How to express a coprime relation as a Congruence relation Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there a way to show that $sqrtp_n < n$?The myth of no prime formula?Finding a better approximation to a prime number relationCounter example for a divisibility relationPrime Number Theorem Related functions Divisibility Relation Counter SoughtBeatty Sequences And Binary Logarithmic IdentitiesCongruence Relation proof over two integer variablesPrime Number Congruence Modulo 8 ProofProof for Natural number IdentitiesCoprime number pair lemma counter example soughtReal number integer and fractional part separation
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How to express a coprime relation as a Congruence relation
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there a way to show that $sqrtp_n < n$?The myth of no prime formula?Finding a better approximation to a prime number relationCounter example for a divisibility relationPrime Number Theorem Related functions Divisibility Relation Counter SoughtBeatty Sequences And Binary Logarithmic IdentitiesCongruence Relation proof over two integer variablesPrime Number Congruence Modulo 8 ProofProof for Natural number IdentitiesCoprime number pair lemma counter example soughtReal number integer and fractional part separation
$begingroup$
When $(n,k) in mathbb P^2 $, the following coprime relations appear to hold:
$$gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$$
$$gcdBigl(n^k,Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$$
So the first part of my question is a request for a counter example for these, which are proving to be more and more elusive. As far I know thus far, they can also be extended to $(n,k)$ for which $k$ is any natural number.
however the later term in the greatest common divisor expression above $Bigllfloor fracp_nn Bigrrfloor^k$, is not always coprime to $Bigllfloor fracp_n^kn^k Bigrrfloor$ and the strictly prime domain of $(n,k)$ is paritioned into two equivalence classes, the predicates for which are the coprimality of this pair of quantities and it's negation:
$$mathbb P^2=mathbb P_1 cup mathbb P_2$$
$$mathbb P_1=Biggl(n,k):gcdBigl(Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1Biggr$$
$$mathbb P_2=Biggl(n,k):gcdBigl(Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr) gt 1Biggr$$
A sample of my numerical observations thus far are:
$$mathbb P_1=(2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7), (5, 5), (5, 7), (7, 2), (11, 2), (11, 5), (11, 7)$$
$$mathbb P_2=(5, 2), (5, 3), (7, 3), (7, 5), (7, 7), (11, 3), (13, 2), (13, 5), (17, 2), (19, 2), (19, 7), (29, 5)$$
likewise,$Bigllfloor fracp_n^kn^k Bigrrfloor$ and $n^k$ are not always coprime, and the strictly prime domain of $(n,k)$ is paritioned into two equivalence classes, the predicates for which are the coprimality of this pair of quantities and it's negation:
$$mathbb P^2=mathbb P_1 cup mathbb P_2$$
$$mathbb P_1=(n,k):gcdBigl(n^k,Bigllfloor fracp_n^kn^k BigrrfloorBigr)=1$$
$$mathbb P_2=(n,k):gcdBigl(n^k,Bigllfloor fracp_n^kn^k BigrrfloorBigr) gt 1$$
Numerical observations thus far are:
$$mathbb P_1=(2, 3), (2, 5), (3, 2), (3, 3), (5, 2), (5, 5), (7, 2), (11, 2), (11, 5), (13, 2), (13, 3), (17, 2), (17, 3)$$
$$mathbb P_2=(2, 2), (3, 5), (5, 3), (7, 3), (7, 5), (11, 3), (13, 5), (23, 3)$$
So the second part of my question, is how, and using what theorem, can I express the above as a congruence relation?
I have tried looking for the sequence of values of $n$ for various fixed values of $k$ on OEIS, but nothing has shown up. I'm not sure if my problem is important enough to put these sequences on there or not, it asks me to do so if I think they are of interest, but that's pretty subjective, I mean if it really is like face book and sequence popularity matters, anyway end of question.
number-theory prime-numbers
asked Apr 1 at 8:11
AdamAdam
560114
$endgroup$
|
show 1 more comment
$begingroup$
When $(n,k) in mathbb P^2 $, the following coprime relations appear to hold:
$$gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$$
$$gcdBigl(n^k,Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$$
So the first part of my question is a request for a counter example for these, which are proving to be more and more elusive. As far I know thus far, they can also be extended to $(n,k)$ for which $k$ is any natural number.
however the later term in the greatest common divisor expression above $Bigllfloor fracp_nn Bigrrfloor^k$, is not always coprime to $Bigllfloor fracp_n^kn^k Bigrrfloor$ and the strictly prime domain of $(n,k)$ is paritioned into two equivalence classes, the predicates for which are the coprimality of this pair of quantities and it's negation:
$$mathbb P^2=mathbb P_1 cup mathbb P_2$$
$$mathbb P_1=Biggl(n,k):gcdBigl(Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1Biggr$$
$$mathbb P_2=Biggl(n,k):gcdBigl(Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr) gt 1Biggr$$
A sample of my numerical observations thus far are:
$$mathbb P_1=(2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7), (5, 5), (5, 7), (7, 2), (11, 2), (11, 5), (11, 7)$$
$$mathbb P_2=(5, 2), (5, 3), (7, 3), (7, 5), (7, 7), (11, 3), (13, 2), (13, 5), (17, 2), (19, 2), (19, 7), (29, 5)$$
likewise,$Bigllfloor fracp_n^kn^k Bigrrfloor$ and $n^k$ are not always coprime, and the strictly prime domain of $(n,k)$ is paritioned into two equivalence classes, the predicates for which are the coprimality of this pair of quantities and it's negation:
$$mathbb P^2=mathbb P_1 cup mathbb P_2$$
$$mathbb P_1=(n,k):gcdBigl(n^k,Bigllfloor fracp_n^kn^k BigrrfloorBigr)=1$$
$$mathbb P_2=(n,k):gcdBigl(n^k,Bigllfloor fracp_n^kn^k BigrrfloorBigr) gt 1$$
Numerical observations thus far are:
$$mathbb P_1=(2, 3), (2, 5), (3, 2), (3, 3), (5, 2), (5, 5), (7, 2), (11, 2), (11, 5), (13, 2), (13, 3), (17, 2), (17, 3)$$
$$mathbb P_2=(2, 2), (3, 5), (5, 3), (7, 3), (7, 5), (11, 3), (13, 5), (23, 3)$$
So the second part of my question, is how, and using what theorem, can I express the above as a congruence relation?
I have tried looking for the sequence of values of $n$ for various fixed values of $k$ on OEIS, but nothing has shown up. I'm not sure if my problem is important enough to put these sequences on there or not, it asks me to do so if I think they are of interest, but that's pretty subjective, I mean if it really is like face book and sequence popularity matters, anyway end of question.
number-theory prime-numbers
asked Apr 1 at 8:11
AdamAdam
560114
$endgroup$
2
$begingroup$
Note that the first relation implies the second. Also, $gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$ for some $k$ if and only if $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$. Thus it suffices to show that if $n$ is prime, then $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$.
$endgroup$
– rogerl
Apr 1 at 11:50
$begingroup$
Sure thanks. is there some kind of identity that enables us to express the greatest common divisor of three quantities in terms of products of the combinations of pairs that can be made from the three? ie $gcd(a,b,c)$ in terms of $gcd(a,b)$,$gcd(b,c)$ &$gcd(a,c)$?
$endgroup$
– Adam
Apr 1 at 12:14
$begingroup$
with regards to proof for the second I was thinking
$endgroup$
– Adam
Apr 1 at 12:17
$begingroup$
because I can see what you mean, by the first implying the second, I just want to perhaps use some kind of identity involving the second to establish more concise conditions for when some of the combinatoric pairs from it are not coprime and are coprime
$endgroup$
– Adam
Apr 1 at 12:23
2
$begingroup$
$gcd(n,lfloorfracp_nnrfloor)=1$ if $n$ is prime because $p_n lt n^2$ for $ngt 1$.
$endgroup$
– FredH
Apr 1 at 13:20
|
show 1 more comment
$begingroup$
When $(n,k) in mathbb P^2 $, the following coprime relations appear to hold:
$$gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$$
$$gcdBigl(n^k,Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$$
So the first part of my question is a request for a counter example for these, which are proving to be more and more elusive. As far I know thus far, they can also be extended to $(n,k)$ for which $k$ is any natural number.
however the later term in the greatest common divisor expression above $Bigllfloor fracp_nn Bigrrfloor^k$, is not always coprime to $Bigllfloor fracp_n^kn^k Bigrrfloor$ and the strictly prime domain of $(n,k)$ is paritioned into two equivalence classes, the predicates for which are the coprimality of this pair of quantities and it's negation:
$$mathbb P^2=mathbb P_1 cup mathbb P_2$$
$$mathbb P_1=Biggl(n,k):gcdBigl(Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1Biggr$$
$$mathbb P_2=Biggl(n,k):gcdBigl(Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr) gt 1Biggr$$
A sample of my numerical observations thus far are:
$$mathbb P_1=(2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7), (5, 5), (5, 7), (7, 2), (11, 2), (11, 5), (11, 7)$$
$$mathbb P_2=(5, 2), (5, 3), (7, 3), (7, 5), (7, 7), (11, 3), (13, 2), (13, 5), (17, 2), (19, 2), (19, 7), (29, 5)$$
likewise,$Bigllfloor fracp_n^kn^k Bigrrfloor$ and $n^k$ are not always coprime, and the strictly prime domain of $(n,k)$ is paritioned into two equivalence classes, the predicates for which are the coprimality of this pair of quantities and it's negation:
$$mathbb P^2=mathbb P_1 cup mathbb P_2$$
$$mathbb P_1=(n,k):gcdBigl(n^k,Bigllfloor fracp_n^kn^k BigrrfloorBigr)=1$$
$$mathbb P_2=(n,k):gcdBigl(n^k,Bigllfloor fracp_n^kn^k BigrrfloorBigr) gt 1$$
Numerical observations thus far are:
$$mathbb P_1=(2, 3), (2, 5), (3, 2), (3, 3), (5, 2), (5, 5), (7, 2), (11, 2), (11, 5), (13, 2), (13, 3), (17, 2), (17, 3)$$
$$mathbb P_2=(2, 2), (3, 5), (5, 3), (7, 3), (7, 5), (11, 3), (13, 5), (23, 3)$$
So the second part of my question, is how, and using what theorem, can I express the above as a congruence relation?
I have tried looking for the sequence of values of $n$ for various fixed values of $k$ on OEIS, but nothing has shown up. I'm not sure if my problem is important enough to put these sequences on there or not, it asks me to do so if I think they are of interest, but that's pretty subjective, I mean if it really is like face book and sequence popularity matters, anyway end of question.
number-theory prime-numbers
asked Apr 1 at 8:11
AdamAdam
560114
$endgroup$
When $(n,k) in mathbb P^2 $, the following coprime relations appear to hold:
$$gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$$
$$gcdBigl(n^k,Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$$
So the first part of my question is a request for a counter example for these, which are proving to be more and more elusive. As far I know thus far, they can also be extended to $(n,k)$ for which $k$ is any natural number.
however the later term in the greatest common divisor expression above $Bigllfloor fracp_nn Bigrrfloor^k$, is not always coprime to $Bigllfloor fracp_n^kn^k Bigrrfloor$ and the strictly prime domain of $(n,k)$ is paritioned into two equivalence classes, the predicates for which are the coprimality of this pair of quantities and it's negation:
$$mathbb P^2=mathbb P_1 cup mathbb P_2$$
$$mathbb P_1=Biggl(n,k):gcdBigl(Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1Biggr$$
$$mathbb P_2=Biggl(n,k):gcdBigl(Bigllfloor fracp_n^kn^k Bigrrfloor,Bigllfloor fracp_nn Bigrrfloor^kBigr) gt 1Biggr$$
A sample of my numerical observations thus far are:
$$mathbb P_1=(2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7), (5, 5), (5, 7), (7, 2), (11, 2), (11, 5), (11, 7)$$
$$mathbb P_2=(5, 2), (5, 3), (7, 3), (7, 5), (7, 7), (11, 3), (13, 2), (13, 5), (17, 2), (19, 2), (19, 7), (29, 5)$$
likewise,$Bigllfloor fracp_n^kn^k Bigrrfloor$ and $n^k$ are not always coprime, and the strictly prime domain of $(n,k)$ is paritioned into two equivalence classes, the predicates for which are the coprimality of this pair of quantities and it's negation:
$$mathbb P^2=mathbb P_1 cup mathbb P_2$$
$$mathbb P_1=(n,k):gcdBigl(n^k,Bigllfloor fracp_n^kn^k BigrrfloorBigr)=1$$
$$mathbb P_2=(n,k):gcdBigl(n^k,Bigllfloor fracp_n^kn^k BigrrfloorBigr) gt 1$$
Numerical observations thus far are:
$$mathbb P_1=(2, 3), (2, 5), (3, 2), (3, 3), (5, 2), (5, 5), (7, 2), (11, 2), (11, 5), (13, 2), (13, 3), (17, 2), (17, 3)$$
$$mathbb P_2=(2, 2), (3, 5), (5, 3), (7, 3), (7, 5), (11, 3), (13, 5), (23, 3)$$
So the second part of my question, is how, and using what theorem, can I express the above as a congruence relation?
I have tried looking for the sequence of values of $n$ for various fixed values of $k$ on OEIS, but nothing has shown up. I'm not sure if my problem is important enough to put these sequences on there or not, it asks me to do so if I think they are of interest, but that's pretty subjective, I mean if it really is like face book and sequence popularity matters, anyway end of question.
number-theory prime-numbers
number-theory prime-numbers
asked Apr 1 at 8:11
AdamAdam
560114
asked Apr 1 at 8:11
AdamAdam
560114
edited Apr 1 at 10:42
Adam
asked Apr 1 at 8:11
AdamAdam
560114
asked Apr 1 at 8:11
AdamAdam
560114
asked Apr 1 at 8:11
AdamAdam
560114
560114
2
$begingroup$
Note that the first relation implies the second. Also, $gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$ for some $k$ if and only if $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$. Thus it suffices to show that if $n$ is prime, then $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$.
$endgroup$
– rogerl
Apr 1 at 11:50
$begingroup$
Sure thanks. is there some kind of identity that enables us to express the greatest common divisor of three quantities in terms of products of the combinations of pairs that can be made from the three? ie $gcd(a,b,c)$ in terms of $gcd(a,b)$,$gcd(b,c)$ &$gcd(a,c)$?
$endgroup$
– Adam
Apr 1 at 12:14
$begingroup$
with regards to proof for the second I was thinking
$endgroup$
– Adam
Apr 1 at 12:17
$begingroup$
because I can see what you mean, by the first implying the second, I just want to perhaps use some kind of identity involving the second to establish more concise conditions for when some of the combinatoric pairs from it are not coprime and are coprime
$endgroup$
– Adam
Apr 1 at 12:23
2
$begingroup$
$gcd(n,lfloorfracp_nnrfloor)=1$ if $n$ is prime because $p_n lt n^2$ for $ngt 1$.
$endgroup$
– FredH
Apr 1 at 13:20
|
show 1 more comment
2
$begingroup$
Note that the first relation implies the second. Also, $gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$ for some $k$ if and only if $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$. Thus it suffices to show that if $n$ is prime, then $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$.
$endgroup$
– rogerl
Apr 1 at 11:50
$begingroup$
Sure thanks. is there some kind of identity that enables us to express the greatest common divisor of three quantities in terms of products of the combinations of pairs that can be made from the three? ie $gcd(a,b,c)$ in terms of $gcd(a,b)$,$gcd(b,c)$ &$gcd(a,c)$?
$endgroup$
– Adam
Apr 1 at 12:14
$begingroup$
with regards to proof for the second I was thinking
$endgroup$
– Adam
Apr 1 at 12:17
$begingroup$
because I can see what you mean, by the first implying the second, I just want to perhaps use some kind of identity involving the second to establish more concise conditions for when some of the combinatoric pairs from it are not coprime and are coprime
$endgroup$
– Adam
Apr 1 at 12:23
2
$begingroup$
$gcd(n,lfloorfracp_nnrfloor)=1$ if $n$ is prime because $p_n lt n^2$ for $ngt 1$.
$endgroup$
– FredH
Apr 1 at 13:20
2
2
$begingroup$
Note that the first relation implies the second. Also, $gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$ for some $k$ if and only if $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$. Thus it suffices to show that if $n$ is prime, then $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$.
$endgroup$
– rogerl
Apr 1 at 11:50
$begingroup$
Note that the first relation implies the second. Also, $gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$ for some $k$ if and only if $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$. Thus it suffices to show that if $n$ is prime, then $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$.
$endgroup$
– rogerl
Apr 1 at 11:50
$begingroup$
Sure thanks. is there some kind of identity that enables us to express the greatest common divisor of three quantities in terms of products of the combinations of pairs that can be made from the three? ie $gcd(a,b,c)$ in terms of $gcd(a,b)$,$gcd(b,c)$ &$gcd(a,c)$?
$endgroup$
– Adam
Apr 1 at 12:14
$begingroup$
Sure thanks. is there some kind of identity that enables us to express the greatest common divisor of three quantities in terms of products of the combinations of pairs that can be made from the three? ie $gcd(a,b,c)$ in terms of $gcd(a,b)$,$gcd(b,c)$ &$gcd(a,c)$?
$endgroup$
– Adam
Apr 1 at 12:14
$begingroup$
with regards to proof for the second I was thinking
$endgroup$
– Adam
Apr 1 at 12:17
$begingroup$
with regards to proof for the second I was thinking
$endgroup$
– Adam
Apr 1 at 12:17
$begingroup$
because I can see what you mean, by the first implying the second, I just want to perhaps use some kind of identity involving the second to establish more concise conditions for when some of the combinatoric pairs from it are not coprime and are coprime
$endgroup$
– Adam
Apr 1 at 12:23
$begingroup$
because I can see what you mean, by the first implying the second, I just want to perhaps use some kind of identity involving the second to establish more concise conditions for when some of the combinatoric pairs from it are not coprime and are coprime
$endgroup$
– Adam
Apr 1 at 12:23
2
2
$begingroup$
$gcd(n,lfloorfracp_nnrfloor)=1$ if $n$ is prime because $p_n lt n^2$ for $ngt 1$.
$endgroup$
– FredH
Apr 1 at 13:20
$begingroup$
$gcd(n,lfloorfracp_nnrfloor)=1$ if $n$ is prime because $p_n lt n^2$ for $ngt 1$.
$endgroup$
– FredH
Apr 1 at 13:20
|
show 1 more comment
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2
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Note that the first relation implies the second. Also, $gcdBigl(n^k,Bigllfloor fracp_nn Bigrrfloor^kBigr)=1$ for some $k$ if and only if $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$. Thus it suffices to show that if $n$ is prime, then $gcdBigl(n,Bigllfloor fracp_nn BigrrfloorBigr)=1$.
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– rogerl
Apr 1 at 11:50
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Sure thanks. is there some kind of identity that enables us to express the greatest common divisor of three quantities in terms of products of the combinations of pairs that can be made from the three? ie $gcd(a,b,c)$ in terms of $gcd(a,b)$,$gcd(b,c)$ &$gcd(a,c)$?
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– Adam
Apr 1 at 12:14
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with regards to proof for the second I was thinking
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– Adam
Apr 1 at 12:17
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because I can see what you mean, by the first implying the second, I just want to perhaps use some kind of identity involving the second to establish more concise conditions for when some of the combinatoric pairs from it are not coprime and are coprime
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– Adam
Apr 1 at 12:23
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$gcd(n,lfloorfracp_nnrfloor)=1$ if $n$ is prime because $p_n lt n^2$ for $ngt 1$.
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– FredH
Apr 1 at 13:20