Method of Proof concerning Prime Gap congruence Relation over two integer variables
$begingroup$
$$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0 Rightarrow m=2$$
$$quadquad(operatorname{tooth1})$$
$$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}not=0 Rightarrow mnot=2$$
$$quadquad(operatorname{tooth2})$$
$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0$ if and only if $m=2$.
$$quadquad(operatorname{tooth3})$$
If the first two teeth are true, then it is obvious that our third tooth also is true.
So my problem is reduced to seeking $(n,m)$ such that $mnot=0$ but $p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0$ as the necessary conditions for a counter example to the initial conjecture made with the stated teeth, therefore we can conclude it is false and no futher investigation is required, or a proof that such a counter example existing is an impossibility over the infinitude of primes for which the prime gap stated is taken, assuring us that the congruence of $p_n$ and $p_{lfloor n^{frac{1}{m}}rfloor^m}$ modulo unity can only be true in the case that $m=2$, so here a I am looking for a means of determining these truth values and taking these considerations into account for another problem.
number-theory prime-numbers modular-arithmetic prime-gaps
asked Jan 18 at 2:12
AdamAdam
560114
$endgroup$
|
show 3 more comments
$begingroup$
$$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0 Rightarrow m=2$$
$$quadquad(operatorname{tooth1})$$
$$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}not=0 Rightarrow mnot=2$$
$$quadquad(operatorname{tooth2})$$
$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0$ if and only if $m=2$.
$$quadquad(operatorname{tooth3})$$
If the first two teeth are true, then it is obvious that our third tooth also is true.
So my problem is reduced to seeking $(n,m)$ such that $mnot=0$ but $p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0$ as the necessary conditions for a counter example to the initial conjecture made with the stated teeth, therefore we can conclude it is false and no futher investigation is required, or a proof that such a counter example existing is an impossibility over the infinitude of primes for which the prime gap stated is taken, assuring us that the congruence of $p_n$ and $p_{lfloor n^{frac{1}{m}}rfloor^m}$ modulo unity can only be true in the case that $m=2$, so here a I am looking for a means of determining these truth values and taking these considerations into account for another problem.
number-theory prime-numbers modular-arithmetic prime-gaps
asked Jan 18 at 2:12
AdamAdam
560114
$endgroup$
1
$begingroup$
This baffles me. I assume that $p$ is a prime and $n$ is an index, i.e. $p_1=2, p_2=3$ etc. The index can run over all positive integers (correct?), so we might have the index $n=27$. In that case, if we look at $m=3$, then the difference of the indexed primes is $0$, contrary to your claim. The same would be true for any $n=a^m$, regardless of the value of $m$. Or have I misunderstood your nomenclature?
$endgroup$
– Keith Backman
Jan 18 at 20:29
$begingroup$
Ok thanks I will have to take a look back as I was distracted by other matters. But the result baffled me, which is why the question was generated immediately upon inspection of the computation I intended to make, but as you know human error is ever present in the work of the analytical number theorist, there is no escaping it's humiliation, nor is there any point in feeling embarrassment in the first place
$endgroup$
– Adam
Jan 19 at 12:18
$begingroup$
I'd say you have to get in line behind me on points such as human error; I've made more than my share
$endgroup$
– Keith Backman
Jan 19 at 15:58
$begingroup$
haha no worries mate I just lost the ability to be upset about any number of people see me drop all of my shit along the moving subway one day, and that was the final seal there without the desire to put the microscope of every passenger having a giggle, no honour in extracting the truth that 99.9 percent of them have so little substance in their ambition and path to self fulfilment that a good lot of them return home to hang themselves after a brief conversation with me. That is my least favourite dream achievement sorry failure
$endgroup$
– Adam
Jan 20 at 12:18
$begingroup$
Ok yes I'm on board now I'm on board now with what I am saying, I know when I start It was on the context of if we have a fixed $m$ value, and evaluate over $mathbb N$ for $n$, and we have the result stated, this can only be the case for $m=2$ but what you have shown is a pretty obvious defiance of the formally stated rule as ive decide to put it on this night, not to worry I will probably fix this up if it becomes something worth while, having said that worthwhile to others is never unless you have one of $%2 bfedgfds#$^%$#T bloodlines or nobility, in which case no you wont even be allowed
$endgroup$
– Adam
Jan 20 at 15:40
|
show 3 more comments
$begingroup$
$$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0 Rightarrow m=2$$
$$quadquad(operatorname{tooth1})$$
$$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}not=0 Rightarrow mnot=2$$
$$quadquad(operatorname{tooth2})$$
$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0$ if and only if $m=2$.
$$quadquad(operatorname{tooth3})$$
If the first two teeth are true, then it is obvious that our third tooth also is true.
So my problem is reduced to seeking $(n,m)$ such that $mnot=0$ but $p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0$ as the necessary conditions for a counter example to the initial conjecture made with the stated teeth, therefore we can conclude it is false and no futher investigation is required, or a proof that such a counter example existing is an impossibility over the infinitude of primes for which the prime gap stated is taken, assuring us that the congruence of $p_n$ and $p_{lfloor n^{frac{1}{m}}rfloor^m}$ modulo unity can only be true in the case that $m=2$, so here a I am looking for a means of determining these truth values and taking these considerations into account for another problem.
number-theory prime-numbers modular-arithmetic prime-gaps
asked Jan 18 at 2:12
AdamAdam
560114
$endgroup$
$$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0 Rightarrow m=2$$
$$quadquad(operatorname{tooth1})$$
$$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}not=0 Rightarrow mnot=2$$
$$quadquad(operatorname{tooth2})$$
$p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0$ if and only if $m=2$.
$$quadquad(operatorname{tooth3})$$
If the first two teeth are true, then it is obvious that our third tooth also is true.
So my problem is reduced to seeking $(n,m)$ such that $mnot=0$ but $p_n-p_{lfloor n^{frac{1}{m}}rfloor^m}=0$ as the necessary conditions for a counter example to the initial conjecture made with the stated teeth, therefore we can conclude it is false and no futher investigation is required, or a proof that such a counter example existing is an impossibility over the infinitude of primes for which the prime gap stated is taken, assuring us that the congruence of $p_n$ and $p_{lfloor n^{frac{1}{m}}rfloor^m}$ modulo unity can only be true in the case that $m=2$, so here a I am looking for a means of determining these truth values and taking these considerations into account for another problem.
number-theory prime-numbers modular-arithmetic prime-gaps
number-theory prime-numbers modular-arithmetic prime-gaps
asked Jan 18 at 2:12
AdamAdam
560114
asked Jan 18 at 2:12
AdamAdam
560114
asked Jan 18 at 2:12
AdamAdam
560114
asked Jan 18 at 2:12
AdamAdam
560114
asked Jan 18 at 2:12
AdamAdam
560114
560114
1
$begingroup$
This baffles me. I assume that $p$ is a prime and $n$ is an index, i.e. $p_1=2, p_2=3$ etc. The index can run over all positive integers (correct?), so we might have the index $n=27$. In that case, if we look at $m=3$, then the difference of the indexed primes is $0$, contrary to your claim. The same would be true for any $n=a^m$, regardless of the value of $m$. Or have I misunderstood your nomenclature?
$endgroup$
– Keith Backman
Jan 18 at 20:29
$begingroup$
Ok thanks I will have to take a look back as I was distracted by other matters. But the result baffled me, which is why the question was generated immediately upon inspection of the computation I intended to make, but as you know human error is ever present in the work of the analytical number theorist, there is no escaping it's humiliation, nor is there any point in feeling embarrassment in the first place
$endgroup$
– Adam
Jan 19 at 12:18
$begingroup$
I'd say you have to get in line behind me on points such as human error; I've made more than my share
$endgroup$
– Keith Backman
Jan 19 at 15:58
$begingroup$
haha no worries mate I just lost the ability to be upset about any number of people see me drop all of my shit along the moving subway one day, and that was the final seal there without the desire to put the microscope of every passenger having a giggle, no honour in extracting the truth that 99.9 percent of them have so little substance in their ambition and path to self fulfilment that a good lot of them return home to hang themselves after a brief conversation with me. That is my least favourite dream achievement sorry failure
$endgroup$
– Adam
Jan 20 at 12:18
$begingroup$
Ok yes I'm on board now I'm on board now with what I am saying, I know when I start It was on the context of if we have a fixed $m$ value, and evaluate over $mathbb N$ for $n$, and we have the result stated, this can only be the case for $m=2$ but what you have shown is a pretty obvious defiance of the formally stated rule as ive decide to put it on this night, not to worry I will probably fix this up if it becomes something worth while, having said that worthwhile to others is never unless you have one of $%2 bfedgfds#$^%$#T bloodlines or nobility, in which case no you wont even be allowed
$endgroup$
– Adam
Jan 20 at 15:40
|
show 3 more comments
1
$begingroup$
This baffles me. I assume that $p$ is a prime and $n$ is an index, i.e. $p_1=2, p_2=3$ etc. The index can run over all positive integers (correct?), so we might have the index $n=27$. In that case, if we look at $m=3$, then the difference of the indexed primes is $0$, contrary to your claim. The same would be true for any $n=a^m$, regardless of the value of $m$. Or have I misunderstood your nomenclature?
$endgroup$
– Keith Backman
Jan 18 at 20:29
$begingroup$
Ok thanks I will have to take a look back as I was distracted by other matters. But the result baffled me, which is why the question was generated immediately upon inspection of the computation I intended to make, but as you know human error is ever present in the work of the analytical number theorist, there is no escaping it's humiliation, nor is there any point in feeling embarrassment in the first place
$endgroup$
– Adam
Jan 19 at 12:18
$begingroup$
I'd say you have to get in line behind me on points such as human error; I've made more than my share
$endgroup$
– Keith Backman
Jan 19 at 15:58
$begingroup$
haha no worries mate I just lost the ability to be upset about any number of people see me drop all of my shit along the moving subway one day, and that was the final seal there without the desire to put the microscope of every passenger having a giggle, no honour in extracting the truth that 99.9 percent of them have so little substance in their ambition and path to self fulfilment that a good lot of them return home to hang themselves after a brief conversation with me. That is my least favourite dream achievement sorry failure
$endgroup$
– Adam
Jan 20 at 12:18
$begingroup$
Ok yes I'm on board now I'm on board now with what I am saying, I know when I start It was on the context of if we have a fixed $m$ value, and evaluate over $mathbb N$ for $n$, and we have the result stated, this can only be the case for $m=2$ but what you have shown is a pretty obvious defiance of the formally stated rule as ive decide to put it on this night, not to worry I will probably fix this up if it becomes something worth while, having said that worthwhile to others is never unless you have one of $%2 bfedgfds#$^%$#T bloodlines or nobility, in which case no you wont even be allowed
$endgroup$
– Adam
Jan 20 at 15:40
1
1
$begingroup$
This baffles me. I assume that $p$ is a prime and $n$ is an index, i.e. $p_1=2, p_2=3$ etc. The index can run over all positive integers (correct?), so we might have the index $n=27$. In that case, if we look at $m=3$, then the difference of the indexed primes is $0$, contrary to your claim. The same would be true for any $n=a^m$, regardless of the value of $m$. Or have I misunderstood your nomenclature?
$endgroup$
– Keith Backman
Jan 18 at 20:29
$begingroup$
This baffles me. I assume that $p$ is a prime and $n$ is an index, i.e. $p_1=2, p_2=3$ etc. The index can run over all positive integers (correct?), so we might have the index $n=27$. In that case, if we look at $m=3$, then the difference of the indexed primes is $0$, contrary to your claim. The same would be true for any $n=a^m$, regardless of the value of $m$. Or have I misunderstood your nomenclature?
$endgroup$
– Keith Backman
Jan 18 at 20:29
$begingroup$
Ok thanks I will have to take a look back as I was distracted by other matters. But the result baffled me, which is why the question was generated immediately upon inspection of the computation I intended to make, but as you know human error is ever present in the work of the analytical number theorist, there is no escaping it's humiliation, nor is there any point in feeling embarrassment in the first place
$endgroup$
– Adam
Jan 19 at 12:18
$begingroup$
Ok thanks I will have to take a look back as I was distracted by other matters. But the result baffled me, which is why the question was generated immediately upon inspection of the computation I intended to make, but as you know human error is ever present in the work of the analytical number theorist, there is no escaping it's humiliation, nor is there any point in feeling embarrassment in the first place
$endgroup$
– Adam
Jan 19 at 12:18
$begingroup$
I'd say you have to get in line behind me on points such as human error; I've made more than my share
$endgroup$
– Keith Backman
Jan 19 at 15:58
$begingroup$
I'd say you have to get in line behind me on points such as human error; I've made more than my share
$endgroup$
– Keith Backman
Jan 19 at 15:58
$begingroup$
haha no worries mate I just lost the ability to be upset about any number of people see me drop all of my shit along the moving subway one day, and that was the final seal there without the desire to put the microscope of every passenger having a giggle, no honour in extracting the truth that 99.9 percent of them have so little substance in their ambition and path to self fulfilment that a good lot of them return home to hang themselves after a brief conversation with me. That is my least favourite dream achievement sorry failure
$endgroup$
– Adam
Jan 20 at 12:18
$begingroup$
haha no worries mate I just lost the ability to be upset about any number of people see me drop all of my shit along the moving subway one day, and that was the final seal there without the desire to put the microscope of every passenger having a giggle, no honour in extracting the truth that 99.9 percent of them have so little substance in their ambition and path to self fulfilment that a good lot of them return home to hang themselves after a brief conversation with me. That is my least favourite dream achievement sorry failure
$endgroup$
– Adam
Jan 20 at 12:18
$begingroup$
Ok yes I'm on board now I'm on board now with what I am saying, I know when I start It was on the context of if we have a fixed $m$ value, and evaluate over $mathbb N$ for $n$, and we have the result stated, this can only be the case for $m=2$ but what you have shown is a pretty obvious defiance of the formally stated rule as ive decide to put it on this night, not to worry I will probably fix this up if it becomes something worth while, having said that worthwhile to others is never unless you have one of $%2 bfedgfds#$^%$#T bloodlines or nobility, in which case no you wont even be allowed
$endgroup$
– Adam
Jan 20 at 15:40
$begingroup$
Ok yes I'm on board now I'm on board now with what I am saying, I know when I start It was on the context of if we have a fixed $m$ value, and evaluate over $mathbb N$ for $n$, and we have the result stated, this can only be the case for $m=2$ but what you have shown is a pretty obvious defiance of the formally stated rule as ive decide to put it on this night, not to worry I will probably fix this up if it becomes something worth while, having said that worthwhile to others is never unless you have one of $%2 bfedgfds#$^%$#T bloodlines or nobility, in which case no you wont even be allowed
$endgroup$
– Adam
Jan 20 at 15:40
|
show 3 more comments
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1
$begingroup$
This baffles me. I assume that $p$ is a prime and $n$ is an index, i.e. $p_1=2, p_2=3$ etc. The index can run over all positive integers (correct?), so we might have the index $n=27$. In that case, if we look at $m=3$, then the difference of the indexed primes is $0$, contrary to your claim. The same would be true for any $n=a^m$, regardless of the value of $m$. Or have I misunderstood your nomenclature?
$endgroup$
– Keith Backman
Jan 18 at 20:29
$begingroup$
Ok thanks I will have to take a look back as I was distracted by other matters. But the result baffled me, which is why the question was generated immediately upon inspection of the computation I intended to make, but as you know human error is ever present in the work of the analytical number theorist, there is no escaping it's humiliation, nor is there any point in feeling embarrassment in the first place
$endgroup$
– Adam
Jan 19 at 12:18
$begingroup$
I'd say you have to get in line behind me on points such as human error; I've made more than my share
$endgroup$
– Keith Backman
Jan 19 at 15:58
$begingroup$
haha no worries mate I just lost the ability to be upset about any number of people see me drop all of my shit along the moving subway one day, and that was the final seal there without the desire to put the microscope of every passenger having a giggle, no honour in extracting the truth that 99.9 percent of them have so little substance in their ambition and path to self fulfilment that a good lot of them return home to hang themselves after a brief conversation with me. That is my least favourite dream achievement sorry failure
$endgroup$
– Adam
Jan 20 at 12:18
$begingroup$
Ok yes I'm on board now I'm on board now with what I am saying, I know when I start It was on the context of if we have a fixed $m$ value, and evaluate over $mathbb N$ for $n$, and we have the result stated, this can only be the case for $m=2$ but what you have shown is a pretty obvious defiance of the formally stated rule as ive decide to put it on this night, not to worry I will probably fix this up if it becomes something worth while, having said that worthwhile to others is never unless you have one of $%2 bfedgfds#$^%$#T bloodlines or nobility, in which case no you wont even be allowed
$endgroup$
– Adam
Jan 20 at 15:40