Even numbers of the form $frac n(n+1)2$ and twin primes
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If we take some even number of the form $fracn(n+1)2$ and add $1$ to it and also subtract $1$ from it then we have a mapping $fracn(n+1)2toleftfracn(n+1)2-1,fracn(n+1)2+1right$.
From some even numbers of that form that I checked only number $6=frac4cdot 32$ maps to a twin prime pair $6 to 5,7$.
Is there any simple explanation why $leftfracn(n+1)2-1, fracn(n+1)2+1right$ is so rarely a twin prime pair?
elementary-number-theory prime-numbers prime-twins
edited Aug 23 at 9:32
Stefan4024
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asked Aug 23 at 9:25
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up vote
3
down vote
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If we take some even number of the form $fracn(n+1)2$ and add $1$ to it and also subtract $1$ from it then we have a mapping $fracn(n+1)2toleftfracn(n+1)2-1,fracn(n+1)2+1right$.
From some even numbers of that form that I checked only number $6=frac4cdot 32$ maps to a twin prime pair $6 to 5,7$.
Is there any simple explanation why $leftfracn(n+1)2-1, fracn(n+1)2+1right$ is so rarely a twin prime pair?
elementary-number-theory prime-numbers prime-twins
edited Aug 23 at 9:32
Stefan4024
29.3k53377
asked Aug 23 at 9:25
Right
1,026213
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If we take some even number of the form $fracn(n+1)2$ and add $1$ to it and also subtract $1$ from it then we have a mapping $fracn(n+1)2toleftfracn(n+1)2-1,fracn(n+1)2+1right$.
From some even numbers of that form that I checked only number $6=frac4cdot 32$ maps to a twin prime pair $6 to 5,7$.
Is there any simple explanation why $leftfracn(n+1)2-1, fracn(n+1)2+1right$ is so rarely a twin prime pair?
elementary-number-theory prime-numbers prime-twins
edited Aug 23 at 9:32
Stefan4024
29.3k53377
asked Aug 23 at 9:25
Right
1,026213
If we take some even number of the form $fracn(n+1)2$ and add $1$ to it and also subtract $1$ from it then we have a mapping $fracn(n+1)2toleftfracn(n+1)2-1,fracn(n+1)2+1right$.
From some even numbers of that form that I checked only number $6=frac4cdot 32$ maps to a twin prime pair $6 to 5,7$.
Is there any simple explanation why $leftfracn(n+1)2-1, fracn(n+1)2+1right$ is so rarely a twin prime pair?
elementary-number-theory prime-numbers prime-twins
edited Aug 23 at 9:32
Stefan4024
29.3k53377
asked Aug 23 at 9:25
Right
1,026213
edited Aug 23 at 9:32
Stefan4024
29.3k53377
edited Aug 23 at 9:32
Stefan4024
29.3k53377
edited Aug 23 at 9:32
Stefan4024
29.3k53377
29.3k53377
asked Aug 23 at 9:25
Right
1,026213
asked Aug 23 at 9:25
Right
1,026213
asked Aug 23 at 9:25
Right
1,026213
1,026213
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
9
down vote
accepted
Note that: $$fracn(n+1)2 - 1 = fracn^2 + n -22 = frac(n+2)(n-1)2$$
Now if we want this number to be prime the $2$ in the denominator must cancel out one of the factors in the numerator. Thus we must have $n+2 = 2$ or $n-1=2$. This doesn't leave too much possibilities. And you can see that the only such pair is $(5,7)$.
answered Aug 23 at 9:30
Stefan4024
29.3k53377
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Note that: $$fracn(n+1)2 - 1 = fracn^2 + n -22 = frac(n+2)(n-1)2$$
Now if we want this number to be prime the $2$ in the denominator must cancel out one of the factors in the numerator. Thus we must have $n+2 = 2$ or $n-1=2$. This doesn't leave too much possibilities. And you can see that the only such pair is $(5,7)$.
answered Aug 23 at 9:30
Stefan4024
29.3k53377
add a comment |Â
up vote
9
down vote
accepted
Note that: $$fracn(n+1)2 - 1 = fracn^2 + n -22 = frac(n+2)(n-1)2$$
Now if we want this number to be prime the $2$ in the denominator must cancel out one of the factors in the numerator. Thus we must have $n+2 = 2$ or $n-1=2$. This doesn't leave too much possibilities. And you can see that the only such pair is $(5,7)$.
answered Aug 23 at 9:30
Stefan4024
29.3k53377
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Note that: $$fracn(n+1)2 - 1 = fracn^2 + n -22 = frac(n+2)(n-1)2$$
Now if we want this number to be prime the $2$ in the denominator must cancel out one of the factors in the numerator. Thus we must have $n+2 = 2$ or $n-1=2$. This doesn't leave too much possibilities. And you can see that the only such pair is $(5,7)$.
answered Aug 23 at 9:30
Stefan4024
29.3k53377
Note that: $$fracn(n+1)2 - 1 = fracn^2 + n -22 = frac(n+2)(n-1)2$$
Now if we want this number to be prime the $2$ in the denominator must cancel out one of the factors in the numerator. Thus we must have $n+2 = 2$ or $n-1=2$. This doesn't leave too much possibilities. And you can see that the only such pair is $(5,7)$.
answered Aug 23 at 9:30
Stefan4024
29.3k53377
edited Aug 23 at 14:08
answered Aug 23 at 9:30
Stefan4024
29.3k53377
answered Aug 23 at 9:30
Stefan4024
29.3k53377
answered Aug 23 at 9:30
Stefan4024
29.3k53377
29.3k53377
add a comment |Â
add a comment |Â
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