A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal...
$begingroup$
Question
Numbers $n$ of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. For example:
$k=1$ then $n=10$.
$k=2$ then $n=31$.
$k=3$ then $n=73$.
$k=4$ then $n=157.$
Conjecture:
the number $(2^k-1)cdot 10^m+2^{k-1}-1$ where $m$ is the number of decimal digits of $2^{k-1}$ is never prime when it is of the form $7s+6$, that is when it is congruent to $6$ $pmod 7$. Examples: $n=1023511$ ($k=10$)$equiv 6 pmod 7$ and thus it is composite $(1023511=19times103times523)$, $n=20471023$ ($k=11$) $equiv 6 pmod 7$ and thus it is composite ($20471023=479times42737)$. With PFGW we arrived to $k=565000$ and all the $n's$ congruent to $6 pmod 7$ are composite. According to Giovanni Resta's calculations in a post which has been canceled, there should be no probable prime congruent to 6 $pmod 7$ upto k=800.000. The residue $6$ $pmod 7$ occurs when either $m=6t+3$ and $k=3l+1$ or $m=6t+4$ and $k=3l+2$ with $k$ and $l$ some non-negative integers, but amazingly when it occurs the number is not prime. Can you find a counter-example or give a proof for the conjecture? Here a link to other interesting questions: Is there a number of the form $f(n)=7k+6=5p$ with prime p? and Why do all residues occur in this similar sequence?
For primes of this form see:
The on-line Encyclopedia of integer sequences
The following vector contains all the exponents k<=366800 leading to a prime
$[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$
Exponent $541456$ leads to another probable prime with residue 5 mod 7 and 325990 digits, but it need not be the next in increasing order.
Remark: we found five-in-a-row probable primes with res 5 mod 7. Probable primes with residue 5 are now twice frequent than expected.
Exponents of these primes seem to be NOT random at all. Another thing I noticed, i don't know if it has some importance: the exponents leading to a probable prime $215, 69660, 92020, 541456$ are multiples of $43$. I noticed that $frac{215}{41}, frac{69660}{41}, frac{92020}{41}, frac{541456}{41}$ all have a periodic decimal expansion equal to $overline{24390}=29^3+1$. This is equivalent to say that when k is a multiple of 43 and the number $10^{m}(2^{k}−1)+2^{k-1}−1$ is prime, then k is of the form $41s+r$ where r is a number in the set (1,10,16,18,37). Is there some mathematical reason for that?
number-theory integers
asked Feb 4 '18 at 13:13
Enzo CretiEnzo Creti
2041419
$endgroup$
|
show 38 more comments
$begingroup$
Question
Numbers $n$ of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. For example:
$k=1$ then $n=10$.
$k=2$ then $n=31$.
$k=3$ then $n=73$.
$k=4$ then $n=157.$
Conjecture:
the number $(2^k-1)cdot 10^m+2^{k-1}-1$ where $m$ is the number of decimal digits of $2^{k-1}$ is never prime when it is of the form $7s+6$, that is when it is congruent to $6$ $pmod 7$. Examples: $n=1023511$ ($k=10$)$equiv 6 pmod 7$ and thus it is composite $(1023511=19times103times523)$, $n=20471023$ ($k=11$) $equiv 6 pmod 7$ and thus it is composite ($20471023=479times42737)$. With PFGW we arrived to $k=565000$ and all the $n's$ congruent to $6 pmod 7$ are composite. According to Giovanni Resta's calculations in a post which has been canceled, there should be no probable prime congruent to 6 $pmod 7$ upto k=800.000. The residue $6$ $pmod 7$ occurs when either $m=6t+3$ and $k=3l+1$ or $m=6t+4$ and $k=3l+2$ with $k$ and $l$ some non-negative integers, but amazingly when it occurs the number is not prime. Can you find a counter-example or give a proof for the conjecture? Here a link to other interesting questions: Is there a number of the form $f(n)=7k+6=5p$ with prime p? and Why do all residues occur in this similar sequence?
For primes of this form see:
The on-line Encyclopedia of integer sequences
The following vector contains all the exponents k<=366800 leading to a prime
$[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$
Exponent $541456$ leads to another probable prime with residue 5 mod 7 and 325990 digits, but it need not be the next in increasing order.
Remark: we found five-in-a-row probable primes with res 5 mod 7. Probable primes with residue 5 are now twice frequent than expected.
Exponents of these primes seem to be NOT random at all. Another thing I noticed, i don't know if it has some importance: the exponents leading to a probable prime $215, 69660, 92020, 541456$ are multiples of $43$. I noticed that $frac{215}{41}, frac{69660}{41}, frac{92020}{41}, frac{541456}{41}$ all have a periodic decimal expansion equal to $overline{24390}=29^3+1$. This is equivalent to say that when k is a multiple of 43 and the number $10^{m}(2^{k}−1)+2^{k-1}−1$ is prime, then k is of the form $41s+r$ where r is a number in the set (1,10,16,18,37). Is there some mathematical reason for that?
number-theory integers
asked Feb 4 '18 at 13:13
Enzo CretiEnzo Creti
2041419
$endgroup$
2
$begingroup$
github.com/gnufinder/special-prime/issues
$endgroup$
– Peter
Feb 22 '18 at 16:56
4
$begingroup$
How can you give a bounty of $200$ when you only have $116$ reputation?
$endgroup$
– user477343
Mar 5 '18 at 6:52
1
$begingroup$
mathoverflow.net/questions/294527/…
$endgroup$
– Peter
Mar 6 '18 at 12:02
3
$begingroup$
It's worth to mention that if $n = (2^k-1)cdot 10^m + 2^{k-1}-1$, then $2n+1=(2cdot 10^m + 1)(2^k - 1)$. In particular, $n$ can never be a Sophie Germain prime.
$endgroup$
– Max Alekseyev
Mar 6 '18 at 17:22
3
$begingroup$
Please reduce your editing of this post. It is not necessary to keep the search limit quite up to date in the post. If you want to have the current search limit on the page, post a comment (and delete it when you post the next). Updating the post once or twice a week is plenty enough.
$endgroup$
– Daniel Fischer♦
Mar 9 '18 at 11:48
|
show 38 more comments
$begingroup$
Question
Numbers $n$ of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. For example:
$k=1$ then $n=10$.
$k=2$ then $n=31$.
$k=3$ then $n=73$.
$k=4$ then $n=157.$
Conjecture:
the number $(2^k-1)cdot 10^m+2^{k-1}-1$ where $m$ is the number of decimal digits of $2^{k-1}$ is never prime when it is of the form $7s+6$, that is when it is congruent to $6$ $pmod 7$. Examples: $n=1023511$ ($k=10$)$equiv 6 pmod 7$ and thus it is composite $(1023511=19times103times523)$, $n=20471023$ ($k=11$) $equiv 6 pmod 7$ and thus it is composite ($20471023=479times42737)$. With PFGW we arrived to $k=565000$ and all the $n's$ congruent to $6 pmod 7$ are composite. According to Giovanni Resta's calculations in a post which has been canceled, there should be no probable prime congruent to 6 $pmod 7$ upto k=800.000. The residue $6$ $pmod 7$ occurs when either $m=6t+3$ and $k=3l+1$ or $m=6t+4$ and $k=3l+2$ with $k$ and $l$ some non-negative integers, but amazingly when it occurs the number is not prime. Can you find a counter-example or give a proof for the conjecture? Here a link to other interesting questions: Is there a number of the form $f(n)=7k+6=5p$ with prime p? and Why do all residues occur in this similar sequence?
For primes of this form see:
The on-line Encyclopedia of integer sequences
The following vector contains all the exponents k<=366800 leading to a prime
$[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$
Exponent $541456$ leads to another probable prime with residue 5 mod 7 and 325990 digits, but it need not be the next in increasing order.
Remark: we found five-in-a-row probable primes with res 5 mod 7. Probable primes with residue 5 are now twice frequent than expected.
Exponents of these primes seem to be NOT random at all. Another thing I noticed, i don't know if it has some importance: the exponents leading to a probable prime $215, 69660, 92020, 541456$ are multiples of $43$. I noticed that $frac{215}{41}, frac{69660}{41}, frac{92020}{41}, frac{541456}{41}$ all have a periodic decimal expansion equal to $overline{24390}=29^3+1$. This is equivalent to say that when k is a multiple of 43 and the number $10^{m}(2^{k}−1)+2^{k-1}−1$ is prime, then k is of the form $41s+r$ where r is a number in the set (1,10,16,18,37). Is there some mathematical reason for that?
number-theory integers
asked Feb 4 '18 at 13:13
Enzo CretiEnzo Creti
2041419
$endgroup$
Question
Numbers $n$ of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. For example:
$k=1$ then $n=10$.
$k=2$ then $n=31$.
$k=3$ then $n=73$.
$k=4$ then $n=157.$
Conjecture:
the number $(2^k-1)cdot 10^m+2^{k-1}-1$ where $m$ is the number of decimal digits of $2^{k-1}$ is never prime when it is of the form $7s+6$, that is when it is congruent to $6$ $pmod 7$. Examples: $n=1023511$ ($k=10$)$equiv 6 pmod 7$ and thus it is composite $(1023511=19times103times523)$, $n=20471023$ ($k=11$) $equiv 6 pmod 7$ and thus it is composite ($20471023=479times42737)$. With PFGW we arrived to $k=565000$ and all the $n's$ congruent to $6 pmod 7$ are composite. According to Giovanni Resta's calculations in a post which has been canceled, there should be no probable prime congruent to 6 $pmod 7$ upto k=800.000. The residue $6$ $pmod 7$ occurs when either $m=6t+3$ and $k=3l+1$ or $m=6t+4$ and $k=3l+2$ with $k$ and $l$ some non-negative integers, but amazingly when it occurs the number is not prime. Can you find a counter-example or give a proof for the conjecture? Here a link to other interesting questions: Is there a number of the form $f(n)=7k+6=5p$ with prime p? and Why do all residues occur in this similar sequence?
For primes of this form see:
The on-line Encyclopedia of integer sequences
The following vector contains all the exponents k<=366800 leading to a prime
$[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$
Exponent $541456$ leads to another probable prime with residue 5 mod 7 and 325990 digits, but it need not be the next in increasing order.
Remark: we found five-in-a-row probable primes with res 5 mod 7. Probable primes with residue 5 are now twice frequent than expected.
Exponents of these primes seem to be NOT random at all. Another thing I noticed, i don't know if it has some importance: the exponents leading to a probable prime $215, 69660, 92020, 541456$ are multiples of $43$. I noticed that $frac{215}{41}, frac{69660}{41}, frac{92020}{41}, frac{541456}{41}$ all have a periodic decimal expansion equal to $overline{24390}=29^3+1$. This is equivalent to say that when k is a multiple of 43 and the number $10^{m}(2^{k}−1)+2^{k-1}−1$ is prime, then k is of the form $41s+r$ where r is a number in the set (1,10,16,18,37). Is there some mathematical reason for that?
number-theory integers
number-theory integers
asked Feb 4 '18 at 13:13
Enzo CretiEnzo Creti
2041419
asked Feb 4 '18 at 13:13
Enzo CretiEnzo Creti
2041419
edited Jan 17 at 10:03
Enzo Creti
asked Feb 4 '18 at 13:13
Enzo CretiEnzo Creti
2041419
asked Feb 4 '18 at 13:13
Enzo CretiEnzo Creti
2041419
asked Feb 4 '18 at 13:13
Enzo CretiEnzo Creti
2041419
2041419
2
$begingroup$
github.com/gnufinder/special-prime/issues
$endgroup$
– Peter
Feb 22 '18 at 16:56
4
$begingroup$
How can you give a bounty of $200$ when you only have $116$ reputation?
$endgroup$
– user477343
Mar 5 '18 at 6:52
1
$begingroup$
mathoverflow.net/questions/294527/…
$endgroup$
– Peter
Mar 6 '18 at 12:02
3
$begingroup$
It's worth to mention that if $n = (2^k-1)cdot 10^m + 2^{k-1}-1$, then $2n+1=(2cdot 10^m + 1)(2^k - 1)$. In particular, $n$ can never be a Sophie Germain prime.
$endgroup$
– Max Alekseyev
Mar 6 '18 at 17:22
3
$begingroup$
Please reduce your editing of this post. It is not necessary to keep the search limit quite up to date in the post. If you want to have the current search limit on the page, post a comment (and delete it when you post the next). Updating the post once or twice a week is plenty enough.
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– Daniel Fischer♦
Mar 9 '18 at 11:48
|
show 38 more comments
2
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github.com/gnufinder/special-prime/issues
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– Peter
Feb 22 '18 at 16:56
4
$begingroup$
How can you give a bounty of $200$ when you only have $116$ reputation?
$endgroup$
– user477343
Mar 5 '18 at 6:52
1
$begingroup$
mathoverflow.net/questions/294527/…
$endgroup$
– Peter
Mar 6 '18 at 12:02
3
$begingroup$
It's worth to mention that if $n = (2^k-1)cdot 10^m + 2^{k-1}-1$, then $2n+1=(2cdot 10^m + 1)(2^k - 1)$. In particular, $n$ can never be a Sophie Germain prime.
$endgroup$
– Max Alekseyev
Mar 6 '18 at 17:22
3
$begingroup$
Please reduce your editing of this post. It is not necessary to keep the search limit quite up to date in the post. If you want to have the current search limit on the page, post a comment (and delete it when you post the next). Updating the post once or twice a week is plenty enough.
$endgroup$
– Daniel Fischer♦
Mar 9 '18 at 11:48
2
2
$begingroup$
github.com/gnufinder/special-prime/issues
$endgroup$
– Peter
Feb 22 '18 at 16:56
$begingroup$
github.com/gnufinder/special-prime/issues
$endgroup$
– Peter
Feb 22 '18 at 16:56
4
4
$begingroup$
How can you give a bounty of $200$ when you only have $116$ reputation?
$endgroup$
– user477343
Mar 5 '18 at 6:52
$begingroup$
How can you give a bounty of $200$ when you only have $116$ reputation?
$endgroup$
– user477343
Mar 5 '18 at 6:52
1
1
$begingroup$
mathoverflow.net/questions/294527/…
$endgroup$
– Peter
Mar 6 '18 at 12:02
$begingroup$
mathoverflow.net/questions/294527/…
$endgroup$
– Peter
Mar 6 '18 at 12:02
3
3
$begingroup$
It's worth to mention that if $n = (2^k-1)cdot 10^m + 2^{k-1}-1$, then $2n+1=(2cdot 10^m + 1)(2^k - 1)$. In particular, $n$ can never be a Sophie Germain prime.
$endgroup$
– Max Alekseyev
Mar 6 '18 at 17:22
$begingroup$
It's worth to mention that if $n = (2^k-1)cdot 10^m + 2^{k-1}-1$, then $2n+1=(2cdot 10^m + 1)(2^k - 1)$. In particular, $n$ can never be a Sophie Germain prime.
$endgroup$
– Max Alekseyev
Mar 6 '18 at 17:22
3
3
$begingroup$
Please reduce your editing of this post. It is not necessary to keep the search limit quite up to date in the post. If you want to have the current search limit on the page, post a comment (and delete it when you post the next). Updating the post once or twice a week is plenty enough.
$endgroup$
– Daniel Fischer♦
Mar 9 '18 at 11:48
$begingroup$
Please reduce your editing of this post. It is not necessary to keep the search limit quite up to date in the post. If you want to have the current search limit on the page, post a comment (and delete it when you post the next). Updating the post once or twice a week is plenty enough.
$endgroup$
– Daniel Fischer♦
Mar 9 '18 at 11:48
|
show 38 more comments
2 Answers
2
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oldest
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According to your list, a counter-example, if it exists, must have more than $60,000$ digits. So, a counterexample would be a quite gigantic prime.
Unfortunately, a proof of the conjecture will almost certainly be out of reach.
The search for a counter-example can be painful as well, it is well possible that the smallest is already too big for current algorithms for primality testing.
edited Feb 19 '18 at 20:26
editor XD
54
answered Feb 4 '18 at 22:05
PeterPeter
47.6k1039131
$endgroup$
2
$begingroup$
@EnzoCreti No, this is only known for arithmetic progressions of the form $an+b$ with coprime $a$ and $b$ (Dirichlet). The numbers here will behave like "random" numbers , hence we won't be able to rule out a large prime.
$endgroup$
– Peter
Feb 5 '18 at 12:33
3
$begingroup$
@Peter Every number is of the form $10^{m}(2^k-1)+2^{k-1}-1$, where $m$ is the number of decimal digits of $2^{k-1}$. As $log_{2}10$ is irrational, the residues of $m$ and $k$ modulo 6, and hence the residues of $10^{m}$ and $2^{k-1}$ modulo 7, are independent. You can find out the distribution simply by tabulating the possible cases.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 16:03
3
$begingroup$
My observation just explains why different residues modulo 7 occur at different frequencies.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 19:34
2
$begingroup$
math.stackexchange.com/questions/2658464/…
$endgroup$
– Peter
Feb 20 '18 at 18:51
2
$begingroup$
For $k=32$, we have the factorization $$131cdot 4463cdot 21601cdot 44623 cdot 76213$$ This is a counterexample with residue $6$
$endgroup$
– Peter
Feb 21 '18 at 10:57
|
show 53 more comments
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@peter told me about this and i found it very interesting.
I made this tool, so anybody can help computing. Simply download and run to participate. Just refresh the page to update stats.
With 30-40 person, getting to $n=10^7$ should not be to long.
answered Jul 21 '18 at 22:54
François HuppéFrançois Huppé
362111
$endgroup$
$begingroup$
How many people participated already ?
$endgroup$
– Peter
Aug 3 '18 at 18:46
$begingroup$
@DanaJ have you a routine for Sage whcih allows to test this conjecture?
$endgroup$
– Enzo Creti
Dec 3 '18 at 13:45
add a comment |
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2 Answers
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2 Answers
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$begingroup$
According to your list, a counter-example, if it exists, must have more than $60,000$ digits. So, a counterexample would be a quite gigantic prime.
Unfortunately, a proof of the conjecture will almost certainly be out of reach.
The search for a counter-example can be painful as well, it is well possible that the smallest is already too big for current algorithms for primality testing.
edited Feb 19 '18 at 20:26
editor XD
54
answered Feb 4 '18 at 22:05
PeterPeter
47.6k1039131
$endgroup$
2
$begingroup$
@EnzoCreti No, this is only known for arithmetic progressions of the form $an+b$ with coprime $a$ and $b$ (Dirichlet). The numbers here will behave like "random" numbers , hence we won't be able to rule out a large prime.
$endgroup$
– Peter
Feb 5 '18 at 12:33
3
$begingroup$
@Peter Every number is of the form $10^{m}(2^k-1)+2^{k-1}-1$, where $m$ is the number of decimal digits of $2^{k-1}$. As $log_{2}10$ is irrational, the residues of $m$ and $k$ modulo 6, and hence the residues of $10^{m}$ and $2^{k-1}$ modulo 7, are independent. You can find out the distribution simply by tabulating the possible cases.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 16:03
3
$begingroup$
My observation just explains why different residues modulo 7 occur at different frequencies.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 19:34
2
$begingroup$
math.stackexchange.com/questions/2658464/…
$endgroup$
– Peter
Feb 20 '18 at 18:51
2
$begingroup$
For $k=32$, we have the factorization $$131cdot 4463cdot 21601cdot 44623 cdot 76213$$ This is a counterexample with residue $6$
$endgroup$
– Peter
Feb 21 '18 at 10:57
|
show 53 more comments
$begingroup$
According to your list, a counter-example, if it exists, must have more than $60,000$ digits. So, a counterexample would be a quite gigantic prime.
Unfortunately, a proof of the conjecture will almost certainly be out of reach.
The search for a counter-example can be painful as well, it is well possible that the smallest is already too big for current algorithms for primality testing.
edited Feb 19 '18 at 20:26
editor XD
54
answered Feb 4 '18 at 22:05
PeterPeter
47.6k1039131
$endgroup$
2
$begingroup$
@EnzoCreti No, this is only known for arithmetic progressions of the form $an+b$ with coprime $a$ and $b$ (Dirichlet). The numbers here will behave like "random" numbers , hence we won't be able to rule out a large prime.
$endgroup$
– Peter
Feb 5 '18 at 12:33
3
$begingroup$
@Peter Every number is of the form $10^{m}(2^k-1)+2^{k-1}-1$, where $m$ is the number of decimal digits of $2^{k-1}$. As $log_{2}10$ is irrational, the residues of $m$ and $k$ modulo 6, and hence the residues of $10^{m}$ and $2^{k-1}$ modulo 7, are independent. You can find out the distribution simply by tabulating the possible cases.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 16:03
3
$begingroup$
My observation just explains why different residues modulo 7 occur at different frequencies.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 19:34
2
$begingroup$
math.stackexchange.com/questions/2658464/…
$endgroup$
– Peter
Feb 20 '18 at 18:51
2
$begingroup$
For $k=32$, we have the factorization $$131cdot 4463cdot 21601cdot 44623 cdot 76213$$ This is a counterexample with residue $6$
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– Peter
Feb 21 '18 at 10:57
|
show 53 more comments
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According to your list, a counter-example, if it exists, must have more than $60,000$ digits. So, a counterexample would be a quite gigantic prime.
Unfortunately, a proof of the conjecture will almost certainly be out of reach.
The search for a counter-example can be painful as well, it is well possible that the smallest is already too big for current algorithms for primality testing.
edited Feb 19 '18 at 20:26
editor XD
54
answered Feb 4 '18 at 22:05
PeterPeter
47.6k1039131
$endgroup$
According to your list, a counter-example, if it exists, must have more than $60,000$ digits. So, a counterexample would be a quite gigantic prime.
Unfortunately, a proof of the conjecture will almost certainly be out of reach.
The search for a counter-example can be painful as well, it is well possible that the smallest is already too big for current algorithms for primality testing.
edited Feb 19 '18 at 20:26
editor XD
54
answered Feb 4 '18 at 22:05
PeterPeter
47.6k1039131
edited Feb 19 '18 at 20:26
editor XD
54
edited Feb 19 '18 at 20:26
editor XD
54
edited Feb 19 '18 at 20:26
editor XD
54
54
answered Feb 4 '18 at 22:05
PeterPeter
47.6k1039131
answered Feb 4 '18 at 22:05
PeterPeter
47.6k1039131
answered Feb 4 '18 at 22:05
PeterPeter
47.6k1039131
47.6k1039131
2
$begingroup$
@EnzoCreti No, this is only known for arithmetic progressions of the form $an+b$ with coprime $a$ and $b$ (Dirichlet). The numbers here will behave like "random" numbers , hence we won't be able to rule out a large prime.
$endgroup$
– Peter
Feb 5 '18 at 12:33
3
$begingroup$
@Peter Every number is of the form $10^{m}(2^k-1)+2^{k-1}-1$, where $m$ is the number of decimal digits of $2^{k-1}$. As $log_{2}10$ is irrational, the residues of $m$ and $k$ modulo 6, and hence the residues of $10^{m}$ and $2^{k-1}$ modulo 7, are independent. You can find out the distribution simply by tabulating the possible cases.
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– Taneli Huuskonen
Feb 6 '18 at 16:03
3
$begingroup$
My observation just explains why different residues modulo 7 occur at different frequencies.
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– Taneli Huuskonen
Feb 6 '18 at 19:34
2
$begingroup$
math.stackexchange.com/questions/2658464/…
$endgroup$
– Peter
Feb 20 '18 at 18:51
2
$begingroup$
For $k=32$, we have the factorization $$131cdot 4463cdot 21601cdot 44623 cdot 76213$$ This is a counterexample with residue $6$
$endgroup$
– Peter
Feb 21 '18 at 10:57
|
show 53 more comments
2
$begingroup$
@EnzoCreti No, this is only known for arithmetic progressions of the form $an+b$ with coprime $a$ and $b$ (Dirichlet). The numbers here will behave like "random" numbers , hence we won't be able to rule out a large prime.
$endgroup$
– Peter
Feb 5 '18 at 12:33
3
$begingroup$
@Peter Every number is of the form $10^{m}(2^k-1)+2^{k-1}-1$, where $m$ is the number of decimal digits of $2^{k-1}$. As $log_{2}10$ is irrational, the residues of $m$ and $k$ modulo 6, and hence the residues of $10^{m}$ and $2^{k-1}$ modulo 7, are independent. You can find out the distribution simply by tabulating the possible cases.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 16:03
3
$begingroup$
My observation just explains why different residues modulo 7 occur at different frequencies.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 19:34
2
$begingroup$
math.stackexchange.com/questions/2658464/…
$endgroup$
– Peter
Feb 20 '18 at 18:51
2
$begingroup$
For $k=32$, we have the factorization $$131cdot 4463cdot 21601cdot 44623 cdot 76213$$ This is a counterexample with residue $6$
$endgroup$
– Peter
Feb 21 '18 at 10:57
2
2
$begingroup$
@EnzoCreti No, this is only known for arithmetic progressions of the form $an+b$ with coprime $a$ and $b$ (Dirichlet). The numbers here will behave like "random" numbers , hence we won't be able to rule out a large prime.
$endgroup$
– Peter
Feb 5 '18 at 12:33
$begingroup$
@EnzoCreti No, this is only known for arithmetic progressions of the form $an+b$ with coprime $a$ and $b$ (Dirichlet). The numbers here will behave like "random" numbers , hence we won't be able to rule out a large prime.
$endgroup$
– Peter
Feb 5 '18 at 12:33
3
3
$begingroup$
@Peter Every number is of the form $10^{m}(2^k-1)+2^{k-1}-1$, where $m$ is the number of decimal digits of $2^{k-1}$. As $log_{2}10$ is irrational, the residues of $m$ and $k$ modulo 6, and hence the residues of $10^{m}$ and $2^{k-1}$ modulo 7, are independent. You can find out the distribution simply by tabulating the possible cases.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 16:03
$begingroup$
@Peter Every number is of the form $10^{m}(2^k-1)+2^{k-1}-1$, where $m$ is the number of decimal digits of $2^{k-1}$. As $log_{2}10$ is irrational, the residues of $m$ and $k$ modulo 6, and hence the residues of $10^{m}$ and $2^{k-1}$ modulo 7, are independent. You can find out the distribution simply by tabulating the possible cases.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 16:03
3
3
$begingroup$
My observation just explains why different residues modulo 7 occur at different frequencies.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 19:34
$begingroup$
My observation just explains why different residues modulo 7 occur at different frequencies.
$endgroup$
– Taneli Huuskonen
Feb 6 '18 at 19:34
2
2
$begingroup$
math.stackexchange.com/questions/2658464/…
$endgroup$
– Peter
Feb 20 '18 at 18:51
$begingroup$
math.stackexchange.com/questions/2658464/…
$endgroup$
– Peter
Feb 20 '18 at 18:51
2
2
$begingroup$
For $k=32$, we have the factorization $$131cdot 4463cdot 21601cdot 44623 cdot 76213$$ This is a counterexample with residue $6$
$endgroup$
– Peter
Feb 21 '18 at 10:57
$begingroup$
For $k=32$, we have the factorization $$131cdot 4463cdot 21601cdot 44623 cdot 76213$$ This is a counterexample with residue $6$
$endgroup$
– Peter
Feb 21 '18 at 10:57
|
show 53 more comments
$begingroup$
@peter told me about this and i found it very interesting.
I made this tool, so anybody can help computing. Simply download and run to participate. Just refresh the page to update stats.
With 30-40 person, getting to $n=10^7$ should not be to long.
answered Jul 21 '18 at 22:54
François HuppéFrançois Huppé
362111
$endgroup$
$begingroup$
How many people participated already ?
$endgroup$
– Peter
Aug 3 '18 at 18:46
$begingroup$
@DanaJ have you a routine for Sage whcih allows to test this conjecture?
$endgroup$
– Enzo Creti
Dec 3 '18 at 13:45
add a comment |
$begingroup$
@peter told me about this and i found it very interesting.
I made this tool, so anybody can help computing. Simply download and run to participate. Just refresh the page to update stats.
With 30-40 person, getting to $n=10^7$ should not be to long.
answered Jul 21 '18 at 22:54
François HuppéFrançois Huppé
362111
$endgroup$
$begingroup$
How many people participated already ?
$endgroup$
– Peter
Aug 3 '18 at 18:46
$begingroup$
@DanaJ have you a routine for Sage whcih allows to test this conjecture?
$endgroup$
– Enzo Creti
Dec 3 '18 at 13:45
add a comment |
$begingroup$
@peter told me about this and i found it very interesting.
I made this tool, so anybody can help computing. Simply download and run to participate. Just refresh the page to update stats.
With 30-40 person, getting to $n=10^7$ should not be to long.
answered Jul 21 '18 at 22:54
François HuppéFrançois Huppé
362111
$endgroup$
@peter told me about this and i found it very interesting.
I made this tool, so anybody can help computing. Simply download and run to participate. Just refresh the page to update stats.
With 30-40 person, getting to $n=10^7$ should not be to long.
answered Jul 21 '18 at 22:54
François HuppéFrançois Huppé
362111
answered Jul 21 '18 at 22:54
François HuppéFrançois Huppé
362111
answered Jul 21 '18 at 22:54
François HuppéFrançois Huppé
362111
answered Jul 21 '18 at 22:54
François HuppéFrançois Huppé
362111
362111
$begingroup$
How many people participated already ?
$endgroup$
– Peter
Aug 3 '18 at 18:46
$begingroup$
@DanaJ have you a routine for Sage whcih allows to test this conjecture?
$endgroup$
– Enzo Creti
Dec 3 '18 at 13:45
add a comment |
$begingroup$
How many people participated already ?
$endgroup$
– Peter
Aug 3 '18 at 18:46
$begingroup$
@DanaJ have you a routine for Sage whcih allows to test this conjecture?
$endgroup$
– Enzo Creti
Dec 3 '18 at 13:45
$begingroup$
How many people participated already ?
$endgroup$
– Peter
Aug 3 '18 at 18:46
$begingroup$
How many people participated already ?
$endgroup$
– Peter
Aug 3 '18 at 18:46
$begingroup$
@DanaJ have you a routine for Sage whcih allows to test this conjecture?
$endgroup$
– Enzo Creti
Dec 3 '18 at 13:45
$begingroup$
@DanaJ have you a routine for Sage whcih allows to test this conjecture?
$endgroup$
– Enzo Creti
Dec 3 '18 at 13:45
add a comment |
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2
$begingroup$
github.com/gnufinder/special-prime/issues
$endgroup$
– Peter
Feb 22 '18 at 16:56
4
$begingroup$
How can you give a bounty of $200$ when you only have $116$ reputation?
$endgroup$
– user477343
Mar 5 '18 at 6:52
1
$begingroup$
mathoverflow.net/questions/294527/…
$endgroup$
– Peter
Mar 6 '18 at 12:02
3
$begingroup$
It's worth to mention that if $n = (2^k-1)cdot 10^m + 2^{k-1}-1$, then $2n+1=(2cdot 10^m + 1)(2^k - 1)$. In particular, $n$ can never be a Sophie Germain prime.
$endgroup$
– Max Alekseyev
Mar 6 '18 at 17:22
3
$begingroup$
Please reduce your editing of this post. It is not necessary to keep the search limit quite up to date in the post. If you want to have the current search limit on the page, post a comment (and delete it when you post the next). Updating the post once or twice a week is plenty enough.
$endgroup$
– Daniel Fischer♦
Mar 9 '18 at 11:48