Bounds on primes with Gilbreath's Conjecture
$begingroup$
Gilbreath's conjecture is a famous conjecture on prime numbers that states that when we take the prime number sequence and perform iterative subtraction, we repeatedly get the number $1$. A link to Wikipedia is provided below:
https://en.wikipedia.org/wiki/Gilbreath%27s_conjecture
Is it possible to get any bound on primes conditionally on this conjecture, that is strong? It seems that the bounds look alike the statement of Bertrand's Postulate, if we substitute the values of powers of $2$, instead of primes, where the statement of the conjecture holds trivially. Thus, it might be likely that some good bound can be obtained.
This bound looks quite strong as when we add the coefficients of the primes, we get $0$. What is the bound we obtain on the $n$th prime? Can this bound be converted to closed form? And if so, does the bound actually hold? (i.e. if the bound is $p_{n+1} < 2p_n$, the bound holds unconditionally.)
number-theory prime-numbers
asked Dec 30 '18 at 16:36
HaranHaran
1,149424
$endgroup$
add a comment |
$begingroup$
Gilbreath's conjecture is a famous conjecture on prime numbers that states that when we take the prime number sequence and perform iterative subtraction, we repeatedly get the number $1$. A link to Wikipedia is provided below:
https://en.wikipedia.org/wiki/Gilbreath%27s_conjecture
Is it possible to get any bound on primes conditionally on this conjecture, that is strong? It seems that the bounds look alike the statement of Bertrand's Postulate, if we substitute the values of powers of $2$, instead of primes, where the statement of the conjecture holds trivially. Thus, it might be likely that some good bound can be obtained.
This bound looks quite strong as when we add the coefficients of the primes, we get $0$. What is the bound we obtain on the $n$th prime? Can this bound be converted to closed form? And if so, does the bound actually hold? (i.e. if the bound is $p_{n+1} < 2p_n$, the bound holds unconditionally.)
number-theory prime-numbers
asked Dec 30 '18 at 16:36
HaranHaran
1,149424
$endgroup$
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
add a comment |
$begingroup$
Gilbreath's conjecture is a famous conjecture on prime numbers that states that when we take the prime number sequence and perform iterative subtraction, we repeatedly get the number $1$. A link to Wikipedia is provided below:
https://en.wikipedia.org/wiki/Gilbreath%27s_conjecture
Is it possible to get any bound on primes conditionally on this conjecture, that is strong? It seems that the bounds look alike the statement of Bertrand's Postulate, if we substitute the values of powers of $2$, instead of primes, where the statement of the conjecture holds trivially. Thus, it might be likely that some good bound can be obtained.
This bound looks quite strong as when we add the coefficients of the primes, we get $0$. What is the bound we obtain on the $n$th prime? Can this bound be converted to closed form? And if so, does the bound actually hold? (i.e. if the bound is $p_{n+1} < 2p_n$, the bound holds unconditionally.)
number-theory prime-numbers
asked Dec 30 '18 at 16:36
HaranHaran
1,149424
$endgroup$
Gilbreath's conjecture is a famous conjecture on prime numbers that states that when we take the prime number sequence and perform iterative subtraction, we repeatedly get the number $1$. A link to Wikipedia is provided below:
https://en.wikipedia.org/wiki/Gilbreath%27s_conjecture
Is it possible to get any bound on primes conditionally on this conjecture, that is strong? It seems that the bounds look alike the statement of Bertrand's Postulate, if we substitute the values of powers of $2$, instead of primes, where the statement of the conjecture holds trivially. Thus, it might be likely that some good bound can be obtained.
This bound looks quite strong as when we add the coefficients of the primes, we get $0$. What is the bound we obtain on the $n$th prime? Can this bound be converted to closed form? And if so, does the bound actually hold? (i.e. if the bound is $p_{n+1} < 2p_n$, the bound holds unconditionally.)
number-theory prime-numbers
number-theory prime-numbers
asked Dec 30 '18 at 16:36
HaranHaran
1,149424
asked Dec 30 '18 at 16:36
HaranHaran
1,149424
asked Dec 30 '18 at 16:36
HaranHaran
1,149424
asked Dec 30 '18 at 16:36
HaranHaran
1,149424
asked Dec 30 '18 at 16:36
HaranHaran
1,149424
1,149424
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
add a comment |
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
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%2f3056991%2fbounds-on-primes-with-gilbreaths-conjecture%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.
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%2f3056991%2fbounds-on-primes-with-gilbreaths-conjecture%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
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32