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Exponents of ec primes ending with digit 7
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A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$.
In this question ec-numbers are introduced, obttained by the concatenation of two consecutive Mersenne numbers (12763=ec(7) for example). The values of $k$ for which $ec(k)$ is prime are: $[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$
$7, 67, 360787$ are all the $k's$ leading to a prime ending with digit $7$. $7,67,360787$ are all congruent to $1$ $pmod 6$. My question is: does a k exist such that k ends with digit $7$, $ec(k)$ is prime and $k$ is NOT congruent to 1 $pmod 6$? Or that can be ruled out? In other words, if $k$ is congruent to 7 mod 10 and $ec(k)$ is prime, then necessarly $k$ must be congruent to 1 mod 6?
number-theory
asked 7 hours ago
paolo galli
163
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up vote
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A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$.
In this question ec-numbers are introduced, obttained by the concatenation of two consecutive Mersenne numbers (12763=ec(7) for example). The values of $k$ for which $ec(k)$ is prime are: $[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$
$7, 67, 360787$ are all the $k's$ leading to a prime ending with digit $7$. $7,67,360787$ are all congruent to $1$ $pmod 6$. My question is: does a k exist such that k ends with digit $7$, $ec(k)$ is prime and $k$ is NOT congruent to 1 $pmod 6$? Or that can be ruled out? In other words, if $k$ is congruent to 7 mod 10 and $ec(k)$ is prime, then necessarly $k$ must be congruent to 1 mod 6?
number-theory
asked 7 hours ago
paolo galli
163
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paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I noticed that the largest factor of $6,66,360786$ is a palindrome. The largest factor of 6 is 3, the largest factor of 66 is 11, the largest factor of 360786 is 383.
– paolo galli
6 hours ago
@Giovanni Resta can you find a counter-example?
– paolo galli
6 hours ago
@Peter i think it is hard to find a counterexample, maybe it can be ruled out with a proof.
– paolo galli
6 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$.
In this question ec-numbers are introduced, obttained by the concatenation of two consecutive Mersenne numbers (12763=ec(7) for example). The values of $k$ for which $ec(k)$ is prime are: $[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$
$7, 67, 360787$ are all the $k's$ leading to a prime ending with digit $7$. $7,67,360787$ are all congruent to $1$ $pmod 6$. My question is: does a k exist such that k ends with digit $7$, $ec(k)$ is prime and $k$ is NOT congruent to 1 $pmod 6$? Or that can be ruled out? In other words, if $k$ is congruent to 7 mod 10 and $ec(k)$ is prime, then necessarly $k$ must be congruent to 1 mod 6?
number-theory
asked 7 hours ago
paolo galli
163
New contributor
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$.
In this question ec-numbers are introduced, obttained by the concatenation of two consecutive Mersenne numbers (12763=ec(7) for example). The values of $k$ for which $ec(k)$ is prime are: $[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$
$7, 67, 360787$ are all the $k's$ leading to a prime ending with digit $7$. $7,67,360787$ are all congruent to $1$ $pmod 6$. My question is: does a k exist such that k ends with digit $7$, $ec(k)$ is prime and $k$ is NOT congruent to 1 $pmod 6$? Or that can be ruled out? In other words, if $k$ is congruent to 7 mod 10 and $ec(k)$ is prime, then necessarly $k$ must be congruent to 1 mod 6?
number-theory
number-theory
asked 7 hours ago
paolo galli
163
New contributor
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 7 hours ago
paolo galli
163
New contributor
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 7 hours ago
asked 7 hours ago
paolo galli
163
New contributor
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
paolo galli
163
asked 7 hours ago
paolo galli
163
163
New contributor
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I noticed that the largest factor of $6,66,360786$ is a palindrome. The largest factor of 6 is 3, the largest factor of 66 is 11, the largest factor of 360786 is 383.
– paolo galli
6 hours ago
@Giovanni Resta can you find a counter-example?
– paolo galli
6 hours ago
@Peter i think it is hard to find a counterexample, maybe it can be ruled out with a proof.
– paolo galli
6 hours ago
add a comment |
I noticed that the largest factor of $6,66,360786$ is a palindrome. The largest factor of 6 is 3, the largest factor of 66 is 11, the largest factor of 360786 is 383.
– paolo galli
6 hours ago
@Giovanni Resta can you find a counter-example?
– paolo galli
6 hours ago
@Peter i think it is hard to find a counterexample, maybe it can be ruled out with a proof.
– paolo galli
6 hours ago
I noticed that the largest factor of $6,66,360786$ is a palindrome. The largest factor of 6 is 3, the largest factor of 66 is 11, the largest factor of 360786 is 383.
– paolo galli
6 hours ago
I noticed that the largest factor of $6,66,360786$ is a palindrome. The largest factor of 6 is 3, the largest factor of 66 is 11, the largest factor of 360786 is 383.
– paolo galli
6 hours ago
@Giovanni Resta can you find a counter-example?
– paolo galli
6 hours ago
@Giovanni Resta can you find a counter-example?
– paolo galli
6 hours ago
@Peter i think it is hard to find a counterexample, maybe it can be ruled out with a proof.
– paolo galli
6 hours ago
@Peter i think it is hard to find a counterexample, maybe it can be ruled out with a proof.
– paolo galli
6 hours ago
add a comment |
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paolo galli is a new contributor. Be nice, and check out our Code of Conduct.
paolo galli is a new contributor. Be nice, and check out our Code of Conduct.
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I noticed that the largest factor of $6,66,360786$ is a palindrome. The largest factor of 6 is 3, the largest factor of 66 is 11, the largest factor of 360786 is 383.
– paolo galli
6 hours ago
@Giovanni Resta can you find a counter-example?
– paolo galli
6 hours ago
@Peter i think it is hard to find a counterexample, maybe it can be ruled out with a proof.
– paolo galli
6 hours ago