)+2)/2011, with fixed p, even though we
>
> presently know of no probable prime in any case with p > 5.

Yeah, I'm clearly confused. This happens occasionally.

Phil
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