What is known about the prime factorization of numbers of the form $2^k+1$?
Let $n=2^k+1$, $k$ a positive integer. What is known about the prime factorization of these numbers? For example, consider $J(n)=Omega(n)-omega(n)$, where $Omega(n)$ is the sum of multiplicities in the prime factorization of $n$, and $omega(n)$ is the number of distinct prime factors of $n$. It would seem that $J(2^{3^i}+1)=i$ for $i$ a positive integer. Is $J(2^k+1)<J(2^{3^i})$ when $k<3^i$? Unfortunately, for these numbers we do not have Fermat's little theorem and the resulting arsenal of theorems and propositions that can be derived from it. Hence my question, is there anything we know about the prime factorization of these numbers that might help settle questions like the one above?
number-theory prime-factorization
asked Nov 28 '18 at 15:48
EGME
112
add a comment |
Let $n=2^k+1$, $k$ a positive integer. What is known about the prime factorization of these numbers? For example, consider $J(n)=Omega(n)-omega(n)$, where $Omega(n)$ is the sum of multiplicities in the prime factorization of $n$, and $omega(n)$ is the number of distinct prime factors of $n$. It would seem that $J(2^{3^i}+1)=i$ for $i$ a positive integer. Is $J(2^k+1)<J(2^{3^i})$ when $k<3^i$? Unfortunately, for these numbers we do not have Fermat's little theorem and the resulting arsenal of theorems and propositions that can be derived from it. Hence my question, is there anything we know about the prime factorization of these numbers that might help settle questions like the one above?
number-theory prime-factorization
asked Nov 28 '18 at 15:48
EGME
112
1
This seems like a spectacularly hard problem. In particular, we don't even know how many Fermat Primes there are.
– user3482749
Nov 28 '18 at 16:01
See factorization of Fermat numbers and its links.
– Dietrich Burde
Nov 28 '18 at 16:23
Unfortunately, most of the interest on that page seems focused on Fermat numbers, those of the form $2^{2^i}+1$, so that numbers for which $2^i<k<2^{i+1}$ are missed. But I will look carefully at the references.
– EGME
Nov 28 '18 at 16:39
I think this question is quite broad. For example image $k=q_1cdot q_2$, where both $q_1$ and $q_2$ are odd. Then from $$2^{q_1} equiv -1 pmod{2^{q_1}+1} Rightarrow 2^{k} equiv (-1)^{q_2} pmod{2^{q_1}+1} Rightarrow \ 2^{k} equiv -1 pmod{2^{q_1}+1}$$ In this case, $k=3^i$ and $q_1=3^j$, $q_2=3^t$ such that $i=j+t$.
– rtybase
Nov 28 '18 at 16:39
Googling the cunningham project page should be helpful. The site also explains the algebraic factors that such numbers have.
– Peter
Nov 30 '18 at 23:01
add a comment |
Let $n=2^k+1$, $k$ a positive integer. What is known about the prime factorization of these numbers? For example, consider $J(n)=Omega(n)-omega(n)$, where $Omega(n)$ is the sum of multiplicities in the prime factorization of $n$, and $omega(n)$ is the number of distinct prime factors of $n$. It would seem that $J(2^{3^i}+1)=i$ for $i$ a positive integer. Is $J(2^k+1)<J(2^{3^i})$ when $k<3^i$? Unfortunately, for these numbers we do not have Fermat's little theorem and the resulting arsenal of theorems and propositions that can be derived from it. Hence my question, is there anything we know about the prime factorization of these numbers that might help settle questions like the one above?
number-theory prime-factorization
asked Nov 28 '18 at 15:48
EGME
112
Let $n=2^k+1$, $k$ a positive integer. What is known about the prime factorization of these numbers? For example, consider $J(n)=Omega(n)-omega(n)$, where $Omega(n)$ is the sum of multiplicities in the prime factorization of $n$, and $omega(n)$ is the number of distinct prime factors of $n$. It would seem that $J(2^{3^i}+1)=i$ for $i$ a positive integer. Is $J(2^k+1)<J(2^{3^i})$ when $k<3^i$? Unfortunately, for these numbers we do not have Fermat's little theorem and the resulting arsenal of theorems and propositions that can be derived from it. Hence my question, is there anything we know about the prime factorization of these numbers that might help settle questions like the one above?
number-theory prime-factorization
number-theory prime-factorization
asked Nov 28 '18 at 15:48
EGME
112
asked Nov 28 '18 at 15:48
EGME
112
asked Nov 28 '18 at 15:48
EGME
112
asked Nov 28 '18 at 15:48
EGME
112
asked Nov 28 '18 at 15:48
EGME
112
112
1
This seems like a spectacularly hard problem. In particular, we don't even know how many Fermat Primes there are.
– user3482749
Nov 28 '18 at 16:01
See factorization of Fermat numbers and its links.
– Dietrich Burde
Nov 28 '18 at 16:23
Unfortunately, most of the interest on that page seems focused on Fermat numbers, those of the form $2^{2^i}+1$, so that numbers for which $2^i<k<2^{i+1}$ are missed. But I will look carefully at the references.
– EGME
Nov 28 '18 at 16:39
I think this question is quite broad. For example image $k=q_1cdot q_2$, where both $q_1$ and $q_2$ are odd. Then from $$2^{q_1} equiv -1 pmod{2^{q_1}+1} Rightarrow 2^{k} equiv (-1)^{q_2} pmod{2^{q_1}+1} Rightarrow \ 2^{k} equiv -1 pmod{2^{q_1}+1}$$ In this case, $k=3^i$ and $q_1=3^j$, $q_2=3^t$ such that $i=j+t$.
– rtybase
Nov 28 '18 at 16:39
Googling the cunningham project page should be helpful. The site also explains the algebraic factors that such numbers have.
– Peter
Nov 30 '18 at 23:01
add a comment |
1
This seems like a spectacularly hard problem. In particular, we don't even know how many Fermat Primes there are.
– user3482749
Nov 28 '18 at 16:01
See factorization of Fermat numbers and its links.
– Dietrich Burde
Nov 28 '18 at 16:23
Unfortunately, most of the interest on that page seems focused on Fermat numbers, those of the form $2^{2^i}+1$, so that numbers for which $2^i<k<2^{i+1}$ are missed. But I will look carefully at the references.
– EGME
Nov 28 '18 at 16:39
I think this question is quite broad. For example image $k=q_1cdot q_2$, where both $q_1$ and $q_2$ are odd. Then from $$2^{q_1} equiv -1 pmod{2^{q_1}+1} Rightarrow 2^{k} equiv (-1)^{q_2} pmod{2^{q_1}+1} Rightarrow \ 2^{k} equiv -1 pmod{2^{q_1}+1}$$ In this case, $k=3^i$ and $q_1=3^j$, $q_2=3^t$ such that $i=j+t$.
– rtybase
Nov 28 '18 at 16:39
Googling the cunningham project page should be helpful. The site also explains the algebraic factors that such numbers have.
– Peter
Nov 30 '18 at 23:01
1
1
This seems like a spectacularly hard problem. In particular, we don't even know how many Fermat Primes there are.
– user3482749
Nov 28 '18 at 16:01
This seems like a spectacularly hard problem. In particular, we don't even know how many Fermat Primes there are.
– user3482749
Nov 28 '18 at 16:01
See factorization of Fermat numbers and its links.
– Dietrich Burde
Nov 28 '18 at 16:23
See factorization of Fermat numbers and its links.
– Dietrich Burde
Nov 28 '18 at 16:23
Unfortunately, most of the interest on that page seems focused on Fermat numbers, those of the form $2^{2^i}+1$, so that numbers for which $2^i<k<2^{i+1}$ are missed. But I will look carefully at the references.
– EGME
Nov 28 '18 at 16:39
Unfortunately, most of the interest on that page seems focused on Fermat numbers, those of the form $2^{2^i}+1$, so that numbers for which $2^i<k<2^{i+1}$ are missed. But I will look carefully at the references.
– EGME
Nov 28 '18 at 16:39
I think this question is quite broad. For example image $k=q_1cdot q_2$, where both $q_1$ and $q_2$ are odd. Then from $$2^{q_1} equiv -1 pmod{2^{q_1}+1} Rightarrow 2^{k} equiv (-1)^{q_2} pmod{2^{q_1}+1} Rightarrow \ 2^{k} equiv -1 pmod{2^{q_1}+1}$$ In this case, $k=3^i$ and $q_1=3^j$, $q_2=3^t$ such that $i=j+t$.
– rtybase
Nov 28 '18 at 16:39
I think this question is quite broad. For example image $k=q_1cdot q_2$, where both $q_1$ and $q_2$ are odd. Then from $$2^{q_1} equiv -1 pmod{2^{q_1}+1} Rightarrow 2^{k} equiv (-1)^{q_2} pmod{2^{q_1}+1} Rightarrow \ 2^{k} equiv -1 pmod{2^{q_1}+1}$$ In this case, $k=3^i$ and $q_1=3^j$, $q_2=3^t$ such that $i=j+t$.
– rtybase
Nov 28 '18 at 16:39
Googling the cunningham project page should be helpful. The site also explains the algebraic factors that such numbers have.
– Peter
Nov 30 '18 at 23:01
Googling the cunningham project page should be helpful. The site also explains the algebraic factors that such numbers have.
– Peter
Nov 30 '18 at 23:01
add a comment |
0
active
oldest
votes
Your Answer
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%2f3017301%2fwhat-is-known-about-the-prime-factorization-of-numbers-of-the-form-2k1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3017301%2fwhat-is-known-about-the-prime-factorization-of-numbers-of-the-form-2k1%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
1
This seems like a spectacularly hard problem. In particular, we don't even know how many Fermat Primes there are.
– user3482749
Nov 28 '18 at 16:01
See factorization of Fermat numbers and its links.
– Dietrich Burde
Nov 28 '18 at 16:23
Unfortunately, most of the interest on that page seems focused on Fermat numbers, those of the form $2^{2^i}+1$, so that numbers for which $2^i<k<2^{i+1}$ are missed. But I will look carefully at the references.
– EGME
Nov 28 '18 at 16:39
I think this question is quite broad. For example image $k=q_1cdot q_2$, where both $q_1$ and $q_2$ are odd. Then from $$2^{q_1} equiv -1 pmod{2^{q_1}+1} Rightarrow 2^{k} equiv (-1)^{q_2} pmod{2^{q_1}+1} Rightarrow \ 2^{k} equiv -1 pmod{2^{q_1}+1}$$ In this case, $k=3^i$ and $q_1=3^j$, $q_2=3^t$ such that $i=j+t$.
– rtybase
Nov 28 '18 at 16:39
Googling the cunningham project page should be helpful. The site also explains the algebraic factors that such numbers have.
– Peter
Nov 30 '18 at 23:01