Does there exist a formula to calculate $2.357137939171ldots$?Prime powers, patterns similar to $lbrace 0,1,0,2,0,1,0,3ldots rbrace$ and formulas for $sigma_k(n)$Does this polynomial exist?Does there exist a polynomial over integers possesing certain property?Does there exist an explicit formula for the coefficient of $x^k$ in the square of a polynomials?does there exist a prime such that…Can we capture the definition to be a Mersenne prime in an identity only involving the arithmetic function $S(n)=sum_k=1^ntextnmod k$?Does there exist an algebraic solvability algorithm?Does there exist such a positive integer?last digit of four consecutive primesPalindromic Numbers - Pattern “inside” Prime Numbers?
Why didn’t Eve recognize the little cockroach as a living organism?
Origin of pigs as a species
Why can't the Brexit deadlock in the UK parliament be solved with a plurality vote?
SOQL query causes internal Salesforce error
Why does the Persian emissary display a string of crowned skulls?
Overlapping circles covering polygon
Can I run 125kHz RF circuit on a breadboard?
Do I have to know the General Relativity theory to understand the concept of inertial frame?
Why do Radio Buttons not fill the entire outer circle?
How do I fix the group tension caused by my character stealing and possibly killing without provocation?
Telemetry for feature health
What should be the ideal length of sentences in a blog post for ease of reading?
Language involving irrational number is not a CFL
Isometric embedding of a genus g surface
Echo with obfuscation
How to leave product feedback on macOS?
Why would five hundred and five be same as one?
Make a Bowl of Alphabet Soup
Can I cause damage to electrical appliances by unplugging them when they are turned on?
Do you waste sorcery points if you try to apply metamagic to a spell from a scroll but fail to cast it?
How do I tell my boss that I'm quitting in 15 days (a colleague left this week)
Would a primitive species be able to learn English from reading books alone?
Do I have to take mana from my deck or hand when tapping a dual land?
How to preserve electronics (computers, iPads and phones) for hundreds of years
Does there exist a formula to calculate $2.357137939171ldots$?
Prime powers, patterns similar to $lbrace 0,1,0,2,0,1,0,3ldots rbrace$ and formulas for $sigma_k(n)$Does this polynomial exist?Does there exist a polynomial over integers possesing certain property?Does there exist an explicit formula for the coefficient of $x^k$ in the square of a polynomials?does there exist a prime such that…Can we capture the definition to be a Mersenne prime in an identity only involving the arithmetic function $S(n)=sum_k=1^ntextnmod k$?Does there exist an algebraic solvability algorithm?Does there exist such a positive integer?last digit of four consecutive primesPalindromic Numbers - Pattern “inside” Prime Numbers?
$begingroup$
So I was messing with polynomials and I encountered the following equation: $$26214x^3 - 27761x^2 - 71019x - 21667 = 0.$$ Solving for $x$ using the cubic formula, I got three solutions (as expected, pursuant to the FTOA, namely, the Fundamental Theorem of Algebra).
Let's call the equation $p(x)$ and I will denote by $p(x)_n$ the $n^textth$ root of $p(x)$.
$p(x)_1,2<0$ but $p(x)_3>0$. In fact, $$p(x)_3 = 2.3571379391713739171440ldots$$ Notice that we begin with the first four primes $2,3,5,7$ and then we go to $1,3,7,9$. Also notice that the next four primes after $7$ are $11,13,17,19$. That's when I realised that $p(x)_3$ has its decimal places being the last digit of primes, apart from $4,4,0$.
Question:
Let $d_n$ be the last digit of the $n^textth$ prime, then is the decimal $2.d_2d_3d_4ldots$ transcendental? Does it have a formula? Can a formula be constructed?
Thank you in advance.
sequences-and-series polynomials prime-numbers transcendence-theory
edited Mar 14 at 5:47
Martin Sleziak
44.9k10121274
asked May 10 '18 at 2:57
user477343user477343
3,59831243
$endgroup$
add a comment |
$begingroup$
So I was messing with polynomials and I encountered the following equation: $$26214x^3 - 27761x^2 - 71019x - 21667 = 0.$$ Solving for $x$ using the cubic formula, I got three solutions (as expected, pursuant to the FTOA, namely, the Fundamental Theorem of Algebra).
Let's call the equation $p(x)$ and I will denote by $p(x)_n$ the $n^textth$ root of $p(x)$.
$p(x)_1,2<0$ but $p(x)_3>0$. In fact, $$p(x)_3 = 2.3571379391713739171440ldots$$ Notice that we begin with the first four primes $2,3,5,7$ and then we go to $1,3,7,9$. Also notice that the next four primes after $7$ are $11,13,17,19$. That's when I realised that $p(x)_3$ has its decimal places being the last digit of primes, apart from $4,4,0$.
Question:
Let $d_n$ be the last digit of the $n^textth$ prime, then is the decimal $2.d_2d_3d_4ldots$ transcendental? Does it have a formula? Can a formula be constructed?
Thank you in advance.
sequences-and-series polynomials prime-numbers transcendence-theory
edited Mar 14 at 5:47
Martin Sleziak
44.9k10121274
asked May 10 '18 at 2:57
user477343user477343
3,59831243
$endgroup$
$begingroup$
@Tyler6 hahah thank you for he edit. Me thinking about prime numbers has confused myself :)
$endgroup$
– user477343
May 10 '18 at 3:04
1
$begingroup$
No worries, we’ve all been there ahah
$endgroup$
– Tyler6
May 10 '18 at 3:05
$begingroup$
Here's the continued fraction for $0.23571379dots$: oeis.org/A071775
$endgroup$
– Gerry Myerson
May 20 '18 at 8:54
$begingroup$
@GerryMyerson thanks for that, but both decimals are not quite the same :)
$endgroup$
– user477343
May 20 '18 at 8:56
$begingroup$
I don't know what you mean. We're talking about the decimal made from the last digit of the primes, right? That's what the OEIS reference is about, too.
$endgroup$
– Gerry Myerson
May 20 '18 at 9:05
add a comment |
$begingroup$
So I was messing with polynomials and I encountered the following equation: $$26214x^3 - 27761x^2 - 71019x - 21667 = 0.$$ Solving for $x$ using the cubic formula, I got three solutions (as expected, pursuant to the FTOA, namely, the Fundamental Theorem of Algebra).
Let's call the equation $p(x)$ and I will denote by $p(x)_n$ the $n^textth$ root of $p(x)$.
$p(x)_1,2<0$ but $p(x)_3>0$. In fact, $$p(x)_3 = 2.3571379391713739171440ldots$$ Notice that we begin with the first four primes $2,3,5,7$ and then we go to $1,3,7,9$. Also notice that the next four primes after $7$ are $11,13,17,19$. That's when I realised that $p(x)_3$ has its decimal places being the last digit of primes, apart from $4,4,0$.
Question:
Let $d_n$ be the last digit of the $n^textth$ prime, then is the decimal $2.d_2d_3d_4ldots$ transcendental? Does it have a formula? Can a formula be constructed?
Thank you in advance.
sequences-and-series polynomials prime-numbers transcendence-theory
edited Mar 14 at 5:47
Martin Sleziak
44.9k10121274
asked May 10 '18 at 2:57
user477343user477343
3,59831243
$endgroup$
So I was messing with polynomials and I encountered the following equation: $$26214x^3 - 27761x^2 - 71019x - 21667 = 0.$$ Solving for $x$ using the cubic formula, I got three solutions (as expected, pursuant to the FTOA, namely, the Fundamental Theorem of Algebra).
Let's call the equation $p(x)$ and I will denote by $p(x)_n$ the $n^textth$ root of $p(x)$.
$p(x)_1,2<0$ but $p(x)_3>0$. In fact, $$p(x)_3 = 2.3571379391713739171440ldots$$ Notice that we begin with the first four primes $2,3,5,7$ and then we go to $1,3,7,9$. Also notice that the next four primes after $7$ are $11,13,17,19$. That's when I realised that $p(x)_3$ has its decimal places being the last digit of primes, apart from $4,4,0$.
Question:
Let $d_n$ be the last digit of the $n^textth$ prime, then is the decimal $2.d_2d_3d_4ldots$ transcendental? Does it have a formula? Can a formula be constructed?
Thank you in advance.
sequences-and-series polynomials prime-numbers transcendence-theory
sequences-and-series polynomials prime-numbers transcendence-theory
edited Mar 14 at 5:47
Martin Sleziak
44.9k10121274
asked May 10 '18 at 2:57
user477343user477343
3,59831243
edited Mar 14 at 5:47
Martin Sleziak
44.9k10121274
asked May 10 '18 at 2:57
user477343user477343
3,59831243
edited Mar 14 at 5:47
Martin Sleziak
44.9k10121274
edited Mar 14 at 5:47
Martin Sleziak
44.9k10121274
edited Mar 14 at 5:47
Martin Sleziak
44.9k10121274
44.9k10121274
asked May 10 '18 at 2:57
user477343user477343
3,59831243
asked May 10 '18 at 2:57
user477343user477343
3,59831243
asked May 10 '18 at 2:57
user477343user477343
3,59831243
3,59831243
$begingroup$
@Tyler6 hahah thank you for he edit. Me thinking about prime numbers has confused myself :)
$endgroup$
– user477343
May 10 '18 at 3:04
1
$begingroup$
No worries, we’ve all been there ahah
$endgroup$
– Tyler6
May 10 '18 at 3:05
$begingroup$
Here's the continued fraction for $0.23571379dots$: oeis.org/A071775
$endgroup$
– Gerry Myerson
May 20 '18 at 8:54
$begingroup$
@GerryMyerson thanks for that, but both decimals are not quite the same :)
$endgroup$
– user477343
May 20 '18 at 8:56
$begingroup$
I don't know what you mean. We're talking about the decimal made from the last digit of the primes, right? That's what the OEIS reference is about, too.
$endgroup$
– Gerry Myerson
May 20 '18 at 9:05
add a comment |
$begingroup$
@Tyler6 hahah thank you for he edit. Me thinking about prime numbers has confused myself :)
$endgroup$
– user477343
May 10 '18 at 3:04
1
$begingroup$
No worries, we’ve all been there ahah
$endgroup$
– Tyler6
May 10 '18 at 3:05
$begingroup$
Here's the continued fraction for $0.23571379dots$: oeis.org/A071775
$endgroup$
– Gerry Myerson
May 20 '18 at 8:54
$begingroup$
@GerryMyerson thanks for that, but both decimals are not quite the same :)
$endgroup$
– user477343
May 20 '18 at 8:56
$begingroup$
I don't know what you mean. We're talking about the decimal made from the last digit of the primes, right? That's what the OEIS reference is about, too.
$endgroup$
– Gerry Myerson
May 20 '18 at 9:05
$begingroup$
@Tyler6 hahah thank you for he edit. Me thinking about prime numbers has confused myself :)
$endgroup$
– user477343
May 10 '18 at 3:04
$begingroup$
@Tyler6 hahah thank you for he edit. Me thinking about prime numbers has confused myself :)
$endgroup$
– user477343
May 10 '18 at 3:04
1
1
$begingroup$
No worries, we’ve all been there ahah
$endgroup$
– Tyler6
May 10 '18 at 3:05
$begingroup$
No worries, we’ve all been there ahah
$endgroup$
– Tyler6
May 10 '18 at 3:05
$begingroup$
Here's the continued fraction for $0.23571379dots$: oeis.org/A071775
$endgroup$
– Gerry Myerson
May 20 '18 at 8:54
$begingroup$
Here's the continued fraction for $0.23571379dots$: oeis.org/A071775
$endgroup$
– Gerry Myerson
May 20 '18 at 8:54
$begingroup$
@GerryMyerson thanks for that, but both decimals are not quite the same :)
$endgroup$
– user477343
May 20 '18 at 8:56
$begingroup$
@GerryMyerson thanks for that, but both decimals are not quite the same :)
$endgroup$
– user477343
May 20 '18 at 8:56
$begingroup$
I don't know what you mean. We're talking about the decimal made from the last digit of the primes, right? That's what the OEIS reference is about, too.
$endgroup$
– Gerry Myerson
May 20 '18 at 9:05
$begingroup$
I don't know what you mean. We're talking about the decimal made from the last digit of the primes, right? That's what the OEIS reference is about, too.
$endgroup$
– Gerry Myerson
May 20 '18 at 9:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Transcendence beats me, but I have a proof for irrationality:
To begin, notice that other than the primes $2$ or $5$, no prime can end in any of $0$, $2$, $4$, $5$, $6$ or $8$, since this would imply divisibility by $2$ or $5$. Therefore we see that for all primes other than those two, they must end with either $1$, $3$, $7$, or $9$.
Shiu proved (https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/S0024610799007863) that if $a$ and $q$ are coprime integers, there exist arbitrary long strings of prime congruent to $a bmod q$. For our purpose, this implies that in our number, $2.a_1a_2...$, there exist arbitrary long strings of $1$s, $3$s, $7$s, and $9$s. This is sufficient to prove it cannot be rational.
answered May 10 '18 at 3:39
Tyler6Tyler6
718414
$endgroup$
1
$begingroup$
I appreciate your efforts. I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places. And, given some formulas by Hardy and Littlewood, we know that there are not infinitely many primes continuing with a certain digit from in a row, so the decimal would not be terminating. Nonetheless, $(+1)$ :)
$endgroup$
– user477343
May 10 '18 at 3:44
1
$begingroup$
@NilotpalKantiSinha the Copeland Erdös Constant is the concatenation of all primes, this is the concatenation of only their last digits
$endgroup$
– Tyler6
May 10 '18 at 6:29
1
$begingroup$
"I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places." By that logic, $1/3=.333dots$ is irrational, since it has infinitely many decimal places.
$endgroup$
– Gerry Myerson
May 20 '18 at 8:49
1
$begingroup$
@user, OK. In the first place, Hardy & Littlewood didn't prove, or even "essentially" prove, anything of the kind: they made a conjecture, a conjecture which remains unproved to this day. And if the conjecture is correct, then a prime ending in a digit $d$ is followed by a prime ending in $d$ less than one-fourth of the time. But how much less than one-fourth? The difference goes to zero as you look at more and more primes. Continued next comment....
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
1
$begingroup$
....But never mind – that has nothing to do with periodicity. The primes could end in 1, 1, 3, 1, 7, 1, 9, 3, 3, 7, 3, 9, 7, 7, 9, 9 repeating that pattern forever, and each digit would be followed by each other digit exactly one-fourth of the time.
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
|
show 13 more comments
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%2f2774575%2fdoes-there-exist-a-formula-to-calculate-2-357137939171-ldots%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
$begingroup$
Transcendence beats me, but I have a proof for irrationality:
To begin, notice that other than the primes $2$ or $5$, no prime can end in any of $0$, $2$, $4$, $5$, $6$ or $8$, since this would imply divisibility by $2$ or $5$. Therefore we see that for all primes other than those two, they must end with either $1$, $3$, $7$, or $9$.
Shiu proved (https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/S0024610799007863) that if $a$ and $q$ are coprime integers, there exist arbitrary long strings of prime congruent to $a bmod q$. For our purpose, this implies that in our number, $2.a_1a_2...$, there exist arbitrary long strings of $1$s, $3$s, $7$s, and $9$s. This is sufficient to prove it cannot be rational.
answered May 10 '18 at 3:39
Tyler6Tyler6
718414
$endgroup$
1
$begingroup$
I appreciate your efforts. I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places. And, given some formulas by Hardy and Littlewood, we know that there are not infinitely many primes continuing with a certain digit from in a row, so the decimal would not be terminating. Nonetheless, $(+1)$ :)
$endgroup$
– user477343
May 10 '18 at 3:44
1
$begingroup$
@NilotpalKantiSinha the Copeland Erdös Constant is the concatenation of all primes, this is the concatenation of only their last digits
$endgroup$
– Tyler6
May 10 '18 at 6:29
1
$begingroup$
"I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places." By that logic, $1/3=.333dots$ is irrational, since it has infinitely many decimal places.
$endgroup$
– Gerry Myerson
May 20 '18 at 8:49
1
$begingroup$
@user, OK. In the first place, Hardy & Littlewood didn't prove, or even "essentially" prove, anything of the kind: they made a conjecture, a conjecture which remains unproved to this day. And if the conjecture is correct, then a prime ending in a digit $d$ is followed by a prime ending in $d$ less than one-fourth of the time. But how much less than one-fourth? The difference goes to zero as you look at more and more primes. Continued next comment....
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
1
$begingroup$
....But never mind – that has nothing to do with periodicity. The primes could end in 1, 1, 3, 1, 7, 1, 9, 3, 3, 7, 3, 9, 7, 7, 9, 9 repeating that pattern forever, and each digit would be followed by each other digit exactly one-fourth of the time.
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
|
show 13 more comments
$begingroup$
Transcendence beats me, but I have a proof for irrationality:
To begin, notice that other than the primes $2$ or $5$, no prime can end in any of $0$, $2$, $4$, $5$, $6$ or $8$, since this would imply divisibility by $2$ or $5$. Therefore we see that for all primes other than those two, they must end with either $1$, $3$, $7$, or $9$.
Shiu proved (https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/S0024610799007863) that if $a$ and $q$ are coprime integers, there exist arbitrary long strings of prime congruent to $a bmod q$. For our purpose, this implies that in our number, $2.a_1a_2...$, there exist arbitrary long strings of $1$s, $3$s, $7$s, and $9$s. This is sufficient to prove it cannot be rational.
answered May 10 '18 at 3:39
Tyler6Tyler6
718414
$endgroup$
1
$begingroup$
I appreciate your efforts. I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places. And, given some formulas by Hardy and Littlewood, we know that there are not infinitely many primes continuing with a certain digit from in a row, so the decimal would not be terminating. Nonetheless, $(+1)$ :)
$endgroup$
– user477343
May 10 '18 at 3:44
1
$begingroup$
@NilotpalKantiSinha the Copeland Erdös Constant is the concatenation of all primes, this is the concatenation of only their last digits
$endgroup$
– Tyler6
May 10 '18 at 6:29
1
$begingroup$
"I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places." By that logic, $1/3=.333dots$ is irrational, since it has infinitely many decimal places.
$endgroup$
– Gerry Myerson
May 20 '18 at 8:49
1
$begingroup$
@user, OK. In the first place, Hardy & Littlewood didn't prove, or even "essentially" prove, anything of the kind: they made a conjecture, a conjecture which remains unproved to this day. And if the conjecture is correct, then a prime ending in a digit $d$ is followed by a prime ending in $d$ less than one-fourth of the time. But how much less than one-fourth? The difference goes to zero as you look at more and more primes. Continued next comment....
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
1
$begingroup$
....But never mind – that has nothing to do with periodicity. The primes could end in 1, 1, 3, 1, 7, 1, 9, 3, 3, 7, 3, 9, 7, 7, 9, 9 repeating that pattern forever, and each digit would be followed by each other digit exactly one-fourth of the time.
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
|
show 13 more comments
$begingroup$
Transcendence beats me, but I have a proof for irrationality:
To begin, notice that other than the primes $2$ or $5$, no prime can end in any of $0$, $2$, $4$, $5$, $6$ or $8$, since this would imply divisibility by $2$ or $5$. Therefore we see that for all primes other than those two, they must end with either $1$, $3$, $7$, or $9$.
Shiu proved (https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/S0024610799007863) that if $a$ and $q$ are coprime integers, there exist arbitrary long strings of prime congruent to $a bmod q$. For our purpose, this implies that in our number, $2.a_1a_2...$, there exist arbitrary long strings of $1$s, $3$s, $7$s, and $9$s. This is sufficient to prove it cannot be rational.
answered May 10 '18 at 3:39
Tyler6Tyler6
718414
$endgroup$
Transcendence beats me, but I have a proof for irrationality:
To begin, notice that other than the primes $2$ or $5$, no prime can end in any of $0$, $2$, $4$, $5$, $6$ or $8$, since this would imply divisibility by $2$ or $5$. Therefore we see that for all primes other than those two, they must end with either $1$, $3$, $7$, or $9$.
Shiu proved (https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/S0024610799007863) that if $a$ and $q$ are coprime integers, there exist arbitrary long strings of prime congruent to $a bmod q$. For our purpose, this implies that in our number, $2.a_1a_2...$, there exist arbitrary long strings of $1$s, $3$s, $7$s, and $9$s. This is sufficient to prove it cannot be rational.
answered May 10 '18 at 3:39
Tyler6Tyler6
718414
answered May 10 '18 at 3:39
Tyler6Tyler6
718414
answered May 10 '18 at 3:39
Tyler6Tyler6
718414
answered May 10 '18 at 3:39
Tyler6Tyler6
718414
718414
1
$begingroup$
I appreciate your efforts. I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places. And, given some formulas by Hardy and Littlewood, we know that there are not infinitely many primes continuing with a certain digit from in a row, so the decimal would not be terminating. Nonetheless, $(+1)$ :)
$endgroup$
– user477343
May 10 '18 at 3:44
1
$begingroup$
@NilotpalKantiSinha the Copeland Erdös Constant is the concatenation of all primes, this is the concatenation of only their last digits
$endgroup$
– Tyler6
May 10 '18 at 6:29
1
$begingroup$
"I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places." By that logic, $1/3=.333dots$ is irrational, since it has infinitely many decimal places.
$endgroup$
– Gerry Myerson
May 20 '18 at 8:49
1
$begingroup$
@user, OK. In the first place, Hardy & Littlewood didn't prove, or even "essentially" prove, anything of the kind: they made a conjecture, a conjecture which remains unproved to this day. And if the conjecture is correct, then a prime ending in a digit $d$ is followed by a prime ending in $d$ less than one-fourth of the time. But how much less than one-fourth? The difference goes to zero as you look at more and more primes. Continued next comment....
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
1
$begingroup$
....But never mind – that has nothing to do with periodicity. The primes could end in 1, 1, 3, 1, 7, 1, 9, 3, 3, 7, 3, 9, 7, 7, 9, 9 repeating that pattern forever, and each digit would be followed by each other digit exactly one-fourth of the time.
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
|
show 13 more comments
1
$begingroup$
I appreciate your efforts. I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places. And, given some formulas by Hardy and Littlewood, we know that there are not infinitely many primes continuing with a certain digit from in a row, so the decimal would not be terminating. Nonetheless, $(+1)$ :)
$endgroup$
– user477343
May 10 '18 at 3:44
1
$begingroup$
@NilotpalKantiSinha the Copeland Erdös Constant is the concatenation of all primes, this is the concatenation of only their last digits
$endgroup$
– Tyler6
May 10 '18 at 6:29
1
$begingroup$
"I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places." By that logic, $1/3=.333dots$ is irrational, since it has infinitely many decimal places.
$endgroup$
– Gerry Myerson
May 20 '18 at 8:49
1
$begingroup$
@user, OK. In the first place, Hardy & Littlewood didn't prove, or even "essentially" prove, anything of the kind: they made a conjecture, a conjecture which remains unproved to this day. And if the conjecture is correct, then a prime ending in a digit $d$ is followed by a prime ending in $d$ less than one-fourth of the time. But how much less than one-fourth? The difference goes to zero as you look at more and more primes. Continued next comment....
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
1
$begingroup$
....But never mind – that has nothing to do with periodicity. The primes could end in 1, 1, 3, 1, 7, 1, 9, 3, 3, 7, 3, 9, 7, 7, 9, 9 repeating that pattern forever, and each digit would be followed by each other digit exactly one-fourth of the time.
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
1
1
$begingroup$
I appreciate your efforts. I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places. And, given some formulas by Hardy and Littlewood, we know that there are not infinitely many primes continuing with a certain digit from in a row, so the decimal would not be terminating. Nonetheless, $(+1)$ :)
$endgroup$
– user477343
May 10 '18 at 3:44
$begingroup$
I appreciate your efforts. I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places. And, given some formulas by Hardy and Littlewood, we know that there are not infinitely many primes continuing with a certain digit from in a row, so the decimal would not be terminating. Nonetheless, $(+1)$ :)
$endgroup$
– user477343
May 10 '18 at 3:44
1
1
$begingroup$
@NilotpalKantiSinha the Copeland Erdös Constant is the concatenation of all primes, this is the concatenation of only their last digits
$endgroup$
– Tyler6
May 10 '18 at 6:29
$begingroup$
@NilotpalKantiSinha the Copeland Erdös Constant is the concatenation of all primes, this is the concatenation of only their last digits
$endgroup$
– Tyler6
May 10 '18 at 6:29
1
1
$begingroup$
"I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places." By that logic, $1/3=.333dots$ is irrational, since it has infinitely many decimal places.
$endgroup$
– Gerry Myerson
May 20 '18 at 8:49
$begingroup$
"I found it rather trivial it was irrational since there are infinitely many primes, so infinitely many decimal places." By that logic, $1/3=.333dots$ is irrational, since it has infinitely many decimal places.
$endgroup$
– Gerry Myerson
May 20 '18 at 8:49
1
1
$begingroup$
@user, OK. In the first place, Hardy & Littlewood didn't prove, or even "essentially" prove, anything of the kind: they made a conjecture, a conjecture which remains unproved to this day. And if the conjecture is correct, then a prime ending in a digit $d$ is followed by a prime ending in $d$ less than one-fourth of the time. But how much less than one-fourth? The difference goes to zero as you look at more and more primes. Continued next comment....
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
$begingroup$
@user, OK. In the first place, Hardy & Littlewood didn't prove, or even "essentially" prove, anything of the kind: they made a conjecture, a conjecture which remains unproved to this day. And if the conjecture is correct, then a prime ending in a digit $d$ is followed by a prime ending in $d$ less than one-fourth of the time. But how much less than one-fourth? The difference goes to zero as you look at more and more primes. Continued next comment....
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
1
1
$begingroup$
....But never mind – that has nothing to do with periodicity. The primes could end in 1, 1, 3, 1, 7, 1, 9, 3, 3, 7, 3, 9, 7, 7, 9, 9 repeating that pattern forever, and each digit would be followed by each other digit exactly one-fourth of the time.
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
$begingroup$
....But never mind – that has nothing to do with periodicity. The primes could end in 1, 1, 3, 1, 7, 1, 9, 3, 3, 7, 3, 9, 7, 7, 9, 9 repeating that pattern forever, and each digit would be followed by each other digit exactly one-fourth of the time.
$endgroup$
– Gerry Myerson
Jun 15 '18 at 6:37
|
show 13 more comments
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%2f2774575%2fdoes-there-exist-a-formula-to-calculate-2-357137939171-ldots%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
$begingroup$
@Tyler6 hahah thank you for he edit. Me thinking about prime numbers has confused myself :)
$endgroup$
– user477343
May 10 '18 at 3:04
1
$begingroup$
No worries, we’ve all been there ahah
$endgroup$
– Tyler6
May 10 '18 at 3:05
$begingroup$
Here's the continued fraction for $0.23571379dots$: oeis.org/A071775
$endgroup$
– Gerry Myerson
May 20 '18 at 8:54
$begingroup$
@GerryMyerson thanks for that, but both decimals are not quite the same :)
$endgroup$
– user477343
May 20 '18 at 8:56
$begingroup$
I don't know what you mean. We're talking about the decimal made from the last digit of the primes, right? That's what the OEIS reference is about, too.
$endgroup$
– Gerry Myerson
May 20 '18 at 9:05