Coprime number pair lemma counter example soughtFinding a better approximation to a prime number relationNumber theory proof by counter exampleAsymptotics of the lower approximation of a pair of natural numbers by a coprime pairA Number Theoretic Argument. Proof or counter-example!About a mysterious sequence who seems to follow some patternsGive a counter exampleCounter example for a divisibility relationDivisibility Relation Counter Sought 2Congruence Relation proof over two integer variablesProof of infinite limitIs there a counter-example to these number theoretic conjectures?
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Coprime number pair lemma counter example sought
Finding a better approximation to a prime number relationNumber theory proof by counter exampleAsymptotics of the lower approximation of a pair of natural numbers by a coprime pairA Number Theoretic Argument. Proof or counter-example!About a mysterious sequence who seems to follow some patternsGive a counter exampleCounter example for a divisibility relationDivisibility Relation Counter Sought 2Congruence Relation proof over two integer variablesProof of infinite limitIs there a counter-example to these number theoretic conjectures?
$begingroup$
Because I am yet to see a concise theoretical foundation as to why it may well be true for all natural number pairs $(n,k)$ greater than $1$, I have been seeking a counter example to disprove that:
$$gcdBigl(BigllfloorBigl(fracp_nnBigr)^frac1k+2Bigrrfloor,lfloor n^frac1k+2rfloorBigr)=1 quadquadforall,n,k in mathbb N ,backslash,1$$
Where $p_n$ is the $n^th$ prime number.
However this has proved to be quite infrequent and only possible at considerably large values of $n$.
Does anyone have any advice as to another approach in disproving the above lemma?
As far as proof is concerned, well, it only makes intuitive sense that it may be true in consideration of identities I have previously established which can be reviewed here
But as you will see, there is clearly some huge gaps if indeed a concise derivation exists.
thankyou for helping in advance.
number-theory
asked Mar 29 at 2:48
AdamAdam
553114
$endgroup$
add a comment |
$begingroup$
Because I am yet to see a concise theoretical foundation as to why it may well be true for all natural number pairs $(n,k)$ greater than $1$, I have been seeking a counter example to disprove that:
$$gcdBigl(BigllfloorBigl(fracp_nnBigr)^frac1k+2Bigrrfloor,lfloor n^frac1k+2rfloorBigr)=1 quadquadforall,n,k in mathbb N ,backslash,1$$
Where $p_n$ is the $n^th$ prime number.
However this has proved to be quite infrequent and only possible at considerably large values of $n$.
Does anyone have any advice as to another approach in disproving the above lemma?
As far as proof is concerned, well, it only makes intuitive sense that it may be true in consideration of identities I have previously established which can be reviewed here
But as you will see, there is clearly some huge gaps if indeed a concise derivation exists.
thankyou for helping in advance.
number-theory
asked Mar 29 at 2:48
AdamAdam
553114
$endgroup$
add a comment |
$begingroup$
Because I am yet to see a concise theoretical foundation as to why it may well be true for all natural number pairs $(n,k)$ greater than $1$, I have been seeking a counter example to disprove that:
$$gcdBigl(BigllfloorBigl(fracp_nnBigr)^frac1k+2Bigrrfloor,lfloor n^frac1k+2rfloorBigr)=1 quadquadforall,n,k in mathbb N ,backslash,1$$
Where $p_n$ is the $n^th$ prime number.
However this has proved to be quite infrequent and only possible at considerably large values of $n$.
Does anyone have any advice as to another approach in disproving the above lemma?
As far as proof is concerned, well, it only makes intuitive sense that it may be true in consideration of identities I have previously established which can be reviewed here
But as you will see, there is clearly some huge gaps if indeed a concise derivation exists.
thankyou for helping in advance.
number-theory
asked Mar 29 at 2:48
AdamAdam
553114
$endgroup$
Because I am yet to see a concise theoretical foundation as to why it may well be true for all natural number pairs $(n,k)$ greater than $1$, I have been seeking a counter example to disprove that:
$$gcdBigl(BigllfloorBigl(fracp_nnBigr)^frac1k+2Bigrrfloor,lfloor n^frac1k+2rfloorBigr)=1 quadquadforall,n,k in mathbb N ,backslash,1$$
Where $p_n$ is the $n^th$ prime number.
However this has proved to be quite infrequent and only possible at considerably large values of $n$.
Does anyone have any advice as to another approach in disproving the above lemma?
As far as proof is concerned, well, it only makes intuitive sense that it may be true in consideration of identities I have previously established which can be reviewed here
But as you will see, there is clearly some huge gaps if indeed a concise derivation exists.
thankyou for helping in advance.
number-theory
number-theory
asked Mar 29 at 2:48
AdamAdam
553114
asked Mar 29 at 2:48
AdamAdam
553114
edited Mar 29 at 2:56
Adam
asked Mar 29 at 2:48
AdamAdam
553114
asked Mar 29 at 2:48
AdamAdam
553114
asked Mar 29 at 2:48
AdamAdam
553114
553114
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Even if $k$ is only $2$, then $bigg(dfracp_nnbigg)^frac1k+2$ will be less than $2$ until $dfracp_nn ge 16$. That doesn't happen until $n$ is well over a million, so a large value of $n$ is indeed required.
A value of $n$ that will work is $6^8 = 1679616$; then $p_n = 26941241$ and $dfracp_nn approx 16.040119$. Thus, with $k=2$,
$$
gcdBigl(BigllfloorBigl(fracp_nnBigr)^frac1k+2Bigrrfloor,lfloor n^frac1k+2rfloorBigr)=gcd(2,36)=2.
$$
answered Mar 29 at 3:30
FredHFredH
3,6851023
$endgroup$
$begingroup$
did your computation there involve something to do with having expressed your value as $6^8$ out of curiosity?
$endgroup$
– Adam
Mar 29 at 3:41
1
$begingroup$
For slightly smaller $n$, $lfloor n^frac1k+2rfloor$ was $35$; I chose $6^8$ to push it up to an even value.
$endgroup$
– FredH
Mar 29 at 3:45
add a comment |
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1 Answer
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$begingroup$
Even if $k$ is only $2$, then $bigg(dfracp_nnbigg)^frac1k+2$ will be less than $2$ until $dfracp_nn ge 16$. That doesn't happen until $n$ is well over a million, so a large value of $n$ is indeed required.
A value of $n$ that will work is $6^8 = 1679616$; then $p_n = 26941241$ and $dfracp_nn approx 16.040119$. Thus, with $k=2$,
$$
gcdBigl(BigllfloorBigl(fracp_nnBigr)^frac1k+2Bigrrfloor,lfloor n^frac1k+2rfloorBigr)=gcd(2,36)=2.
$$
answered Mar 29 at 3:30
FredHFredH
3,6851023
$endgroup$
$begingroup$
did your computation there involve something to do with having expressed your value as $6^8$ out of curiosity?
$endgroup$
– Adam
Mar 29 at 3:41
1
$begingroup$
For slightly smaller $n$, $lfloor n^frac1k+2rfloor$ was $35$; I chose $6^8$ to push it up to an even value.
$endgroup$
– FredH
Mar 29 at 3:45
add a comment |
$begingroup$
Even if $k$ is only $2$, then $bigg(dfracp_nnbigg)^frac1k+2$ will be less than $2$ until $dfracp_nn ge 16$. That doesn't happen until $n$ is well over a million, so a large value of $n$ is indeed required.
A value of $n$ that will work is $6^8 = 1679616$; then $p_n = 26941241$ and $dfracp_nn approx 16.040119$. Thus, with $k=2$,
$$
gcdBigl(BigllfloorBigl(fracp_nnBigr)^frac1k+2Bigrrfloor,lfloor n^frac1k+2rfloorBigr)=gcd(2,36)=2.
$$
answered Mar 29 at 3:30
FredHFredH
3,6851023
$endgroup$
$begingroup$
did your computation there involve something to do with having expressed your value as $6^8$ out of curiosity?
$endgroup$
– Adam
Mar 29 at 3:41
1
$begingroup$
For slightly smaller $n$, $lfloor n^frac1k+2rfloor$ was $35$; I chose $6^8$ to push it up to an even value.
$endgroup$
– FredH
Mar 29 at 3:45
add a comment |
$begingroup$
Even if $k$ is only $2$, then $bigg(dfracp_nnbigg)^frac1k+2$ will be less than $2$ until $dfracp_nn ge 16$. That doesn't happen until $n$ is well over a million, so a large value of $n$ is indeed required.
A value of $n$ that will work is $6^8 = 1679616$; then $p_n = 26941241$ and $dfracp_nn approx 16.040119$. Thus, with $k=2$,
$$
gcdBigl(BigllfloorBigl(fracp_nnBigr)^frac1k+2Bigrrfloor,lfloor n^frac1k+2rfloorBigr)=gcd(2,36)=2.
$$
answered Mar 29 at 3:30
FredHFredH
3,6851023
$endgroup$
Even if $k$ is only $2$, then $bigg(dfracp_nnbigg)^frac1k+2$ will be less than $2$ until $dfracp_nn ge 16$. That doesn't happen until $n$ is well over a million, so a large value of $n$ is indeed required.
A value of $n$ that will work is $6^8 = 1679616$; then $p_n = 26941241$ and $dfracp_nn approx 16.040119$. Thus, with $k=2$,
$$
gcdBigl(BigllfloorBigl(fracp_nnBigr)^frac1k+2Bigrrfloor,lfloor n^frac1k+2rfloorBigr)=gcd(2,36)=2.
$$
answered Mar 29 at 3:30
FredHFredH
3,6851023
answered Mar 29 at 3:30
FredHFredH
3,6851023
answered Mar 29 at 3:30
FredHFredH
3,6851023
answered Mar 29 at 3:30
FredHFredH
3,6851023
3,6851023
$begingroup$
did your computation there involve something to do with having expressed your value as $6^8$ out of curiosity?
$endgroup$
– Adam
Mar 29 at 3:41
1
$begingroup$
For slightly smaller $n$, $lfloor n^frac1k+2rfloor$ was $35$; I chose $6^8$ to push it up to an even value.
$endgroup$
– FredH
Mar 29 at 3:45
add a comment |
$begingroup$
did your computation there involve something to do with having expressed your value as $6^8$ out of curiosity?
$endgroup$
– Adam
Mar 29 at 3:41
1
$begingroup$
For slightly smaller $n$, $lfloor n^frac1k+2rfloor$ was $35$; I chose $6^8$ to push it up to an even value.
$endgroup$
– FredH
Mar 29 at 3:45
$begingroup$
did your computation there involve something to do with having expressed your value as $6^8$ out of curiosity?
$endgroup$
– Adam
Mar 29 at 3:41
$begingroup$
did your computation there involve something to do with having expressed your value as $6^8$ out of curiosity?
$endgroup$
– Adam
Mar 29 at 3:41
1
1
$begingroup$
For slightly smaller $n$, $lfloor n^frac1k+2rfloor$ was $35$; I chose $6^8$ to push it up to an even value.
$endgroup$
– FredH
Mar 29 at 3:45
$begingroup$
For slightly smaller $n$, $lfloor n^frac1k+2rfloor$ was $35$; I chose $6^8$ to push it up to an even value.
$endgroup$
– FredH
Mar 29 at 3:45
add a comment |
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