Mixing
Improving the bound for $sigma(q^k)/q^k$ where $q^k n^2$ is an odd perfect number given in Eulerian form
$begingroup$
Let $x$ be a positive integer. (That is, let $x in mathbb{N}$.)
We denote the sum of divisors of $x$ as
$$sigma(x) = sum_{d mid x}{d}.$$
We also denote the abundancy index of $x$ as $I(x)=sigma(x)/x$.
If $N$ is odd and $sigma(N)=2N$, then $N$ is called an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$, i.e. $q$ is the special prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Since $k equiv 1 pmod 4$, then we have
$$frac{q+1}{q} = I(q) leq I(q^k) = frac{q^{k+1} - 1}{q^k (q - 1)} < frac{q^{k+1}}{q^k (q - 1)} = frac{q}{q - 1}.$$
Since $q$ is prime and satisfies $q equiv 1 pmod 4$, we have $q geq 5$, from which we obtain
$$frac{1}{q} leq frac{1}{5} implies -frac{1}{q} geq - frac{1}{5} implies frac{q-1}{q} = 1 - frac{1}{q} geq 1 - frac{1}{5} = frac{4}{5}.$$
We get that
$$I(q^k) < frac{q}{q - 1} leq frac{5}{4}$$
from which we conclude that
$$I(q^k) < frac{5}{4}.$$
We also obtain
$$I(q^k) < frac{5}{4} < sqrt{frac{8}{5}} < sqrt{I(n^2)} = sqrt{frac{2}{I(q^k)}},$$
from which we get
$$bigg(I(q^k)bigg)^2 < frac{2}{I(q^k)} implies bigg(I(q^k)bigg)^3 < 2 implies I(q^k) < sqrt[3]{2}.$$
Here is my question:
Is it possible to improve on the inequality
$$I(q^k) < sqrt{frac{2}{I(q^k)}},$$
to something like (say)
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}?$$
Note that $1/3 = 0.overline{333}$.
MOTIVATION
Here is the reason why I think it might be possible to improve on the bound for $I(q^k)$.
Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.
The Descartes-Frenicle-Sorli Conjecture predicts that $k=1$. Suppose that this Conjecture holds.
Then
$$I(q^k) = I(q) = frac{q+1}{q} = 1+frac{1}{q} leq 1+frac{1}{5}=frac{6}{5}$$
and
$$I(n^2) = frac{2}{I(q^k)} = frac{2}{I(q)} geq frac{2}{frac{6}{5}} = frac{5}{3}.$$
Let
$$frac{6}{5} = bigg(frac{5}{3}bigg)^y.$$
Note that we then have that
$$I(q) leq frac{6}{5} = bigg(frac{5}{3}bigg)^y leq bigg(I(n^2)bigg)^y = bigg(frac{2}{I(q)}bigg)^y$$
where
$$y = frac{logbigg(frac{6}{5}bigg)}{logbigg(frac{5}{3}bigg)} approx 0.356915448856724.$$
WLOG, if we assume that $k>1$ and let
$$frac{5}{4} = bigg(frac{8}{5}bigg)^z,$$
then we have that
$$I(q^k) < frac{5}{4} = bigg(frac{8}{5}bigg)^z < bigg(I(n^2)bigg)^z = bigg(frac{2}{I(q^k)}bigg)^z$$
where
$$z = frac{logbigg(frac{5}{4}bigg)}{logbigg(frac{8}{5}bigg)} approx 0.474769847356948651282146696312271.$$
number-theory inequality upper-lower-bounds divisor-sum perfect-numbers
edited Jan 29 at 12:09
YuiTo Cheng
2,1862937
asked Jan 29 at 6:46
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
$endgroup$
add a comment |
$begingroup$
Let $x$ be a positive integer. (That is, let $x in mathbb{N}$.)
We denote the sum of divisors of $x$ as
$$sigma(x) = sum_{d mid x}{d}.$$
We also denote the abundancy index of $x$ as $I(x)=sigma(x)/x$.
If $N$ is odd and $sigma(N)=2N$, then $N$ is called an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$, i.e. $q$ is the special prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Since $k equiv 1 pmod 4$, then we have
$$frac{q+1}{q} = I(q) leq I(q^k) = frac{q^{k+1} - 1}{q^k (q - 1)} < frac{q^{k+1}}{q^k (q - 1)} = frac{q}{q - 1}.$$
Since $q$ is prime and satisfies $q equiv 1 pmod 4$, we have $q geq 5$, from which we obtain
$$frac{1}{q} leq frac{1}{5} implies -frac{1}{q} geq - frac{1}{5} implies frac{q-1}{q} = 1 - frac{1}{q} geq 1 - frac{1}{5} = frac{4}{5}.$$
We get that
$$I(q^k) < frac{q}{q - 1} leq frac{5}{4}$$
from which we conclude that
$$I(q^k) < frac{5}{4}.$$
We also obtain
$$I(q^k) < frac{5}{4} < sqrt{frac{8}{5}} < sqrt{I(n^2)} = sqrt{frac{2}{I(q^k)}},$$
from which we get
$$bigg(I(q^k)bigg)^2 < frac{2}{I(q^k)} implies bigg(I(q^k)bigg)^3 < 2 implies I(q^k) < sqrt[3]{2}.$$
Here is my question:
Is it possible to improve on the inequality
$$I(q^k) < sqrt{frac{2}{I(q^k)}},$$
to something like (say)
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}?$$
Note that $1/3 = 0.overline{333}$.
MOTIVATION
Here is the reason why I think it might be possible to improve on the bound for $I(q^k)$.
Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.
The Descartes-Frenicle-Sorli Conjecture predicts that $k=1$. Suppose that this Conjecture holds.
Then
$$I(q^k) = I(q) = frac{q+1}{q} = 1+frac{1}{q} leq 1+frac{1}{5}=frac{6}{5}$$
and
$$I(n^2) = frac{2}{I(q^k)} = frac{2}{I(q)} geq frac{2}{frac{6}{5}} = frac{5}{3}.$$
Let
$$frac{6}{5} = bigg(frac{5}{3}bigg)^y.$$
Note that we then have that
$$I(q) leq frac{6}{5} = bigg(frac{5}{3}bigg)^y leq bigg(I(n^2)bigg)^y = bigg(frac{2}{I(q)}bigg)^y$$
where
$$y = frac{logbigg(frac{6}{5}bigg)}{logbigg(frac{5}{3}bigg)} approx 0.356915448856724.$$
WLOG, if we assume that $k>1$ and let
$$frac{5}{4} = bigg(frac{8}{5}bigg)^z,$$
then we have that
$$I(q^k) < frac{5}{4} = bigg(frac{8}{5}bigg)^z < bigg(I(n^2)bigg)^z = bigg(frac{2}{I(q^k)}bigg)^z$$
where
$$z = frac{logbigg(frac{5}{4}bigg)}{logbigg(frac{8}{5}bigg)} approx 0.474769847356948651282146696312271.$$
number-theory inequality upper-lower-bounds divisor-sum perfect-numbers
edited Jan 29 at 12:09
YuiTo Cheng
2,1862937
asked Jan 29 at 6:46
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
$endgroup$
add a comment |
$begingroup$
Let $x$ be a positive integer. (That is, let $x in mathbb{N}$.)
We denote the sum of divisors of $x$ as
$$sigma(x) = sum_{d mid x}{d}.$$
We also denote the abundancy index of $x$ as $I(x)=sigma(x)/x$.
If $N$ is odd and $sigma(N)=2N$, then $N$ is called an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$, i.e. $q$ is the special prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Since $k equiv 1 pmod 4$, then we have
$$frac{q+1}{q} = I(q) leq I(q^k) = frac{q^{k+1} - 1}{q^k (q - 1)} < frac{q^{k+1}}{q^k (q - 1)} = frac{q}{q - 1}.$$
Since $q$ is prime and satisfies $q equiv 1 pmod 4$, we have $q geq 5$, from which we obtain
$$frac{1}{q} leq frac{1}{5} implies -frac{1}{q} geq - frac{1}{5} implies frac{q-1}{q} = 1 - frac{1}{q} geq 1 - frac{1}{5} = frac{4}{5}.$$
We get that
$$I(q^k) < frac{q}{q - 1} leq frac{5}{4}$$
from which we conclude that
$$I(q^k) < frac{5}{4}.$$
We also obtain
$$I(q^k) < frac{5}{4} < sqrt{frac{8}{5}} < sqrt{I(n^2)} = sqrt{frac{2}{I(q^k)}},$$
from which we get
$$bigg(I(q^k)bigg)^2 < frac{2}{I(q^k)} implies bigg(I(q^k)bigg)^3 < 2 implies I(q^k) < sqrt[3]{2}.$$
Here is my question:
Is it possible to improve on the inequality
$$I(q^k) < sqrt{frac{2}{I(q^k)}},$$
to something like (say)
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}?$$
Note that $1/3 = 0.overline{333}$.
MOTIVATION
Here is the reason why I think it might be possible to improve on the bound for $I(q^k)$.
Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.
The Descartes-Frenicle-Sorli Conjecture predicts that $k=1$. Suppose that this Conjecture holds.
Then
$$I(q^k) = I(q) = frac{q+1}{q} = 1+frac{1}{q} leq 1+frac{1}{5}=frac{6}{5}$$
and
$$I(n^2) = frac{2}{I(q^k)} = frac{2}{I(q)} geq frac{2}{frac{6}{5}} = frac{5}{3}.$$
Let
$$frac{6}{5} = bigg(frac{5}{3}bigg)^y.$$
Note that we then have that
$$I(q) leq frac{6}{5} = bigg(frac{5}{3}bigg)^y leq bigg(I(n^2)bigg)^y = bigg(frac{2}{I(q)}bigg)^y$$
where
$$y = frac{logbigg(frac{6}{5}bigg)}{logbigg(frac{5}{3}bigg)} approx 0.356915448856724.$$
WLOG, if we assume that $k>1$ and let
$$frac{5}{4} = bigg(frac{8}{5}bigg)^z,$$
then we have that
$$I(q^k) < frac{5}{4} = bigg(frac{8}{5}bigg)^z < bigg(I(n^2)bigg)^z = bigg(frac{2}{I(q^k)}bigg)^z$$
where
$$z = frac{logbigg(frac{5}{4}bigg)}{logbigg(frac{8}{5}bigg)} approx 0.474769847356948651282146696312271.$$
number-theory inequality upper-lower-bounds divisor-sum perfect-numbers
edited Jan 29 at 12:09
YuiTo Cheng
2,1862937
asked Jan 29 at 6:46
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
$endgroup$
Let $x$ be a positive integer. (That is, let $x in mathbb{N}$.)
We denote the sum of divisors of $x$ as
$$sigma(x) = sum_{d mid x}{d}.$$
We also denote the abundancy index of $x$ as $I(x)=sigma(x)/x$.
If $N$ is odd and $sigma(N)=2N$, then $N$ is called an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$, i.e. $q$ is the special prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.
Since $k equiv 1 pmod 4$, then we have
$$frac{q+1}{q} = I(q) leq I(q^k) = frac{q^{k+1} - 1}{q^k (q - 1)} < frac{q^{k+1}}{q^k (q - 1)} = frac{q}{q - 1}.$$
Since $q$ is prime and satisfies $q equiv 1 pmod 4$, we have $q geq 5$, from which we obtain
$$frac{1}{q} leq frac{1}{5} implies -frac{1}{q} geq - frac{1}{5} implies frac{q-1}{q} = 1 - frac{1}{q} geq 1 - frac{1}{5} = frac{4}{5}.$$
We get that
$$I(q^k) < frac{q}{q - 1} leq frac{5}{4}$$
from which we conclude that
$$I(q^k) < frac{5}{4}.$$
We also obtain
$$I(q^k) < frac{5}{4} < sqrt{frac{8}{5}} < sqrt{I(n^2)} = sqrt{frac{2}{I(q^k)}},$$
from which we get
$$bigg(I(q^k)bigg)^2 < frac{2}{I(q^k)} implies bigg(I(q^k)bigg)^3 < 2 implies I(q^k) < sqrt[3]{2}.$$
Here is my question:
Is it possible to improve on the inequality
$$I(q^k) < sqrt{frac{2}{I(q^k)}},$$
to something like (say)
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}?$$
Note that $1/3 = 0.overline{333}$.
MOTIVATION
Here is the reason why I think it might be possible to improve on the bound for $I(q^k)$.
Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.
The Descartes-Frenicle-Sorli Conjecture predicts that $k=1$. Suppose that this Conjecture holds.
Then
$$I(q^k) = I(q) = frac{q+1}{q} = 1+frac{1}{q} leq 1+frac{1}{5}=frac{6}{5}$$
and
$$I(n^2) = frac{2}{I(q^k)} = frac{2}{I(q)} geq frac{2}{frac{6}{5}} = frac{5}{3}.$$
Let
$$frac{6}{5} = bigg(frac{5}{3}bigg)^y.$$
Note that we then have that
$$I(q) leq frac{6}{5} = bigg(frac{5}{3}bigg)^y leq bigg(I(n^2)bigg)^y = bigg(frac{2}{I(q)}bigg)^y$$
where
$$y = frac{logbigg(frac{6}{5}bigg)}{logbigg(frac{5}{3}bigg)} approx 0.356915448856724.$$
WLOG, if we assume that $k>1$ and let
$$frac{5}{4} = bigg(frac{8}{5}bigg)^z,$$
then we have that
$$I(q^k) < frac{5}{4} = bigg(frac{8}{5}bigg)^z < bigg(I(n^2)bigg)^z = bigg(frac{2}{I(q^k)}bigg)^z$$
where
$$z = frac{logbigg(frac{5}{4}bigg)}{logbigg(frac{8}{5}bigg)} approx 0.474769847356948651282146696312271.$$
number-theory inequality upper-lower-bounds divisor-sum perfect-numbers
number-theory inequality upper-lower-bounds divisor-sum perfect-numbers
edited Jan 29 at 12:09
YuiTo Cheng
2,1862937
asked Jan 29 at 6:46
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
edited Jan 29 at 12:09
YuiTo Cheng
2,1862937
asked Jan 29 at 6:46
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
edited Jan 29 at 12:09
YuiTo Cheng
2,1862937
edited Jan 29 at 12:09
YuiTo Cheng
2,1862937
edited Jan 29 at 12:09
YuiTo Cheng
2,1862937
2,1862937
asked Jan 29 at 6:46
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
asked Jan 29 at 6:46
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
asked Jan 29 at 6:46
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
5,51741945
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It turns out that
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
implies
$$1 + frac{1}{q} = I(q) leq I(q^k) < sqrt[4]{2}$$
from which we obtain
$$q > bigg(sqrt[4]{2} - 1bigg)^{-1} approx 5.2852135$$
thereby giving
$$q geq 13$$
since $q$ is a prime satisfying $q equiv 1 pmod 4$. Thus, the implication
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} implies q geq 13$$
holds.
If the reverse inequality
$$I(q^k) > sqrt[3]{frac{2}{I(q^k)}}$$
holds, then we get
$$frac{q}{q - 1} > I(q^k) > sqrt[4]{2}$$
which implies that
$$1 < q < frac{sqrt[4]{2}}{sqrt[4]{2} - 1} approx 6.28521$$
from which we conclude that $q = 5$.
Since $q geq 5$ and $q$ is a prime satisfying $q equiv 1 pmod 4$, by the contrapositive of the last implication, we get the implication
$$q geq 13 implies I(q^k) < sqrt[3]{frac{2}{I(q^k)}}.$$
We therefore have the biconditional
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} iff q geq 13.$$
(Note that we cannot have
$$I(q^k) = sqrt[3]{frac{2}{I(q^k)}}$$
as equality implies that $I(q^k) = sqrt[4]{2}$, contradicting the fact that
$$I(q^k) = frac{q^{k+1} - 1}{q^k (q - 1)}$$
is rational.)
Thus, to prove the inequality
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
we need to rule out $q=5$. This is currently open, and is also unknown even if we assume the Descartes-Frenicle-Sorli Conjecture that $k=1$.
answered Jan 29 at 11:51
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091843%2fimproving-the-bound-for-sigmaqk-qk-where-qk-n2-is-an-odd-perfect-numb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It turns out that
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
implies
$$1 + frac{1}{q} = I(q) leq I(q^k) < sqrt[4]{2}$$
from which we obtain
$$q > bigg(sqrt[4]{2} - 1bigg)^{-1} approx 5.2852135$$
thereby giving
$$q geq 13$$
since $q$ is a prime satisfying $q equiv 1 pmod 4$. Thus, the implication
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} implies q geq 13$$
holds.
If the reverse inequality
$$I(q^k) > sqrt[3]{frac{2}{I(q^k)}}$$
holds, then we get
$$frac{q}{q - 1} > I(q^k) > sqrt[4]{2}$$
which implies that
$$1 < q < frac{sqrt[4]{2}}{sqrt[4]{2} - 1} approx 6.28521$$
from which we conclude that $q = 5$.
Since $q geq 5$ and $q$ is a prime satisfying $q equiv 1 pmod 4$, by the contrapositive of the last implication, we get the implication
$$q geq 13 implies I(q^k) < sqrt[3]{frac{2}{I(q^k)}}.$$
We therefore have the biconditional
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} iff q geq 13.$$
(Note that we cannot have
$$I(q^k) = sqrt[3]{frac{2}{I(q^k)}}$$
as equality implies that $I(q^k) = sqrt[4]{2}$, contradicting the fact that
$$I(q^k) = frac{q^{k+1} - 1}{q^k (q - 1)}$$
is rational.)
Thus, to prove the inequality
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
we need to rule out $q=5$. This is currently open, and is also unknown even if we assume the Descartes-Frenicle-Sorli Conjecture that $k=1$.
answered Jan 29 at 11:51
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
$endgroup$
add a comment |
$begingroup$
It turns out that
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
implies
$$1 + frac{1}{q} = I(q) leq I(q^k) < sqrt[4]{2}$$
from which we obtain
$$q > bigg(sqrt[4]{2} - 1bigg)^{-1} approx 5.2852135$$
thereby giving
$$q geq 13$$
since $q$ is a prime satisfying $q equiv 1 pmod 4$. Thus, the implication
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} implies q geq 13$$
holds.
If the reverse inequality
$$I(q^k) > sqrt[3]{frac{2}{I(q^k)}}$$
holds, then we get
$$frac{q}{q - 1} > I(q^k) > sqrt[4]{2}$$
which implies that
$$1 < q < frac{sqrt[4]{2}}{sqrt[4]{2} - 1} approx 6.28521$$
from which we conclude that $q = 5$.
Since $q geq 5$ and $q$ is a prime satisfying $q equiv 1 pmod 4$, by the contrapositive of the last implication, we get the implication
$$q geq 13 implies I(q^k) < sqrt[3]{frac{2}{I(q^k)}}.$$
We therefore have the biconditional
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} iff q geq 13.$$
(Note that we cannot have
$$I(q^k) = sqrt[3]{frac{2}{I(q^k)}}$$
as equality implies that $I(q^k) = sqrt[4]{2}$, contradicting the fact that
$$I(q^k) = frac{q^{k+1} - 1}{q^k (q - 1)}$$
is rational.)
Thus, to prove the inequality
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
we need to rule out $q=5$. This is currently open, and is also unknown even if we assume the Descartes-Frenicle-Sorli Conjecture that $k=1$.
answered Jan 29 at 11:51
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
$endgroup$
add a comment |
$begingroup$
It turns out that
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
implies
$$1 + frac{1}{q} = I(q) leq I(q^k) < sqrt[4]{2}$$
from which we obtain
$$q > bigg(sqrt[4]{2} - 1bigg)^{-1} approx 5.2852135$$
thereby giving
$$q geq 13$$
since $q$ is a prime satisfying $q equiv 1 pmod 4$. Thus, the implication
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} implies q geq 13$$
holds.
If the reverse inequality
$$I(q^k) > sqrt[3]{frac{2}{I(q^k)}}$$
holds, then we get
$$frac{q}{q - 1} > I(q^k) > sqrt[4]{2}$$
which implies that
$$1 < q < frac{sqrt[4]{2}}{sqrt[4]{2} - 1} approx 6.28521$$
from which we conclude that $q = 5$.
Since $q geq 5$ and $q$ is a prime satisfying $q equiv 1 pmod 4$, by the contrapositive of the last implication, we get the implication
$$q geq 13 implies I(q^k) < sqrt[3]{frac{2}{I(q^k)}}.$$
We therefore have the biconditional
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} iff q geq 13.$$
(Note that we cannot have
$$I(q^k) = sqrt[3]{frac{2}{I(q^k)}}$$
as equality implies that $I(q^k) = sqrt[4]{2}$, contradicting the fact that
$$I(q^k) = frac{q^{k+1} - 1}{q^k (q - 1)}$$
is rational.)
Thus, to prove the inequality
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
we need to rule out $q=5$. This is currently open, and is also unknown even if we assume the Descartes-Frenicle-Sorli Conjecture that $k=1$.
answered Jan 29 at 11:51
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
$endgroup$
It turns out that
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
implies
$$1 + frac{1}{q} = I(q) leq I(q^k) < sqrt[4]{2}$$
from which we obtain
$$q > bigg(sqrt[4]{2} - 1bigg)^{-1} approx 5.2852135$$
thereby giving
$$q geq 13$$
since $q$ is a prime satisfying $q equiv 1 pmod 4$. Thus, the implication
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} implies q geq 13$$
holds.
If the reverse inequality
$$I(q^k) > sqrt[3]{frac{2}{I(q^k)}}$$
holds, then we get
$$frac{q}{q - 1} > I(q^k) > sqrt[4]{2}$$
which implies that
$$1 < q < frac{sqrt[4]{2}}{sqrt[4]{2} - 1} approx 6.28521$$
from which we conclude that $q = 5$.
Since $q geq 5$ and $q$ is a prime satisfying $q equiv 1 pmod 4$, by the contrapositive of the last implication, we get the implication
$$q geq 13 implies I(q^k) < sqrt[3]{frac{2}{I(q^k)}}.$$
We therefore have the biconditional
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}} iff q geq 13.$$
(Note that we cannot have
$$I(q^k) = sqrt[3]{frac{2}{I(q^k)}}$$
as equality implies that $I(q^k) = sqrt[4]{2}$, contradicting the fact that
$$I(q^k) = frac{q^{k+1} - 1}{q^k (q - 1)}$$
is rational.)
Thus, to prove the inequality
$$I(q^k) < sqrt[3]{frac{2}{I(q^k)}}$$
we need to rule out $q=5$. This is currently open, and is also unknown even if we assume the Descartes-Frenicle-Sorli Conjecture that $k=1$.
answered Jan 29 at 11:51
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
answered Jan 29 at 11:51
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
answered Jan 29 at 11:51
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
answered Jan 29 at 11:51
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,51741945
5,51741945
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091843%2fimproving-the-bound-for-sigmaqk-qk-where-qk-n2-is-an-odd-perfect-numb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
