Mixing
Does this sequence of product of two primes always exist
$begingroup$
Consider a sequence $A: a_1, a_2, a_3 .. a_n$ such that each element of this sequence is product of two primes from $p_1, p_2 .. p_m$ so that:
Each $a_i$ is distinct.
$m$ is the least integer so that $frac{m(m-1)}{2} ge n$
Any three consecutive elements of sequence are coprime, but any two are not.
For example,
for $n = 3$, we have $m = 3$ and so taking the 3 primes as $a,b,c$, we can say the sequence $A$ is $ab, bc, ac$
For $n = 4$, $m = 4$ and $A: ab, bc, cd, da$
For $n = 6$, $m = 4$ and $A: ab, bd, da, ac, cd, db$
The question is to show that such a sequence exists for any $n$.
If this sequence exists for all $n$, is there a simple pattern for $A$ ?
Edit
There was an error in example 3, no such sequence exists for $n=6$.
sequences-and-series algebra-precalculus discrete-mathematics prime-numbers
asked Jan 8 at 15:01
jeeajeea
57314
$endgroup$
add a comment |
$begingroup$
Consider a sequence $A: a_1, a_2, a_3 .. a_n$ such that each element of this sequence is product of two primes from $p_1, p_2 .. p_m$ so that:
Each $a_i$ is distinct.
$m$ is the least integer so that $frac{m(m-1)}{2} ge n$
Any three consecutive elements of sequence are coprime, but any two are not.
For example,
for $n = 3$, we have $m = 3$ and so taking the 3 primes as $a,b,c$, we can say the sequence $A$ is $ab, bc, ac$
For $n = 4$, $m = 4$ and $A: ab, bc, cd, da$
For $n = 6$, $m = 4$ and $A: ab, bd, da, ac, cd, db$
The question is to show that such a sequence exists for any $n$.
If this sequence exists for all $n$, is there a simple pattern for $A$ ?
Edit
There was an error in example 3, no such sequence exists for $n=6$.
sequences-and-series algebra-precalculus discrete-mathematics prime-numbers
asked Jan 8 at 15:01
jeeajeea
57314
$endgroup$
1
$begingroup$
There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
$endgroup$
– ajotatxe
Jan 8 at 15:13
$begingroup$
@ajotatxe sorry I overlooked this one!
$endgroup$
– jeea
Jan 8 at 15:23
add a comment |
$begingroup$
Consider a sequence $A: a_1, a_2, a_3 .. a_n$ such that each element of this sequence is product of two primes from $p_1, p_2 .. p_m$ so that:
Each $a_i$ is distinct.
$m$ is the least integer so that $frac{m(m-1)}{2} ge n$
Any three consecutive elements of sequence are coprime, but any two are not.
For example,
for $n = 3$, we have $m = 3$ and so taking the 3 primes as $a,b,c$, we can say the sequence $A$ is $ab, bc, ac$
For $n = 4$, $m = 4$ and $A: ab, bc, cd, da$
For $n = 6$, $m = 4$ and $A: ab, bd, da, ac, cd, db$
The question is to show that such a sequence exists for any $n$.
If this sequence exists for all $n$, is there a simple pattern for $A$ ?
Edit
There was an error in example 3, no such sequence exists for $n=6$.
sequences-and-series algebra-precalculus discrete-mathematics prime-numbers
asked Jan 8 at 15:01
jeeajeea
57314
$endgroup$
Consider a sequence $A: a_1, a_2, a_3 .. a_n$ such that each element of this sequence is product of two primes from $p_1, p_2 .. p_m$ so that:
Each $a_i$ is distinct.
$m$ is the least integer so that $frac{m(m-1)}{2} ge n$
Any three consecutive elements of sequence are coprime, but any two are not.
For example,
for $n = 3$, we have $m = 3$ and so taking the 3 primes as $a,b,c$, we can say the sequence $A$ is $ab, bc, ac$
For $n = 4$, $m = 4$ and $A: ab, bc, cd, da$
For $n = 6$, $m = 4$ and $A: ab, bd, da, ac, cd, db$
The question is to show that such a sequence exists for any $n$.
If this sequence exists for all $n$, is there a simple pattern for $A$ ?
Edit
There was an error in example 3, no such sequence exists for $n=6$.
sequences-and-series algebra-precalculus discrete-mathematics prime-numbers
sequences-and-series algebra-precalculus discrete-mathematics prime-numbers
asked Jan 8 at 15:01
jeeajeea
57314
asked Jan 8 at 15:01
jeeajeea
57314
edited Jan 8 at 15:22
jeea
asked Jan 8 at 15:01
jeeajeea
57314
asked Jan 8 at 15:01
jeeajeea
57314
asked Jan 8 at 15:01
jeeajeea
57314
57314
1
$begingroup$
There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
$endgroup$
– ajotatxe
Jan 8 at 15:13
$begingroup$
@ajotatxe sorry I overlooked this one!
$endgroup$
– jeea
Jan 8 at 15:23
add a comment |
1
$begingroup$
There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
$endgroup$
– ajotatxe
Jan 8 at 15:13
$begingroup$
@ajotatxe sorry I overlooked this one!
$endgroup$
– jeea
Jan 8 at 15:23
1
1
$begingroup$
There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
$endgroup$
– ajotatxe
Jan 8 at 15:13
$begingroup$
There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
$endgroup$
– ajotatxe
Jan 8 at 15:13
$begingroup$
@ajotatxe sorry I overlooked this one!
$endgroup$
– jeea
Jan 8 at 15:23
$begingroup$
@ajotatxe sorry I overlooked this one!
$endgroup$
– jeea
Jan 8 at 15:23
add a comment |
1 Answer
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$begingroup$
Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.
This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.
answered Jan 8 at 15:15
Especially LimeEspecially Lime
21.9k22858
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add a comment |
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$begingroup$
Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.
This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.
answered Jan 8 at 15:15
Especially LimeEspecially Lime
21.9k22858
$endgroup$
add a comment |
$begingroup$
Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.
This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.
answered Jan 8 at 15:15
Especially LimeEspecially Lime
21.9k22858
$endgroup$
add a comment |
$begingroup$
Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.
This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.
answered Jan 8 at 15:15
Especially LimeEspecially Lime
21.9k22858
$endgroup$
Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.
This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.
answered Jan 8 at 15:15
Especially LimeEspecially Lime
21.9k22858
answered Jan 8 at 15:15
Especially LimeEspecially Lime
21.9k22858
answered Jan 8 at 15:15
Especially LimeEspecially Lime
21.9k22858
answered Jan 8 at 15:15
Especially LimeEspecially Lime
21.9k22858
21.9k22858
add a comment |
add a comment |
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$begingroup$
There is a problem in your example for $n=6$: $bd$ and $db$ are equal.
$endgroup$
– ajotatxe
Jan 8 at 15:13
$begingroup$
@ajotatxe sorry I overlooked this one!
$endgroup$
– jeea
Jan 8 at 15:23