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Primes of the form $((2^k-1)10^m+2^{(k-1)}+10)/42$, where m is the number of decimal digits of $2^{k-1}-1$
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Primes of the form $dfrac{(2^k-1)*10^m+2^{(k-1)}+10}{42}$, where $m$ is the number of decimal digits of $2^{k-1}-1$.
With Pari I found that primes occur for $k=3,, 6, 12, 36, 105, 156,336, 2286, 4272,$ $4427, 11979, 20076, 29343, 29988, 30405$. $:$
The first thing I would ask is this: why there is only one $k=4427$ which is not a multiple of 3, whereas all the other $k$'s $(3, 6, 12, 36, 105, 156, 336, 2286, 4272, 11979, 20076, 29343, 29988, 30405)$ are congruent to $0mod3$.$:$ Is there any mathematical reason?$:$
The second question is: could be $k$ be of the form $3s+1$, i mean is there a $k$ of the form $3s+1$ such that $dfrac{(2^k-1)*10^m+2^{(k-1)}+10}{42}$ is prime?
The question is related to this other question:
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}$.
number-theory
asked Nov 15 at 12:48
paolo galli
213
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up vote
2
down vote
favorite
Primes of the form $dfrac{(2^k-1)*10^m+2^{(k-1)}+10}{42}$, where $m$ is the number of decimal digits of $2^{k-1}-1$.
With Pari I found that primes occur for $k=3,, 6, 12, 36, 105, 156,336, 2286, 4272,$ $4427, 11979, 20076, 29343, 29988, 30405$. $:$
The first thing I would ask is this: why there is only one $k=4427$ which is not a multiple of 3, whereas all the other $k$'s $(3, 6, 12, 36, 105, 156, 336, 2286, 4272, 11979, 20076, 29343, 29988, 30405)$ are congruent to $0mod3$.$:$ Is there any mathematical reason?$:$
The second question is: could be $k$ be of the form $3s+1$, i mean is there a $k$ of the form $3s+1$ such that $dfrac{(2^k-1)*10^m+2^{(k-1)}+10}{42}$ is prime?
The question is related to this other question:
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}$.
number-theory
asked Nov 15 at 12:48
paolo galli
213
New contributor
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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@Taneli Huuskonen@lulu is there some mathematical reason for that?
– paolo galli
Nov 15 at 13:10
(PARI) ec(n)= fromdigits(concat(digits(2^n-1), digits(2^(n-1)-1))) for(j=2, 10^4, s=ec(j); if(Mod(s+11, 42)==0, if(ispseudoprime((s+11)/42)==1, print1(j, ", "))))
– paolo galli
Nov 15 at 13:21
@Especially Lime is there any particular mathematical reason why primes of this type occur more often when k is a multiple of 3?
– paolo galli
Nov 15 at 14:10
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Primes of the form $dfrac{(2^k-1)*10^m+2^{(k-1)}+10}{42}$, where $m$ is the number of decimal digits of $2^{k-1}-1$.
With Pari I found that primes occur for $k=3,, 6, 12, 36, 105, 156,336, 2286, 4272,$ $4427, 11979, 20076, 29343, 29988, 30405$. $:$
The first thing I would ask is this: why there is only one $k=4427$ which is not a multiple of 3, whereas all the other $k$'s $(3, 6, 12, 36, 105, 156, 336, 2286, 4272, 11979, 20076, 29343, 29988, 30405)$ are congruent to $0mod3$.$:$ Is there any mathematical reason?$:$
The second question is: could be $k$ be of the form $3s+1$, i mean is there a $k$ of the form $3s+1$ such that $dfrac{(2^k-1)*10^m+2^{(k-1)}+10}{42}$ is prime?
The question is related to this other question:
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}$.
number-theory
asked Nov 15 at 12:48
paolo galli
213
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.
Primes of the form $dfrac{(2^k-1)*10^m+2^{(k-1)}+10}{42}$, where $m$ is the number of decimal digits of $2^{k-1}-1$.
With Pari I found that primes occur for $k=3,, 6, 12, 36, 105, 156,336, 2286, 4272,$ $4427, 11979, 20076, 29343, 29988, 30405$. $:$
The first thing I would ask is this: why there is only one $k=4427$ which is not a multiple of 3, whereas all the other $k$'s $(3, 6, 12, 36, 105, 156, 336, 2286, 4272, 11979, 20076, 29343, 29988, 30405)$ are congruent to $0mod3$.$:$ Is there any mathematical reason?$:$
The second question is: could be $k$ be of the form $3s+1$, i mean is there a $k$ of the form $3s+1$ such that $dfrac{(2^k-1)*10^m+2^{(k-1)}+10}{42}$ is prime?
The question is related to this other question:
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}$.
number-theory
number-theory
asked Nov 15 at 12:48
paolo galli
213
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 Nov 15 at 12:48
paolo galli
213
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 Nov 16 at 7:39
asked Nov 15 at 12:48
paolo galli
213
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 Nov 15 at 12:48
paolo galli
213
asked Nov 15 at 12:48
paolo galli
213
213
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.
@Taneli Huuskonen@lulu is there some mathematical reason for that?
– paolo galli
Nov 15 at 13:10
(PARI) ec(n)= fromdigits(concat(digits(2^n-1), digits(2^(n-1)-1))) for(j=2, 10^4, s=ec(j); if(Mod(s+11, 42)==0, if(ispseudoprime((s+11)/42)==1, print1(j, ", "))))
– paolo galli
Nov 15 at 13:21
@Especially Lime is there any particular mathematical reason why primes of this type occur more often when k is a multiple of 3?
– paolo galli
Nov 15 at 14:10
add a comment |
@Taneli Huuskonen@lulu is there some mathematical reason for that?
– paolo galli
Nov 15 at 13:10
(PARI) ec(n)= fromdigits(concat(digits(2^n-1), digits(2^(n-1)-1))) for(j=2, 10^4, s=ec(j); if(Mod(s+11, 42)==0, if(ispseudoprime((s+11)/42)==1, print1(j, ", "))))
– paolo galli
Nov 15 at 13:21
@Especially Lime is there any particular mathematical reason why primes of this type occur more often when k is a multiple of 3?
– paolo galli
Nov 15 at 14:10
@Taneli Huuskonen@lulu is there some mathematical reason for that?
– paolo galli
Nov 15 at 13:10
@Taneli Huuskonen@lulu is there some mathematical reason for that?
– paolo galli
Nov 15 at 13:10
(PARI) ec(n)= fromdigits(concat(digits(2^n-1), digits(2^(n-1)-1))) for(j=2, 10^4, s=ec(j); if(Mod(s+11, 42)==0, if(ispseudoprime((s+11)/42)==1, print1(j, ", "))))
– paolo galli
Nov 15 at 13:21
(PARI) ec(n)= fromdigits(concat(digits(2^n-1), digits(2^(n-1)-1))) for(j=2, 10^4, s=ec(j); if(Mod(s+11, 42)==0, if(ispseudoprime((s+11)/42)==1, print1(j, ", "))))
– paolo galli
Nov 15 at 13:21
@Especially Lime is there any particular mathematical reason why primes of this type occur more often when k is a multiple of 3?
– paolo galli
Nov 15 at 14:10
@Especially Lime is there any particular mathematical reason why primes of this type occur more often when k is a multiple of 3?
– paolo galli
Nov 15 at 14:10
add a comment |
1 Answer
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If 3|k then Mod(2^k - 1, 7) = 0 and Mod(2^(k-1) - 1, 7) = 3, so Mod(ec(k),7) = 3 for any m.
If 3 does not divide k, the fact that 10 is a primitive root (mod 7) comes into play. In each case, there will only be one residue class of Mod(m, 6) that makes Mod(ec(k),7) = 3.
If k == 1 (mod 3) then Mod(2^(k-1) - 1, 7) = 0 and Mod(2^k - 1, 7) = 1, so Mod(ec(k), 7) = Mod(10^m, 7). This is 3 when Mod(m, 6) = 1.
If k == 2 (mod 3) then we have Mod(ec(k), 7) = Mod(3*10^m + 1, 7), which is again 3 when Mod(m,6) = 1.
Of the 18 possible pairs (Mod(k,3), Mod(m, 6)) then, the 6 pairs with Mod(k,3) = 0 , the pair (Mod(k,3) = 1, Mod(m,6) = 1), and the pair (Mod(k,3) = 2, Mod(m,6) = 1) allow Mod(pw(k), 7) = 3.
For the remaining ten pairs, Mod(ec(k), 7) is not 3.
So, ec(k) + 4 is divisible by 7 about 6 times as often for 3|k as for k == 1 (mod 3), and about 6 times as often as for k == 2 (mod 3). A numerical check up to k = 10000 confirms this.
answered 12 hours ago
paolo galli
213
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1 Answer
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1 Answer
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If 3|k then Mod(2^k - 1, 7) = 0 and Mod(2^(k-1) - 1, 7) = 3, so Mod(ec(k),7) = 3 for any m.
If 3 does not divide k, the fact that 10 is a primitive root (mod 7) comes into play. In each case, there will only be one residue class of Mod(m, 6) that makes Mod(ec(k),7) = 3.
If k == 1 (mod 3) then Mod(2^(k-1) - 1, 7) = 0 and Mod(2^k - 1, 7) = 1, so Mod(ec(k), 7) = Mod(10^m, 7). This is 3 when Mod(m, 6) = 1.
If k == 2 (mod 3) then we have Mod(ec(k), 7) = Mod(3*10^m + 1, 7), which is again 3 when Mod(m,6) = 1.
Of the 18 possible pairs (Mod(k,3), Mod(m, 6)) then, the 6 pairs with Mod(k,3) = 0 , the pair (Mod(k,3) = 1, Mod(m,6) = 1), and the pair (Mod(k,3) = 2, Mod(m,6) = 1) allow Mod(pw(k), 7) = 3.
For the remaining ten pairs, Mod(ec(k), 7) is not 3.
So, ec(k) + 4 is divisible by 7 about 6 times as often for 3|k as for k == 1 (mod 3), and about 6 times as often as for k == 2 (mod 3). A numerical check up to k = 10000 confirms this.
answered 12 hours ago
paolo galli
213
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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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If 3|k then Mod(2^k - 1, 7) = 0 and Mod(2^(k-1) - 1, 7) = 3, so Mod(ec(k),7) = 3 for any m.
If 3 does not divide k, the fact that 10 is a primitive root (mod 7) comes into play. In each case, there will only be one residue class of Mod(m, 6) that makes Mod(ec(k),7) = 3.
If k == 1 (mod 3) then Mod(2^(k-1) - 1, 7) = 0 and Mod(2^k - 1, 7) = 1, so Mod(ec(k), 7) = Mod(10^m, 7). This is 3 when Mod(m, 6) = 1.
If k == 2 (mod 3) then we have Mod(ec(k), 7) = Mod(3*10^m + 1, 7), which is again 3 when Mod(m,6) = 1.
Of the 18 possible pairs (Mod(k,3), Mod(m, 6)) then, the 6 pairs with Mod(k,3) = 0 , the pair (Mod(k,3) = 1, Mod(m,6) = 1), and the pair (Mod(k,3) = 2, Mod(m,6) = 1) allow Mod(pw(k), 7) = 3.
For the remaining ten pairs, Mod(ec(k), 7) is not 3.
So, ec(k) + 4 is divisible by 7 about 6 times as often for 3|k as for k == 1 (mod 3), and about 6 times as often as for k == 2 (mod 3). A numerical check up to k = 10000 confirms this.
answered 12 hours ago
paolo galli
213
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up vote
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If 3|k then Mod(2^k - 1, 7) = 0 and Mod(2^(k-1) - 1, 7) = 3, so Mod(ec(k),7) = 3 for any m.
If 3 does not divide k, the fact that 10 is a primitive root (mod 7) comes into play. In each case, there will only be one residue class of Mod(m, 6) that makes Mod(ec(k),7) = 3.
If k == 1 (mod 3) then Mod(2^(k-1) - 1, 7) = 0 and Mod(2^k - 1, 7) = 1, so Mod(ec(k), 7) = Mod(10^m, 7). This is 3 when Mod(m, 6) = 1.
If k == 2 (mod 3) then we have Mod(ec(k), 7) = Mod(3*10^m + 1, 7), which is again 3 when Mod(m,6) = 1.
Of the 18 possible pairs (Mod(k,3), Mod(m, 6)) then, the 6 pairs with Mod(k,3) = 0 , the pair (Mod(k,3) = 1, Mod(m,6) = 1), and the pair (Mod(k,3) = 2, Mod(m,6) = 1) allow Mod(pw(k), 7) = 3.
For the remaining ten pairs, Mod(ec(k), 7) is not 3.
So, ec(k) + 4 is divisible by 7 about 6 times as often for 3|k as for k == 1 (mod 3), and about 6 times as often as for k == 2 (mod 3). A numerical check up to k = 10000 confirms this.
answered 12 hours ago
paolo galli
213
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.
If 3|k then Mod(2^k - 1, 7) = 0 and Mod(2^(k-1) - 1, 7) = 3, so Mod(ec(k),7) = 3 for any m.
If 3 does not divide k, the fact that 10 is a primitive root (mod 7) comes into play. In each case, there will only be one residue class of Mod(m, 6) that makes Mod(ec(k),7) = 3.
If k == 1 (mod 3) then Mod(2^(k-1) - 1, 7) = 0 and Mod(2^k - 1, 7) = 1, so Mod(ec(k), 7) = Mod(10^m, 7). This is 3 when Mod(m, 6) = 1.
If k == 2 (mod 3) then we have Mod(ec(k), 7) = Mod(3*10^m + 1, 7), which is again 3 when Mod(m,6) = 1.
Of the 18 possible pairs (Mod(k,3), Mod(m, 6)) then, the 6 pairs with Mod(k,3) = 0 , the pair (Mod(k,3) = 1, Mod(m,6) = 1), and the pair (Mod(k,3) = 2, Mod(m,6) = 1) allow Mod(pw(k), 7) = 3.
For the remaining ten pairs, Mod(ec(k), 7) is not 3.
So, ec(k) + 4 is divisible by 7 about 6 times as often for 3|k as for k == 1 (mod 3), and about 6 times as often as for k == 2 (mod 3). A numerical check up to k = 10000 confirms this.
answered 12 hours ago
paolo galli
213
New contributor
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 12 hours ago
paolo galli
213
New contributor
paolo galli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 12 hours ago
paolo galli
213
answered 12 hours ago
paolo galli
213
213
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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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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.
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add a comment |
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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@Taneli Huuskonen@lulu is there some mathematical reason for that?
– paolo galli
Nov 15 at 13:10
(PARI) ec(n)= fromdigits(concat(digits(2^n-1), digits(2^(n-1)-1))) for(j=2, 10^4, s=ec(j); if(Mod(s+11, 42)==0, if(ispseudoprime((s+11)/42)==1, print1(j, ", "))))
– paolo galli
Nov 15 at 13:21
@Especially Lime is there any particular mathematical reason why primes of this type occur more often when k is a multiple of 3?
– paolo galli
Nov 15 at 14:10