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Is there a reason why the first few primes have hexagonal symmetry when snaked around the plane?
Clash Royale CLAN TAG#URR8PPP
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See the image. I've been playing around with some "space filling snakes" of sequences. This sequence is precisely the sequence of natural numbers. But when you snake it thusly, and color $1$ and the odd prime numbers blue, you get a symmetric figure!
Is there a reason behind it? Does the pattern somehow continue?
elementary-number-theory prime-numbers integers symmetry pattern-recognition
asked Jul 30 at 19:28
EnjoysMath
8,63442153
 |Â
show 1 more comment
up vote
2
down vote
favorite
See the image. I've been playing around with some "space filling snakes" of sequences. This sequence is precisely the sequence of natural numbers. But when you snake it thusly, and color $1$ and the odd prime numbers blue, you get a symmetric figure!
Is there a reason behind it? Does the pattern somehow continue?
elementary-number-theory prime-numbers integers symmetry pattern-recognition
asked Jul 30 at 19:28
EnjoysMath
8,63442153
1
I suppose for any given sequence of numbers, an arrangement in the plane exists for each subsequence such that the subsequence forms a hexagonal pattern.
– packetpacket
Jul 30 at 19:32
1
This is mostly a consequence of the fact that there are a relative abundance of prime pairs $p,q$ such that $p+q=30$.
– Jeffery Opoku-Mensah
Jul 30 at 19:33
@packetpacket I colored precisely the primes though...
– EnjoysMath
Jul 30 at 19:33
1
But 2 is prime.
– Jeffery Opoku-Mensah
Jul 30 at 19:34
2
It looks rather strange (or suspicious to me) that you: a) painted 0 as non-prime and 1 as prime b) started with 0 c) the jump from 5 to 6... i would have jumped to position 9
– leonbloy
Jul 30 at 19:34
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
See the image. I've been playing around with some "space filling snakes" of sequences. This sequence is precisely the sequence of natural numbers. But when you snake it thusly, and color $1$ and the odd prime numbers blue, you get a symmetric figure!
Is there a reason behind it? Does the pattern somehow continue?
elementary-number-theory prime-numbers integers symmetry pattern-recognition
asked Jul 30 at 19:28
EnjoysMath
8,63442153
See the image. I've been playing around with some "space filling snakes" of sequences. This sequence is precisely the sequence of natural numbers. But when you snake it thusly, and color $1$ and the odd prime numbers blue, you get a symmetric figure!
Is there a reason behind it? Does the pattern somehow continue?
elementary-number-theory prime-numbers integers symmetry pattern-recognition
asked Jul 30 at 19:28
EnjoysMath
8,63442153
edited Jul 30 at 19:34
asked Jul 30 at 19:28
EnjoysMath
8,63442153
asked Jul 30 at 19:28
EnjoysMath
8,63442153
asked Jul 30 at 19:28
EnjoysMath
8,63442153
8,63442153
1
I suppose for any given sequence of numbers, an arrangement in the plane exists for each subsequence such that the subsequence forms a hexagonal pattern.
– packetpacket
Jul 30 at 19:32
1
This is mostly a consequence of the fact that there are a relative abundance of prime pairs $p,q$ such that $p+q=30$.
– Jeffery Opoku-Mensah
Jul 30 at 19:33
@packetpacket I colored precisely the primes though...
– EnjoysMath
Jul 30 at 19:33
1
But 2 is prime.
– Jeffery Opoku-Mensah
Jul 30 at 19:34
2
It looks rather strange (or suspicious to me) that you: a) painted 0 as non-prime and 1 as prime b) started with 0 c) the jump from 5 to 6... i would have jumped to position 9
– leonbloy
Jul 30 at 19:34
 |Â
show 1 more comment
1
I suppose for any given sequence of numbers, an arrangement in the plane exists for each subsequence such that the subsequence forms a hexagonal pattern.
– packetpacket
Jul 30 at 19:32
1
This is mostly a consequence of the fact that there are a relative abundance of prime pairs $p,q$ such that $p+q=30$.
– Jeffery Opoku-Mensah
Jul 30 at 19:33
@packetpacket I colored precisely the primes though...
– EnjoysMath
Jul 30 at 19:33
1
But 2 is prime.
– Jeffery Opoku-Mensah
Jul 30 at 19:34
2
It looks rather strange (or suspicious to me) that you: a) painted 0 as non-prime and 1 as prime b) started with 0 c) the jump from 5 to 6... i would have jumped to position 9
– leonbloy
Jul 30 at 19:34
1
1
I suppose for any given sequence of numbers, an arrangement in the plane exists for each subsequence such that the subsequence forms a hexagonal pattern.
– packetpacket
Jul 30 at 19:32
I suppose for any given sequence of numbers, an arrangement in the plane exists for each subsequence such that the subsequence forms a hexagonal pattern.
– packetpacket
Jul 30 at 19:32
1
1
This is mostly a consequence of the fact that there are a relative abundance of prime pairs $p,q$ such that $p+q=30$.
– Jeffery Opoku-Mensah
Jul 30 at 19:33
This is mostly a consequence of the fact that there are a relative abundance of prime pairs $p,q$ such that $p+q=30$.
– Jeffery Opoku-Mensah
Jul 30 at 19:33
@packetpacket I colored precisely the primes though...
– EnjoysMath
Jul 30 at 19:33
@packetpacket I colored precisely the primes though...
– EnjoysMath
Jul 30 at 19:33
1
1
But 2 is prime.
– Jeffery Opoku-Mensah
Jul 30 at 19:34
But 2 is prime.
– Jeffery Opoku-Mensah
Jul 30 at 19:34
2
2
It looks rather strange (or suspicious to me) that you: a) painted 0 as non-prime and 1 as prime b) started with 0 c) the jump from 5 to 6... i would have jumped to position 9
– leonbloy
Jul 30 at 19:34
It looks rather strange (or suspicious to me) that you: a) painted 0 as non-prime and 1 as prime b) started with 0 c) the jump from 5 to 6... i would have jumped to position 9
– leonbloy
Jul 30 at 19:34
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The bilateral symmetry of the outer ring is an easy consequence of two 'coincidences': (a) $2cdot 3cdot 5=30$, and (b) $30lt 49 = 7cdot 7$. Now, (a) means that $p|n$ iff $p|(30-n)$ for $pin2,3,5$; (b) means that $2,3,5notmid nimplies n$ prime for $nlt 30$. Together, they imply that for $6leq nleq 24$, $n$ is prime iff $30-n$ is prime.
Now, this symmetry can't really continue any further, because the next primorial, $210=2cdot3cdot5cdot7$, is larger than the next 'sieve number' $121=11^2$, so it's not necessarily the case that $n$ is prime iff $210-n$ is prime for $30leq nleq 180$; for instance, $67$ is prime but $210-67 = 143 = 11cdot 13$.
The rotational symmetry is a similar 'accident' of small numbers: for $7leq nlt 19=25-6$, $n$ is prime iff $n+6$ is prime. This is because all non-primes less than $25$ must have a factor of $2$ or $3$, and $n$ has such a factor iff $n+6$ does. Again, this can't really continue further, because the size of the products of primes that you need to ensure that $n$ and $n+d$ are divisible by the same set of numbers too rapidly outpaces the size of the squares of primes that you need to bound your $n$ to make sure they can't have 'spurious' prime factors.
answered Jul 30 at 19:42
Steven Stadnicki
40.1k765119
add a comment |Â
up vote
1
down vote
If you think about sieving out the primes, you only need to check $2$ and $3$ as divisors because you haven't gotten to $5^2=25$ yet. That means all numbers of the form $6k+1$ and $6k+5$ are prime, plus $2$ and $3$. Your chart is incorrect in that it shows $1$ as a prime and not $2$. That would spoil the symmetry.
answered Jul 30 at 19:33
Ross Millikan
275k21184351
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The bilateral symmetry of the outer ring is an easy consequence of two 'coincidences': (a) $2cdot 3cdot 5=30$, and (b) $30lt 49 = 7cdot 7$. Now, (a) means that $p|n$ iff $p|(30-n)$ for $pin2,3,5$; (b) means that $2,3,5notmid nimplies n$ prime for $nlt 30$. Together, they imply that for $6leq nleq 24$, $n$ is prime iff $30-n$ is prime.
Now, this symmetry can't really continue any further, because the next primorial, $210=2cdot3cdot5cdot7$, is larger than the next 'sieve number' $121=11^2$, so it's not necessarily the case that $n$ is prime iff $210-n$ is prime for $30leq nleq 180$; for instance, $67$ is prime but $210-67 = 143 = 11cdot 13$.
The rotational symmetry is a similar 'accident' of small numbers: for $7leq nlt 19=25-6$, $n$ is prime iff $n+6$ is prime. This is because all non-primes less than $25$ must have a factor of $2$ or $3$, and $n$ has such a factor iff $n+6$ does. Again, this can't really continue further, because the size of the products of primes that you need to ensure that $n$ and $n+d$ are divisible by the same set of numbers too rapidly outpaces the size of the squares of primes that you need to bound your $n$ to make sure they can't have 'spurious' prime factors.
answered Jul 30 at 19:42
Steven Stadnicki
40.1k765119
add a comment |Â
up vote
2
down vote
accepted
The bilateral symmetry of the outer ring is an easy consequence of two 'coincidences': (a) $2cdot 3cdot 5=30$, and (b) $30lt 49 = 7cdot 7$. Now, (a) means that $p|n$ iff $p|(30-n)$ for $pin2,3,5$; (b) means that $2,3,5notmid nimplies n$ prime for $nlt 30$. Together, they imply that for $6leq nleq 24$, $n$ is prime iff $30-n$ is prime.
Now, this symmetry can't really continue any further, because the next primorial, $210=2cdot3cdot5cdot7$, is larger than the next 'sieve number' $121=11^2$, so it's not necessarily the case that $n$ is prime iff $210-n$ is prime for $30leq nleq 180$; for instance, $67$ is prime but $210-67 = 143 = 11cdot 13$.
The rotational symmetry is a similar 'accident' of small numbers: for $7leq nlt 19=25-6$, $n$ is prime iff $n+6$ is prime. This is because all non-primes less than $25$ must have a factor of $2$ or $3$, and $n$ has such a factor iff $n+6$ does. Again, this can't really continue further, because the size of the products of primes that you need to ensure that $n$ and $n+d$ are divisible by the same set of numbers too rapidly outpaces the size of the squares of primes that you need to bound your $n$ to make sure they can't have 'spurious' prime factors.
answered Jul 30 at 19:42
Steven Stadnicki
40.1k765119
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The bilateral symmetry of the outer ring is an easy consequence of two 'coincidences': (a) $2cdot 3cdot 5=30$, and (b) $30lt 49 = 7cdot 7$. Now, (a) means that $p|n$ iff $p|(30-n)$ for $pin2,3,5$; (b) means that $2,3,5notmid nimplies n$ prime for $nlt 30$. Together, they imply that for $6leq nleq 24$, $n$ is prime iff $30-n$ is prime.
Now, this symmetry can't really continue any further, because the next primorial, $210=2cdot3cdot5cdot7$, is larger than the next 'sieve number' $121=11^2$, so it's not necessarily the case that $n$ is prime iff $210-n$ is prime for $30leq nleq 180$; for instance, $67$ is prime but $210-67 = 143 = 11cdot 13$.
The rotational symmetry is a similar 'accident' of small numbers: for $7leq nlt 19=25-6$, $n$ is prime iff $n+6$ is prime. This is because all non-primes less than $25$ must have a factor of $2$ or $3$, and $n$ has such a factor iff $n+6$ does. Again, this can't really continue further, because the size of the products of primes that you need to ensure that $n$ and $n+d$ are divisible by the same set of numbers too rapidly outpaces the size of the squares of primes that you need to bound your $n$ to make sure they can't have 'spurious' prime factors.
answered Jul 30 at 19:42
Steven Stadnicki
40.1k765119
The bilateral symmetry of the outer ring is an easy consequence of two 'coincidences': (a) $2cdot 3cdot 5=30$, and (b) $30lt 49 = 7cdot 7$. Now, (a) means that $p|n$ iff $p|(30-n)$ for $pin2,3,5$; (b) means that $2,3,5notmid nimplies n$ prime for $nlt 30$. Together, they imply that for $6leq nleq 24$, $n$ is prime iff $30-n$ is prime.
Now, this symmetry can't really continue any further, because the next primorial, $210=2cdot3cdot5cdot7$, is larger than the next 'sieve number' $121=11^2$, so it's not necessarily the case that $n$ is prime iff $210-n$ is prime for $30leq nleq 180$; for instance, $67$ is prime but $210-67 = 143 = 11cdot 13$.
The rotational symmetry is a similar 'accident' of small numbers: for $7leq nlt 19=25-6$, $n$ is prime iff $n+6$ is prime. This is because all non-primes less than $25$ must have a factor of $2$ or $3$, and $n$ has such a factor iff $n+6$ does. Again, this can't really continue further, because the size of the products of primes that you need to ensure that $n$ and $n+d$ are divisible by the same set of numbers too rapidly outpaces the size of the squares of primes that you need to bound your $n$ to make sure they can't have 'spurious' prime factors.
answered Jul 30 at 19:42
Steven Stadnicki
40.1k765119
edited Jul 30 at 19:48
answered Jul 30 at 19:42
Steven Stadnicki
40.1k765119
answered Jul 30 at 19:42
Steven Stadnicki
40.1k765119
answered Jul 30 at 19:42
Steven Stadnicki
40.1k765119
40.1k765119
add a comment |Â
add a comment |Â
up vote
1
down vote
If you think about sieving out the primes, you only need to check $2$ and $3$ as divisors because you haven't gotten to $5^2=25$ yet. That means all numbers of the form $6k+1$ and $6k+5$ are prime, plus $2$ and $3$. Your chart is incorrect in that it shows $1$ as a prime and not $2$. That would spoil the symmetry.
answered Jul 30 at 19:33
Ross Millikan
275k21184351
add a comment |Â
up vote
1
down vote
If you think about sieving out the primes, you only need to check $2$ and $3$ as divisors because you haven't gotten to $5^2=25$ yet. That means all numbers of the form $6k+1$ and $6k+5$ are prime, plus $2$ and $3$. Your chart is incorrect in that it shows $1$ as a prime and not $2$. That would spoil the symmetry.
answered Jul 30 at 19:33
Ross Millikan
275k21184351
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you think about sieving out the primes, you only need to check $2$ and $3$ as divisors because you haven't gotten to $5^2=25$ yet. That means all numbers of the form $6k+1$ and $6k+5$ are prime, plus $2$ and $3$. Your chart is incorrect in that it shows $1$ as a prime and not $2$. That would spoil the symmetry.
answered Jul 30 at 19:33
Ross Millikan
275k21184351
If you think about sieving out the primes, you only need to check $2$ and $3$ as divisors because you haven't gotten to $5^2=25$ yet. That means all numbers of the form $6k+1$ and $6k+5$ are prime, plus $2$ and $3$. Your chart is incorrect in that it shows $1$ as a prime and not $2$. That would spoil the symmetry.
answered Jul 30 at 19:33
Ross Millikan
275k21184351
answered Jul 30 at 19:33
Ross Millikan
275k21184351
answered Jul 30 at 19:33
Ross Millikan
275k21184351
answered Jul 30 at 19:33
Ross Millikan
275k21184351
275k21184351
add a comment |Â
add a comment |Â
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1
I suppose for any given sequence of numbers, an arrangement in the plane exists for each subsequence such that the subsequence forms a hexagonal pattern.
– packetpacket
Jul 30 at 19:32
1
This is mostly a consequence of the fact that there are a relative abundance of prime pairs $p,q$ such that $p+q=30$.
– Jeffery Opoku-Mensah
Jul 30 at 19:33
@packetpacket I colored precisely the primes though...
– EnjoysMath
Jul 30 at 19:33
1
But 2 is prime.
– Jeffery Opoku-Mensah
Jul 30 at 19:34
2
It looks rather strange (or suspicious to me) that you: a) painted 0 as non-prime and 1 as prime b) started with 0 c) the jump from 5 to 6... i would have jumped to position 9
– leonbloy
Jul 30 at 19:34