Is there any other number that has similar properties as $21$?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
It's my observation.
Let
$$n=p_1×p_2×p_3×dots×p_r$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$f(n)=1+2+dots+n$$
And
$$g(n)=p_1+p_2+dots+p_r$$
If we put $n=21$ then
$$g(f(21))=g(231)=21.$$
I checked it upto $n=10000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that other such numbers do not exist?
number-theory elementary-number-theory prime-factorization
edited Mar 1 at 14:12
Asaf Karagila♦
307k33439771
asked Mar 1 at 7:15
Pruthviraj HajariPruthviraj Hajari
19517
$endgroup$
|
show 1 more comment
$begingroup$
It's my observation.
Let
$$n=p_1×p_2×p_3×dots×p_r$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$f(n)=1+2+dots+n$$
And
$$g(n)=p_1+p_2+dots+p_r$$
If we put $n=21$ then
$$g(f(21))=g(231)=21.$$
I checked it upto $n=10000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that other such numbers do not exist?
number-theory elementary-number-theory prime-factorization
edited Mar 1 at 14:12
Asaf Karagila♦
307k33439771
asked Mar 1 at 7:15
Pruthviraj HajariPruthviraj Hajari
19517
$endgroup$
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
Mar 1 at 7:22
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
Mar 1 at 7:35
2
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
Mar 1 at 7:47
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
Mar 1 at 13:37
1
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
Mar 1 at 14:01
|
show 1 more comment
$begingroup$
It's my observation.
Let
$$n=p_1×p_2×p_3×dots×p_r$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$f(n)=1+2+dots+n$$
And
$$g(n)=p_1+p_2+dots+p_r$$
If we put $n=21$ then
$$g(f(21))=g(231)=21.$$
I checked it upto $n=10000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that other such numbers do not exist?
number-theory elementary-number-theory prime-factorization
edited Mar 1 at 14:12
Asaf Karagila♦
307k33439771
asked Mar 1 at 7:15
Pruthviraj HajariPruthviraj Hajari
19517
$endgroup$
It's my observation.
Let
$$n=p_1×p_2×p_3×dots×p_r$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$f(n)=1+2+dots+n$$
And
$$g(n)=p_1+p_2+dots+p_r$$
If we put $n=21$ then
$$g(f(21))=g(231)=21.$$
I checked it upto $n=10000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that other such numbers do not exist?
number-theory elementary-number-theory prime-factorization
number-theory elementary-number-theory prime-factorization
edited Mar 1 at 14:12
Asaf Karagila♦
307k33439771
asked Mar 1 at 7:15
Pruthviraj HajariPruthviraj Hajari
19517
edited Mar 1 at 14:12
Asaf Karagila♦
307k33439771
asked Mar 1 at 7:15
Pruthviraj HajariPruthviraj Hajari
19517
edited Mar 1 at 14:12
Asaf Karagila♦
307k33439771
edited Mar 1 at 14:12
Asaf Karagila♦
307k33439771
edited Mar 1 at 14:12
Asaf Karagila♦
307k33439771
307k33439771
asked Mar 1 at 7:15
Pruthviraj HajariPruthviraj Hajari
19517
asked Mar 1 at 7:15
Pruthviraj HajariPruthviraj Hajari
19517
asked Mar 1 at 7:15
Pruthviraj HajariPruthviraj Hajari
19517
19517
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
Mar 1 at 7:22
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
Mar 1 at 7:35
2
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
Mar 1 at 7:47
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
Mar 1 at 13:37
1
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
Mar 1 at 14:01
|
show 1 more comment
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
Mar 1 at 7:22
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
Mar 1 at 7:35
2
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
Mar 1 at 7:47
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
Mar 1 at 13:37
1
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
Mar 1 at 14:01
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
Mar 1 at 7:22
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
Mar 1 at 7:22
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
Mar 1 at 7:35
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
Mar 1 at 7:35
2
2
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
Mar 1 at 7:47
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
Mar 1 at 7:47
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
Mar 1 at 13:37
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
Mar 1 at 13:37
1
1
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
Mar 1 at 14:01
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
Mar 1 at 14:01
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
This is a very interesting question…
$newcommandsopfroperatornamesopfr$
$f(n)=fracn(n+1)2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(fracn(n+1)2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+fracn+13$ and thus
$$frac n2le3+fracn+13$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(fracn+12right)+sopfr(n)=n$, where $sopfrleft(fracn+12right)lefracn+12$ and so $sopfr(n)gefracn-12$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$fracn-12le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered Mar 1 at 8:13
Parcly TaxelParcly Taxel
44.7k1376110
$endgroup$
1
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatornamesopfr(n)=p+operatornamesopfr(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatornamesopfr(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
Mar 1 at 15:47
add a comment |
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%2f3131137%2fis-there-any-other-number-that-has-similar-properties-as-21%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
$begingroup$
This is a very interesting question…
$newcommandsopfroperatornamesopfr$
$f(n)=fracn(n+1)2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(fracn(n+1)2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+fracn+13$ and thus
$$frac n2le3+fracn+13$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(fracn+12right)+sopfr(n)=n$, where $sopfrleft(fracn+12right)lefracn+12$ and so $sopfr(n)gefracn-12$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$fracn-12le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered Mar 1 at 8:13
Parcly TaxelParcly Taxel
44.7k1376110
$endgroup$
1
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatornamesopfr(n)=p+operatornamesopfr(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatornamesopfr(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
Mar 1 at 15:47
add a comment |
$begingroup$
This is a very interesting question…
$newcommandsopfroperatornamesopfr$
$f(n)=fracn(n+1)2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(fracn(n+1)2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+fracn+13$ and thus
$$frac n2le3+fracn+13$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(fracn+12right)+sopfr(n)=n$, where $sopfrleft(fracn+12right)lefracn+12$ and so $sopfr(n)gefracn-12$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$fracn-12le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered Mar 1 at 8:13
Parcly TaxelParcly Taxel
44.7k1376110
$endgroup$
1
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatornamesopfr(n)=p+operatornamesopfr(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatornamesopfr(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
Mar 1 at 15:47
add a comment |
$begingroup$
This is a very interesting question…
$newcommandsopfroperatornamesopfr$
$f(n)=fracn(n+1)2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(fracn(n+1)2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+fracn+13$ and thus
$$frac n2le3+fracn+13$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(fracn+12right)+sopfr(n)=n$, where $sopfrleft(fracn+12right)lefracn+12$ and so $sopfr(n)gefracn-12$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$fracn-12le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered Mar 1 at 8:13
Parcly TaxelParcly Taxel
44.7k1376110
$endgroup$
This is a very interesting question…
$newcommandsopfroperatornamesopfr$
$f(n)=fracn(n+1)2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(fracn(n+1)2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+fracn+13$ and thus
$$frac n2le3+fracn+13$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(fracn+12right)+sopfr(n)=n$, where $sopfrleft(fracn+12right)lefracn+12$ and so $sopfr(n)gefracn-12$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$fracn-12le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered Mar 1 at 8:13
Parcly TaxelParcly Taxel
44.7k1376110
edited Mar 1 at 8:48
answered Mar 1 at 8:13
Parcly TaxelParcly Taxel
44.7k1376110
answered Mar 1 at 8:13
Parcly TaxelParcly Taxel
44.7k1376110
answered Mar 1 at 8:13
Parcly TaxelParcly Taxel
44.7k1376110
44.7k1376110
1
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatornamesopfr(n)=p+operatornamesopfr(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatornamesopfr(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
Mar 1 at 15:47
add a comment |
1
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatornamesopfr(n)=p+operatornamesopfr(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatornamesopfr(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
Mar 1 at 15:47
1
1
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatornamesopfr(n)=p+operatornamesopfr(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatornamesopfr(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
Mar 1 at 15:47
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatornamesopfr(n)=p+operatornamesopfr(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatornamesopfr(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
Mar 1 at 15:47
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%2f3131137%2fis-there-any-other-number-that-has-similar-properties-as-21%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
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
Mar 1 at 7:22
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
Mar 1 at 7:35
2
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
Mar 1 at 7:47
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
Mar 1 at 13:37
1
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
Mar 1 at 14:01