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Plotting the sum of prime factors of integers with rational exponents [closed]
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In a sense, some numbers other than integers can be written in terms of prime factors.
For example
$$
sqrt[3]frac16^5 = 6^frac-53 = 2^frac-53 times 3^frac-53
$$
We can sum the prime factors we obtain and get some real
Here it's : $2^frac-53 + 3^frac-53 = 0.47523...$
If f(x) is the sum of the "prime factors" of x, how would the graph of f(x) look like ?
One way to do it, as highlighted by Ross Millikan, is to plot for all rationals x. And since the rationals are dense on the number line, although $2^frac-53 + 3^frac-53$ won't give the same result as the sum of the prime factors of an approximation of $sqrt[3]frac16^5$ , the other rationals will "hide it".
I personally think it's quite cruel to the irrational roots and if you can find a way to bring them to the game, that would be very nice. It can be considered as an additional challenge - an optional quest - an extra noble achievement - you get the idea.
Thank you in advance.
number-theory prime-numbers graphing-functions factoring prime-factorization
asked Jul 28 at 14:16
Frousse
92
closed as unclear what you're asking by John B, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, amWhy Jul 29 at 0:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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In a sense, some numbers other than integers can be written in terms of prime factors.
For example
$$
sqrt[3]frac16^5 = 6^frac-53 = 2^frac-53 times 3^frac-53
$$
We can sum the prime factors we obtain and get some real
Here it's : $2^frac-53 + 3^frac-53 = 0.47523...$
If f(x) is the sum of the "prime factors" of x, how would the graph of f(x) look like ?
One way to do it, as highlighted by Ross Millikan, is to plot for all rationals x. And since the rationals are dense on the number line, although $2^frac-53 + 3^frac-53$ won't give the same result as the sum of the prime factors of an approximation of $sqrt[3]frac16^5$ , the other rationals will "hide it".
I personally think it's quite cruel to the irrational roots and if you can find a way to bring them to the game, that would be very nice. It can be considered as an additional challenge - an optional quest - an extra noble achievement - you get the idea.
Thank you in advance.
number-theory prime-numbers graphing-functions factoring prime-factorization
asked Jul 28 at 14:16
Frousse
92
closed as unclear what you're asking by John B, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, amWhy Jul 29 at 0:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
All integers can be factorised uniquely, and if fractional powers are taken, by the laws of exponents if $n=pq$, where $p$, $q$ are primes, then $n^a/b=(n)^a/b=(pq)^a/b=p^a/bq^a/b$. The prime factors stay the same throughout. I'm not sure as to what you mean as to the sum of primes in intervals like $(0,2)$. I mean $2^1/k$ is in $(0,2)$ for all $k>1$, so do you mean to sum $2$ infinitely and plot it. How exactly?
– Daniel Buck
Jul 28 at 14:40
The numbers of the form $a^b/c$ will be dense in the real line greater than $1$. That would be true even if you restrict $a$ to primes. Plotting them would be like plotting the rationals for that reason. Is that what you are looking for?
– Ross Millikan
Jul 28 at 15:02
To Daniel : If $$ n^fracab = p^fracab q^fracab $$ then what we're looking for is : $p^fracab + q^fracab$ Basically, on the x axis you take all numbers that can be written as $n^fracab$ (up to a certain interval) and you look for their image which would be $ p^fracab + q^fracab$. To Millikan : It feels like cheating but yeah, you could plot for the rationals and we cannot tell ^^
– Frousse
Jul 28 at 15:28
This is impossible to plot. The function can take arbitrarily large values, think of $sum_n=1^1000p_n^1/100000000approx1000$.
– Yves Daoust
Jul 30 at 17:10
What if you plot, not for the primes but for the original number ? ex : f(0.1), f(0.2)... f(0.9), f(0.01), f(0.02)...
– Frousse
Jul 30 at 17:19
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In a sense, some numbers other than integers can be written in terms of prime factors.
For example
$$
sqrt[3]frac16^5 = 6^frac-53 = 2^frac-53 times 3^frac-53
$$
We can sum the prime factors we obtain and get some real
Here it's : $2^frac-53 + 3^frac-53 = 0.47523...$
If f(x) is the sum of the "prime factors" of x, how would the graph of f(x) look like ?
One way to do it, as highlighted by Ross Millikan, is to plot for all rationals x. And since the rationals are dense on the number line, although $2^frac-53 + 3^frac-53$ won't give the same result as the sum of the prime factors of an approximation of $sqrt[3]frac16^5$ , the other rationals will "hide it".
I personally think it's quite cruel to the irrational roots and if you can find a way to bring them to the game, that would be very nice. It can be considered as an additional challenge - an optional quest - an extra noble achievement - you get the idea.
Thank you in advance.
number-theory prime-numbers graphing-functions factoring prime-factorization
asked Jul 28 at 14:16
Frousse
92
In a sense, some numbers other than integers can be written in terms of prime factors.
For example
$$
sqrt[3]frac16^5 = 6^frac-53 = 2^frac-53 times 3^frac-53
$$
We can sum the prime factors we obtain and get some real
Here it's : $2^frac-53 + 3^frac-53 = 0.47523...$
If f(x) is the sum of the "prime factors" of x, how would the graph of f(x) look like ?
One way to do it, as highlighted by Ross Millikan, is to plot for all rationals x. And since the rationals are dense on the number line, although $2^frac-53 + 3^frac-53$ won't give the same result as the sum of the prime factors of an approximation of $sqrt[3]frac16^5$ , the other rationals will "hide it".
I personally think it's quite cruel to the irrational roots and if you can find a way to bring them to the game, that would be very nice. It can be considered as an additional challenge - an optional quest - an extra noble achievement - you get the idea.
Thank you in advance.
number-theory prime-numbers graphing-functions factoring prime-factorization
asked Jul 28 at 14:16
Frousse
92
edited Jul 30 at 16:45
asked Jul 28 at 14:16
Frousse
92
asked Jul 28 at 14:16
Frousse
92
asked Jul 28 at 14:16
Frousse
92
92
closed as unclear what you're asking by John B, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, amWhy Jul 29 at 0:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by John B, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, amWhy Jul 29 at 0:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
All integers can be factorised uniquely, and if fractional powers are taken, by the laws of exponents if $n=pq$, where $p$, $q$ are primes, then $n^a/b=(n)^a/b=(pq)^a/b=p^a/bq^a/b$. The prime factors stay the same throughout. I'm not sure as to what you mean as to the sum of primes in intervals like $(0,2)$. I mean $2^1/k$ is in $(0,2)$ for all $k>1$, so do you mean to sum $2$ infinitely and plot it. How exactly?
– Daniel Buck
Jul 28 at 14:40
The numbers of the form $a^b/c$ will be dense in the real line greater than $1$. That would be true even if you restrict $a$ to primes. Plotting them would be like plotting the rationals for that reason. Is that what you are looking for?
– Ross Millikan
Jul 28 at 15:02
To Daniel : If $$ n^fracab = p^fracab q^fracab $$ then what we're looking for is : $p^fracab + q^fracab$ Basically, on the x axis you take all numbers that can be written as $n^fracab$ (up to a certain interval) and you look for their image which would be $ p^fracab + q^fracab$. To Millikan : It feels like cheating but yeah, you could plot for the rationals and we cannot tell ^^
– Frousse
Jul 28 at 15:28
This is impossible to plot. The function can take arbitrarily large values, think of $sum_n=1^1000p_n^1/100000000approx1000$.
– Yves Daoust
Jul 30 at 17:10
What if you plot, not for the primes but for the original number ? ex : f(0.1), f(0.2)... f(0.9), f(0.01), f(0.02)...
– Frousse
Jul 30 at 17:19
add a comment |Â
All integers can be factorised uniquely, and if fractional powers are taken, by the laws of exponents if $n=pq$, where $p$, $q$ are primes, then $n^a/b=(n)^a/b=(pq)^a/b=p^a/bq^a/b$. The prime factors stay the same throughout. I'm not sure as to what you mean as to the sum of primes in intervals like $(0,2)$. I mean $2^1/k$ is in $(0,2)$ for all $k>1$, so do you mean to sum $2$ infinitely and plot it. How exactly?
– Daniel Buck
Jul 28 at 14:40
The numbers of the form $a^b/c$ will be dense in the real line greater than $1$. That would be true even if you restrict $a$ to primes. Plotting them would be like plotting the rationals for that reason. Is that what you are looking for?
– Ross Millikan
Jul 28 at 15:02
To Daniel : If $$ n^fracab = p^fracab q^fracab $$ then what we're looking for is : $p^fracab + q^fracab$ Basically, on the x axis you take all numbers that can be written as $n^fracab$ (up to a certain interval) and you look for their image which would be $ p^fracab + q^fracab$. To Millikan : It feels like cheating but yeah, you could plot for the rationals and we cannot tell ^^
– Frousse
Jul 28 at 15:28
This is impossible to plot. The function can take arbitrarily large values, think of $sum_n=1^1000p_n^1/100000000approx1000$.
– Yves Daoust
Jul 30 at 17:10
What if you plot, not for the primes but for the original number ? ex : f(0.1), f(0.2)... f(0.9), f(0.01), f(0.02)...
– Frousse
Jul 30 at 17:19
All integers can be factorised uniquely, and if fractional powers are taken, by the laws of exponents if $n=pq$, where $p$, $q$ are primes, then $n^a/b=(n)^a/b=(pq)^a/b=p^a/bq^a/b$. The prime factors stay the same throughout. I'm not sure as to what you mean as to the sum of primes in intervals like $(0,2)$. I mean $2^1/k$ is in $(0,2)$ for all $k>1$, so do you mean to sum $2$ infinitely and plot it. How exactly?
– Daniel Buck
Jul 28 at 14:40
All integers can be factorised uniquely, and if fractional powers are taken, by the laws of exponents if $n=pq$, where $p$, $q$ are primes, then $n^a/b=(n)^a/b=(pq)^a/b=p^a/bq^a/b$. The prime factors stay the same throughout. I'm not sure as to what you mean as to the sum of primes in intervals like $(0,2)$. I mean $2^1/k$ is in $(0,2)$ for all $k>1$, so do you mean to sum $2$ infinitely and plot it. How exactly?
– Daniel Buck
Jul 28 at 14:40
The numbers of the form $a^b/c$ will be dense in the real line greater than $1$. That would be true even if you restrict $a$ to primes. Plotting them would be like plotting the rationals for that reason. Is that what you are looking for?
– Ross Millikan
Jul 28 at 15:02
The numbers of the form $a^b/c$ will be dense in the real line greater than $1$. That would be true even if you restrict $a$ to primes. Plotting them would be like plotting the rationals for that reason. Is that what you are looking for?
– Ross Millikan
Jul 28 at 15:02
To Daniel : If $$ n^fracab = p^fracab q^fracab $$ then what we're looking for is : $p^fracab + q^fracab$ Basically, on the x axis you take all numbers that can be written as $n^fracab$ (up to a certain interval) and you look for their image which would be $ p^fracab + q^fracab$. To Millikan : It feels like cheating but yeah, you could plot for the rationals and we cannot tell ^^
– Frousse
Jul 28 at 15:28
To Daniel : If $$ n^fracab = p^fracab q^fracab $$ then what we're looking for is : $p^fracab + q^fracab$ Basically, on the x axis you take all numbers that can be written as $n^fracab$ (up to a certain interval) and you look for their image which would be $ p^fracab + q^fracab$. To Millikan : It feels like cheating but yeah, you could plot for the rationals and we cannot tell ^^
– Frousse
Jul 28 at 15:28
This is impossible to plot. The function can take arbitrarily large values, think of $sum_n=1^1000p_n^1/100000000approx1000$.
– Yves Daoust
Jul 30 at 17:10
This is impossible to plot. The function can take arbitrarily large values, think of $sum_n=1^1000p_n^1/100000000approx1000$.
– Yves Daoust
Jul 30 at 17:10
What if you plot, not for the primes but for the original number ? ex : f(0.1), f(0.2)... f(0.9), f(0.01), f(0.02)...
– Frousse
Jul 30 at 17:19
What if you plot, not for the primes but for the original number ? ex : f(0.1), f(0.2)... f(0.9), f(0.01), f(0.02)...
– Frousse
Jul 30 at 17:19
add a comment |Â
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All integers can be factorised uniquely, and if fractional powers are taken, by the laws of exponents if $n=pq$, where $p$, $q$ are primes, then $n^a/b=(n)^a/b=(pq)^a/b=p^a/bq^a/b$. The prime factors stay the same throughout. I'm not sure as to what you mean as to the sum of primes in intervals like $(0,2)$. I mean $2^1/k$ is in $(0,2)$ for all $k>1$, so do you mean to sum $2$ infinitely and plot it. How exactly?
– Daniel Buck
Jul 28 at 14:40
The numbers of the form $a^b/c$ will be dense in the real line greater than $1$. That would be true even if you restrict $a$ to primes. Plotting them would be like plotting the rationals for that reason. Is that what you are looking for?
– Ross Millikan
Jul 28 at 15:02
To Daniel : If $$ n^fracab = p^fracab q^fracab $$ then what we're looking for is : $p^fracab + q^fracab$ Basically, on the x axis you take all numbers that can be written as $n^fracab$ (up to a certain interval) and you look for their image which would be $ p^fracab + q^fracab$. To Millikan : It feels like cheating but yeah, you could plot for the rationals and we cannot tell ^^
– Frousse
Jul 28 at 15:28
This is impossible to plot. The function can take arbitrarily large values, think of $sum_n=1^1000p_n^1/100000000approx1000$.
– Yves Daoust
Jul 30 at 17:10
What if you plot, not for the primes but for the original number ? ex : f(0.1), f(0.2)... f(0.9), f(0.01), f(0.02)...
– Frousse
Jul 30 at 17:19