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edited Nov 30 2011 at 21:43 |
First, from a more detailed theorem of Robin we have an unconditional result (1984) that says that your ratio of interest is, for $n \geq 3,$ 13,$ smaller than
$$ e^\gamma + \frac{0.6482}{(\log frac{0.64821364942...}{(\log \log n)^2},$$ with the constant in the numerator giving equality for $n=12.$
from which it follows that the your supremum is achieved for some $n,$ perhaps 5040 itself. , but almost certainly with $n \leq 5040.$ Note that $\log \log n$ first exceeds 1 for $n \geq 16$ so we should probably include that as a hypothesis.
Second, it is a virtual certainty that the maximum will be achieved by a colossally abundant number, see COLOSSUS
There is a recipe for these, given some $\delta > 0,$ the prime factorization for the (largest if more than one) number $n$ that maximizes $$ \frac{ \sigma(n)}{n^{1 + \delta}} $$ is given by an explicit formula for each prime's exponent involving the floor function. I will see if I can find that, meanwhile just look at the sequence A004490 in OEIS. The recipe should be written out in Alaoglu and Erdos (1944). Indeed, Ramanujan included these numbers in his original article on highly composite numbers, but that section was not included in the publication owing to shortages of paper at the time(1915). The full manuscript was published in the Ramanujan Journal in 1997.
EDIT: I was able to download Alaoglu and Erdos, given some $\delta > 0,$ the correct exponent for some prime $p$ is
$$ \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1. $$
This is Theorem 10 on page 455. For a fixed $\delta,$ the exponents either stay the same or decrease for increasing $p,$ and eventually the exponent 0 is reached, so there is your complete number. For a fixed $p,$ the exponent either stays the same or increases with decreasing $\delta.$
I'm not seeing any lists that show $\delta$ and the result, so here, if I call $f(\delta)$ the corresponding colossally abundant number for $\delta,$ I calculate $$ f(1) = 1, \; f(1/2) = 2, \; f(1/4) = 6, \; f(1/6) = 12, \; f(1/10) = 60, \; f(1/12) = 120,$$
then
$$ f(1/14) = 360, \; f(1/17) = 2520, \; f(1/25) = 5040, \; f(1/31) = 55440, \; f(1/39) = 720720,$$
and so on as $\delta$ decreases.
If you want the first (largest) $\delta$ for which a favorite prime $p$ gets assigned exponent $k,$ let
$$ \delta = \frac{\log(p^{k+1} - 1) - \log(p^{k+1} - p)}{\log p} $$
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edited Nov 4 2011 at 4:29 |
First, from a more detailed theorem of Robin we have an unconditional result (1984) that says that your ratio of interest is, for $n \geq 3,$ smaller than
$$ e^\gamma + \frac{0.6482}{(\log \log n)^2},$$
from which it follows that the supremum is achieved for some $n,$ perhaps 5040 itself. Note that $\log \log n$ first exceeds 1 for $n \geq 16$ so we should probably include that as a hypothesis.
Second, it is a virtual certainty that the maximum will be achieved by a colossally abundant number, see COLOSSUS
There is a recipe for these, given some $\delta > 0,$ the prime factorization for the (largest if more than one) number $n$ that maximizes $$ \frac{ \sigma(n)}{n^{1 + \delta}} $$ is given by an explicit formula for each prime's exponent involving the floor function. I will see if I can find that, meanwhile just look at the sequence A004490 in OEIS. The recipe should be written out in Alaoglu and Erdos (1944). Indeed, Ramanujan included these numbers in his original article on highly composite numbers, but that section was not included in the publication owing to shortages of paper at the time(1915). The full manuscript was published in the Ramanujan Journal in 1997.
EDIT: I was able to download Alaoglu and Erdos, given some $\delta > 0,$ the correct exponent for some prime $p$ is
$$ \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1. $$
This is Theorem 10 on page 455. For a fixed $\delta,$ the exponents either stay the same or decrease for increasing $p,$ and eventually the exponent 0 is reached, so there is your complete number. For a fixed $p,$ the exponent either stays the same or increases with decreasing $\delta.$
I'm not seeing any lists that show $\delta$ and the result, so here, if I call $f(\delta)$ the corresponding colossally abundant number for $\delta,$ I calculate $$ f(1) = 1, \; f(1/2) = 2, \; f(1/4) = 6, \; f(2/9f(1/6) = 12, \; f(1/9f(1/10) = 60$$ 60, \; f(1/12) = 120,$$
then
$$ f(1/14) = 360, \; f(1/17) = 2520, \; f(1/25) = 5040, \; f(1/31) = 55440, \; f(1/39) = 720720,$$
and so on as $\delta$ decreases.
If you want the first (largest) $\delta$ for which a favorite prime $p$ gets assigned exponent $k,$ let
$$ \delta = \frac{\log(p^{k+1} - 1) - \log(p^{k+1} - p)}{\log p} $$
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edited Nov 4 2011 at 0:56 |
First, from a more detailed theorem of Robin we have an unconditional result (1984) that says that your ratio of interest is, for $n \geq 3,$ smaller than
$$ e^\gamma + \frac{0.6482}{(\log \log n)^2},$$
from which it follows that the supremum is achieved for some $n,$ perhaps 5040 itself. Note that $\log \log n$ first exceeds 1 for $n \geq 16$ so we should probably include that as a hypothesis.
Second, it is a virtual certainty that the maximum will be achieved by a colossally abundant number, see COLOSSUS
There is a recipe for these, given some $\delta > 0,$ the prime factorization for the (largest if more than one) number $n$ that maximizes $$ \frac{ \sigma(n)}{n^{1 + \delta}} $$ is given by an explicit formula for each prime's exponent involving the floor function. I will see if I can find that, meanwhile just look at the sequence A004490 in OEIS. The recipe should be written out in Alaoglu and Erdos (1944). Indeed, Ramanujan included these numbers in his original article on highly composite numbers, but that section was not included in the publication owing to shortages of paper at the time(1915). The full manuscript was published in the Ramanujan Journal in 1997.
EDIT: I was able to download Alaoglu and Erdos, given some $\delta > 0,$ the correct exponent for some prime $p$ is
$$ \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1. $$
This is Theorem 10 on page 455. For a fixed $\delta,$ the exponents either stay the same or decrease for increasing $p,$ and eventually the exponent 0 is reached, so there is your complete number. For a fixed $p,$ the exponent either stays the same or increases with decreasing $\delta.$
I'm not seeing any lists that show $\delta$ and the result, so here, if I call $f(\delta)$ the corresponding colossally abundant number for $\delta,$ I calculate $$ f(1) = 1, \; f(1/2) = 2, \; f(1/4) = 6, \; f(2/9) = 12, \; f(1/9) = 60$$ and so on as $\delta \delta$ decreases.
If you want the first (largest) $\delta$ for which a favorite prime $p$ gets assigned exponent $k,$ let
$$ \delta = \frac{\log(p^{k+1} - 1) - \log(p^{k+1} - p)}{\log p} $$
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edited Nov 4 2011 at 0:51 |
First, from a more detailed theorem of Robin we have an unconditional result (1984) that says that your ratio of interest is, for $n \geq 3,$ smaller than
$$ e^\gamma + \frac{0.6482}{(\log \log n)^2},$$
from which it follows that the supremum is achieved for some $n,$ perhaps 5040 itself. Note that $\log \log n$ first exceeds 1 for $n \geq 16$ so we should probably include that as a hypothesis.
Second, it is a virtual certainty that the maximum will be achieved by a colossally abundant number, see COLOSSUS
There is a recipe for these, given some $\delta > 0,$ the prime factorization for the (largest if more than one) number $n$ that maximizes $$ \frac{ \sigma(n)}{n^{1 + \delta}} $$ is given by an explicit formula for each prime's exponent involving the floor function. I will see if I can find that, meanwhile just look at the sequence A004490 in OEIS. The recipe should be written out in Alaoglu and Erdos (1944). Indeed, Ramanujan included these numbers in his original article on highly composite numbers, but that section was not included in the publication owing to shortages of paper at the time(1915). The full manuscript was published in the Ramanujan Journal in 1997.
EDIT: I was able to download Alaoglu and Erdos, given some $\delta > 0,$ the correct exponent for some prime $p$ is
$$ \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1. $$
This is Theorem 10 on page 455. For a fixed $\delta,$ the exponents either stay the same or decrease for increasing $p,$ and eventually the exponent 0 is reached, so there is your complete number. For a fixed $p,$ the exponent either stays the same or increases with decreasing $\delta.$
I'm not seeing any lists that show $\delta$ and the result, so here, if I call $f(\delta)$ the corresponding colossally abundant number for $\delta,$ I calculate $$ f(1) = 1, \; f(1/2) = 2, \; f(1/4) = 6, \; f(2/9) = 12, \; f(1/9) = 60$$ and so on as $\delta decreases.
If you want the first (largest) $\delta$ for which a favorite prime $p$ gets assigned exponent $k,$ let
$$ \delta = \frac{\log(p^{k+1} - 1) - \log(p^{k+1} - p)}{\log p} $$
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3
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edited Nov 4 2011 at 0:23 |
First, from a more detailed theorem of Robin we have an unconditional result (1984) that says that your ratio of interest is, for $n \geq 3,$ smaller than
$$ e^\gamma + \frac{0.6482}{(\log \log n)^2},$$
from which it follows that the supremum is achieved for some $n,$ perhaps 5040 itself. Note that $\log \log n$ first exceeds 1 for $n \geq 16$ so we should probably include that as a hypothesis.
Second, it is a virtual certainty that the maximum will be achieved by a colossally abundant number, see COLOSSUS
There is a recipe for these, given some $\delta > 0,$ the prime factorization for the (largest if more than one) number $n$ that maximizes $$ \frac{ \sigma(n)}{n^{1 + \delta}} $$ is given by an explicit formula for each prime's exponent involving the floor function. I will see if I can find that, meanwhile just look at the sequence A004490 in OEIS. The recipe should be written out in Alaoglu and Erdos (1944). Indeed, Ramanujan included these numbers in his original article on highly composite numbers, but that section was not included in the publication owing to shortages of paper at the time(1915). The full manuscript was published in the Ramanujan Journal in 1997.
EDIT: I was able to download Alaoglu and Erdos, given some $\delta > 0,$ the correct exponent for some prime $p$ is
$$ \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1. $$
This is Theorem 10 on page 455. For a fixed $\delta,$ the exponents either stay the same or decrease for increasing $p,$ and eventually the exponent 0 is reached, so there is your complete number. For a fixed $p,$ the exponent either stays the same or increases with decreasing $\delta.$
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edited Nov 3 2011 at 23:25 |
First, from a more detailed theorem of Robin we have an unconditional result (1984) that says that your ratio of interest is, for $n \geq 3,$ smaller than
$$ e^\gamma + \frac{0.6482}{(\log \log n)^2},$$
from which it follows that the supremum is achieved for some $n,$ perhaps 5040 itself. Note that $\log \log n$ first exceeds 1 for $n \geq 16$ so we should probably include that as a hypothesis.
Second, it is a virtual certainty that the maximum will be achieved by a colossally abundant number, see COLOSSUS
There is a recipe for these, given some $\delta > 0,$ the prime factorization for the (largest if more than one) number $n$ that maximizes $$ \frac{ \sigma(n)}{n^{1 + \delta}} $$ is given by an explicit formula for each prime prime's exponent involving the floor function. I will see if I can find that, meanwhile just look at the sequence A004490 in OEIS. The recipe should be written ought out in Alaoglu and Erdos (1944). Indeed, Ramanujan included these numbers in his original article on highly composite numbers, but that section was not included in the publication owing to shortages of paper at the time(1915). The full manuscript was published in the Ramanujan Journal in 1997.
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answered Nov 3 2011 at 23:18 |
First, from a more detailed theorem of Robin we have an unconditional result (1984) that says that your ratio of interest is, for $n \geq 3,$ smaller than
$$ e^\gamma + \frac{0.6482}{(\log \log n)^2},$$
from which it follows that the supremum is achieved for some $n,$ perhaps 5040 itself.
Second, it is a virtual certainty that the maximum will be achieved by a colossally abundant number, see COLOSSUS
There is a recipe for these, given some $\delta > 0,$ the prime factorization for the (largest if more than one) number $n$ that maximizes $$ \frac{ \sigma(n)}{n^{1 + \delta}} $$ is given by an explicit formula for each prime involving the floor function. I will see if I can find that, meanwhile just look at the sequence A004490 in OEIS. The recipe should be written ought in Alaoglu and Erdos (1944).
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