Mixing
How many pairs $(x,y)$ such that $x+2=y$ have an $x$ or $y$ divisible by 5 and not 3?
$begingroup$
Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+2=y$ and $y$ is less than or equal to odd integer $n$ and $mathbb{O}$ is set of odd integers > 1.
Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.
Examples: $S_{9} ={(3,5),(5,7),(7,9)}$ and $f(9) = 3$.
$S_{13} = {(3,5), (5,7), (7,9), (9,11),(11,13)}$ and $f(13) = 5$
$S_{37} = {(3,5), (5,7), (7,9), (9,11), (11,13), (13,15), (15,17), (17,19), (19,21), (21,23), (23,25), (25,27), (27,29), (29,31),(31,33),(33,35),(35,37)}$
and $f(37) = 17$
It turns out that $f(n) = ((n-3)/2)$ for odd integers $n$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 5 and $x$ or $y$ is not divisible by 3, and $x neq 5$ and $yneq 5$.
Example: $g(37) = 2$ because there are only 2 pairs where $x$ or $y$ is divisible by 5 and not 3, and $x$ or $y$ is not equal to 5.
They are (23,25) and (35,37).
Note, the pairs (3,5) and (5,7) are not included since they contain 5, the pairs (13,15) and (15,17) are not included because 15 is divisible by 3, the pair (25,27) is not included because 27 is divisible by 3, and the pair (33,35) is not included because 33 is divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ ?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?
Edit:
I believe $g(n)$ approaches $(1/3)(2/5)f(n)$ as $n toinfty$.
Since $f(n)$ approaches $(n/2)$ as $n$ gets large, $g(n)$ approaches $n/15$ as $n toinfty$.
I need somebody to confirm this.
The table below compares the actual value for $g(n)$ and $n/15$ for a few values of $n$ and they look comparable.
$g(n)$ approaches $n/15$ as $n toinfty$">
elementary-number-theory
asked Jan 9 at 19:43
temp wattstemp watts
373
$endgroup$
add a comment |
$begingroup$
Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+2=y$ and $y$ is less than or equal to odd integer $n$ and $mathbb{O}$ is set of odd integers > 1.
Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.
Examples: $S_{9} ={(3,5),(5,7),(7,9)}$ and $f(9) = 3$.
$S_{13} = {(3,5), (5,7), (7,9), (9,11),(11,13)}$ and $f(13) = 5$
$S_{37} = {(3,5), (5,7), (7,9), (9,11), (11,13), (13,15), (15,17), (17,19), (19,21), (21,23), (23,25), (25,27), (27,29), (29,31),(31,33),(33,35),(35,37)}$
and $f(37) = 17$
It turns out that $f(n) = ((n-3)/2)$ for odd integers $n$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 5 and $x$ or $y$ is not divisible by 3, and $x neq 5$ and $yneq 5$.
Example: $g(37) = 2$ because there are only 2 pairs where $x$ or $y$ is divisible by 5 and not 3, and $x$ or $y$ is not equal to 5.
They are (23,25) and (35,37).
Note, the pairs (3,5) and (5,7) are not included since they contain 5, the pairs (13,15) and (15,17) are not included because 15 is divisible by 3, the pair (25,27) is not included because 27 is divisible by 3, and the pair (33,35) is not included because 33 is divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ ?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?
Edit:
I believe $g(n)$ approaches $(1/3)(2/5)f(n)$ as $n toinfty$.
Since $f(n)$ approaches $(n/2)$ as $n$ gets large, $g(n)$ approaches $n/15$ as $n toinfty$.
I need somebody to confirm this.
The table below compares the actual value for $g(n)$ and $n/15$ for a few values of $n$ and they look comparable.
$g(n)$ approaches $n/15$ as $n toinfty$">
elementary-number-theory
asked Jan 9 at 19:43
temp wattstemp watts
373
$endgroup$
add a comment |
$begingroup$
Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+2=y$ and $y$ is less than or equal to odd integer $n$ and $mathbb{O}$ is set of odd integers > 1.
Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.
Examples: $S_{9} ={(3,5),(5,7),(7,9)}$ and $f(9) = 3$.
$S_{13} = {(3,5), (5,7), (7,9), (9,11),(11,13)}$ and $f(13) = 5$
$S_{37} = {(3,5), (5,7), (7,9), (9,11), (11,13), (13,15), (15,17), (17,19), (19,21), (21,23), (23,25), (25,27), (27,29), (29,31),(31,33),(33,35),(35,37)}$
and $f(37) = 17$
It turns out that $f(n) = ((n-3)/2)$ for odd integers $n$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 5 and $x$ or $y$ is not divisible by 3, and $x neq 5$ and $yneq 5$.
Example: $g(37) = 2$ because there are only 2 pairs where $x$ or $y$ is divisible by 5 and not 3, and $x$ or $y$ is not equal to 5.
They are (23,25) and (35,37).
Note, the pairs (3,5) and (5,7) are not included since they contain 5, the pairs (13,15) and (15,17) are not included because 15 is divisible by 3, the pair (25,27) is not included because 27 is divisible by 3, and the pair (33,35) is not included because 33 is divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ ?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?
Edit:
I believe $g(n)$ approaches $(1/3)(2/5)f(n)$ as $n toinfty$.
Since $f(n)$ approaches $(n/2)$ as $n$ gets large, $g(n)$ approaches $n/15$ as $n toinfty$.
I need somebody to confirm this.
The table below compares the actual value for $g(n)$ and $n/15$ for a few values of $n$ and they look comparable.
$g(n)$ approaches $n/15$ as $n toinfty$">
elementary-number-theory
asked Jan 9 at 19:43
temp wattstemp watts
373
$endgroup$
Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+2=y$ and $y$ is less than or equal to odd integer $n$ and $mathbb{O}$ is set of odd integers > 1.
Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.
Examples: $S_{9} ={(3,5),(5,7),(7,9)}$ and $f(9) = 3$.
$S_{13} = {(3,5), (5,7), (7,9), (9,11),(11,13)}$ and $f(13) = 5$
$S_{37} = {(3,5), (5,7), (7,9), (9,11), (11,13), (13,15), (15,17), (17,19), (19,21), (21,23), (23,25), (25,27), (27,29), (29,31),(31,33),(33,35),(35,37)}$
and $f(37) = 17$
It turns out that $f(n) = ((n-3)/2)$ for odd integers $n$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 5 and $x$ or $y$ is not divisible by 3, and $x neq 5$ and $yneq 5$.
Example: $g(37) = 2$ because there are only 2 pairs where $x$ or $y$ is divisible by 5 and not 3, and $x$ or $y$ is not equal to 5.
They are (23,25) and (35,37).
Note, the pairs (3,5) and (5,7) are not included since they contain 5, the pairs (13,15) and (15,17) are not included because 15 is divisible by 3, the pair (25,27) is not included because 27 is divisible by 3, and the pair (33,35) is not included because 33 is divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ ?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?
Edit:
I believe $g(n)$ approaches $(1/3)(2/5)f(n)$ as $n toinfty$.
Since $f(n)$ approaches $(n/2)$ as $n$ gets large, $g(n)$ approaches $n/15$ as $n toinfty$.
I need somebody to confirm this.
The table below compares the actual value for $g(n)$ and $n/15$ for a few values of $n$ and they look comparable.
$g(n)$ approaches $n/15$ as $n toinfty$">
elementary-number-theory
elementary-number-theory
asked Jan 9 at 19:43
temp wattstemp watts
373
asked Jan 9 at 19:43
temp wattstemp watts
373
edited Jan 10 at 15:28
temp watts
asked Jan 9 at 19:43
temp wattstemp watts
373
asked Jan 9 at 19:43
temp wattstemp watts
373
asked Jan 9 at 19:43
temp wattstemp watts
373
373
add a comment |
add a comment |
1 Answer
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$begingroup$
Integers to be avoided are multiples of $15$ so you have the following sequence for $g(infty$): $${color{red}{(23,25),(25,27)},color{blue}{(33,35),(35,37)},color{green}{(53,55),(55,57)},cdots}$$ Suppose that $n$ is odd and is greater than $23$ for non-triviality. The cases $n<23$ can be brute-forced separately, which I will not do so here. By doing some counting, it is possible to obtain the following closed forms modulo $30$.
Note: Please try to derive some of them yourself before asking for further hints.
begin{array}{c|c}npmod{30}&1&3&5&7&9\hline g(n)&frac2{15}(n-1)-2&frac2{15}(n-3)-2&frac1{10}(n-5)&frac2{15}(n-7)&frac2{15}(n-9)end{array}
begin{array}{c|c}npmod{30}&11&13&15&17&19\hline g(n)&frac2{15}(n-11)&frac2{15}(n-13)&frac2{15}(n-15)&frac2{15}(n-17)&frac2{15}(n-19)end{array}
begin{array}{c|c}npmod{30}&21&23&25&27&29\hline g(n)&frac2{15}(n-21)&frac2{15}(n-23)&frac2{15}(n+5)-3&frac2{15}(n+3)-2&frac2{15}(n+1)-2end{array}
Therefore, since $f(n)=frac{n-3}2$ as you have calculated, for odd $n>23$, $$lim_{ntoinfty}frac{g(n)}{f(n)}=begin{cases}frac15,text{for},nequiv5pmod{30}\frac4{15},text{otherwise}end{cases}$$
answered Jan 9 at 20:19
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
$begingroup$
Thank you for your quick reply. Kudos for your excellent analysis in such a short time. Unfortunately I made a mistake in my question. Pairs like (25,27) I wish to ignore since 27 is divisible by 3. I apologize greatly for my mistake. You can call me an idiot if you want. Anyway, I fixed the question. If you can kindly solve the corrected question it would be greatly appreciated.
$endgroup$
– temp watts
Jan 9 at 21:11
$begingroup$
Feel free to modify the argument. It works very similarly! Please post your attempts in your main post as well so that I can see where you've made any mistakes :)
$endgroup$
– TheSimpliFire
Jan 9 at 21:38
add a comment |
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$begingroup$
Integers to be avoided are multiples of $15$ so you have the following sequence for $g(infty$): $${color{red}{(23,25),(25,27)},color{blue}{(33,35),(35,37)},color{green}{(53,55),(55,57)},cdots}$$ Suppose that $n$ is odd and is greater than $23$ for non-triviality. The cases $n<23$ can be brute-forced separately, which I will not do so here. By doing some counting, it is possible to obtain the following closed forms modulo $30$.
Note: Please try to derive some of them yourself before asking for further hints.
begin{array}{c|c}npmod{30}&1&3&5&7&9\hline g(n)&frac2{15}(n-1)-2&frac2{15}(n-3)-2&frac1{10}(n-5)&frac2{15}(n-7)&frac2{15}(n-9)end{array}
begin{array}{c|c}npmod{30}&11&13&15&17&19\hline g(n)&frac2{15}(n-11)&frac2{15}(n-13)&frac2{15}(n-15)&frac2{15}(n-17)&frac2{15}(n-19)end{array}
begin{array}{c|c}npmod{30}&21&23&25&27&29\hline g(n)&frac2{15}(n-21)&frac2{15}(n-23)&frac2{15}(n+5)-3&frac2{15}(n+3)-2&frac2{15}(n+1)-2end{array}
Therefore, since $f(n)=frac{n-3}2$ as you have calculated, for odd $n>23$, $$lim_{ntoinfty}frac{g(n)}{f(n)}=begin{cases}frac15,text{for},nequiv5pmod{30}\frac4{15},text{otherwise}end{cases}$$
answered Jan 9 at 20:19
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
$begingroup$
Thank you for your quick reply. Kudos for your excellent analysis in such a short time. Unfortunately I made a mistake in my question. Pairs like (25,27) I wish to ignore since 27 is divisible by 3. I apologize greatly for my mistake. You can call me an idiot if you want. Anyway, I fixed the question. If you can kindly solve the corrected question it would be greatly appreciated.
$endgroup$
– temp watts
Jan 9 at 21:11
$begingroup$
Feel free to modify the argument. It works very similarly! Please post your attempts in your main post as well so that I can see where you've made any mistakes :)
$endgroup$
– TheSimpliFire
Jan 9 at 21:38
add a comment |
$begingroup$
Integers to be avoided are multiples of $15$ so you have the following sequence for $g(infty$): $${color{red}{(23,25),(25,27)},color{blue}{(33,35),(35,37)},color{green}{(53,55),(55,57)},cdots}$$ Suppose that $n$ is odd and is greater than $23$ for non-triviality. The cases $n<23$ can be brute-forced separately, which I will not do so here. By doing some counting, it is possible to obtain the following closed forms modulo $30$.
Note: Please try to derive some of them yourself before asking for further hints.
begin{array}{c|c}npmod{30}&1&3&5&7&9\hline g(n)&frac2{15}(n-1)-2&frac2{15}(n-3)-2&frac1{10}(n-5)&frac2{15}(n-7)&frac2{15}(n-9)end{array}
begin{array}{c|c}npmod{30}&11&13&15&17&19\hline g(n)&frac2{15}(n-11)&frac2{15}(n-13)&frac2{15}(n-15)&frac2{15}(n-17)&frac2{15}(n-19)end{array}
begin{array}{c|c}npmod{30}&21&23&25&27&29\hline g(n)&frac2{15}(n-21)&frac2{15}(n-23)&frac2{15}(n+5)-3&frac2{15}(n+3)-2&frac2{15}(n+1)-2end{array}
Therefore, since $f(n)=frac{n-3}2$ as you have calculated, for odd $n>23$, $$lim_{ntoinfty}frac{g(n)}{f(n)}=begin{cases}frac15,text{for},nequiv5pmod{30}\frac4{15},text{otherwise}end{cases}$$
answered Jan 9 at 20:19
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
$begingroup$
Thank you for your quick reply. Kudos for your excellent analysis in such a short time. Unfortunately I made a mistake in my question. Pairs like (25,27) I wish to ignore since 27 is divisible by 3. I apologize greatly for my mistake. You can call me an idiot if you want. Anyway, I fixed the question. If you can kindly solve the corrected question it would be greatly appreciated.
$endgroup$
– temp watts
Jan 9 at 21:11
$begingroup$
Feel free to modify the argument. It works very similarly! Please post your attempts in your main post as well so that I can see where you've made any mistakes :)
$endgroup$
– TheSimpliFire
Jan 9 at 21:38
add a comment |
$begingroup$
Integers to be avoided are multiples of $15$ so you have the following sequence for $g(infty$): $${color{red}{(23,25),(25,27)},color{blue}{(33,35),(35,37)},color{green}{(53,55),(55,57)},cdots}$$ Suppose that $n$ is odd and is greater than $23$ for non-triviality. The cases $n<23$ can be brute-forced separately, which I will not do so here. By doing some counting, it is possible to obtain the following closed forms modulo $30$.
Note: Please try to derive some of them yourself before asking for further hints.
begin{array}{c|c}npmod{30}&1&3&5&7&9\hline g(n)&frac2{15}(n-1)-2&frac2{15}(n-3)-2&frac1{10}(n-5)&frac2{15}(n-7)&frac2{15}(n-9)end{array}
begin{array}{c|c}npmod{30}&11&13&15&17&19\hline g(n)&frac2{15}(n-11)&frac2{15}(n-13)&frac2{15}(n-15)&frac2{15}(n-17)&frac2{15}(n-19)end{array}
begin{array}{c|c}npmod{30}&21&23&25&27&29\hline g(n)&frac2{15}(n-21)&frac2{15}(n-23)&frac2{15}(n+5)-3&frac2{15}(n+3)-2&frac2{15}(n+1)-2end{array}
Therefore, since $f(n)=frac{n-3}2$ as you have calculated, for odd $n>23$, $$lim_{ntoinfty}frac{g(n)}{f(n)}=begin{cases}frac15,text{for},nequiv5pmod{30}\frac4{15},text{otherwise}end{cases}$$
answered Jan 9 at 20:19
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
Integers to be avoided are multiples of $15$ so you have the following sequence for $g(infty$): $${color{red}{(23,25),(25,27)},color{blue}{(33,35),(35,37)},color{green}{(53,55),(55,57)},cdots}$$ Suppose that $n$ is odd and is greater than $23$ for non-triviality. The cases $n<23$ can be brute-forced separately, which I will not do so here. By doing some counting, it is possible to obtain the following closed forms modulo $30$.
Note: Please try to derive some of them yourself before asking for further hints.
begin{array}{c|c}npmod{30}&1&3&5&7&9\hline g(n)&frac2{15}(n-1)-2&frac2{15}(n-3)-2&frac1{10}(n-5)&frac2{15}(n-7)&frac2{15}(n-9)end{array}
begin{array}{c|c}npmod{30}&11&13&15&17&19\hline g(n)&frac2{15}(n-11)&frac2{15}(n-13)&frac2{15}(n-15)&frac2{15}(n-17)&frac2{15}(n-19)end{array}
begin{array}{c|c}npmod{30}&21&23&25&27&29\hline g(n)&frac2{15}(n-21)&frac2{15}(n-23)&frac2{15}(n+5)-3&frac2{15}(n+3)-2&frac2{15}(n+1)-2end{array}
Therefore, since $f(n)=frac{n-3}2$ as you have calculated, for odd $n>23$, $$lim_{ntoinfty}frac{g(n)}{f(n)}=begin{cases}frac15,text{for},nequiv5pmod{30}\frac4{15},text{otherwise}end{cases}$$
answered Jan 9 at 20:19
TheSimpliFireTheSimpliFire
12.4k62460
edited Jan 9 at 20:25
answered Jan 9 at 20:19
TheSimpliFireTheSimpliFire
12.4k62460
answered Jan 9 at 20:19
TheSimpliFireTheSimpliFire
12.4k62460
answered Jan 9 at 20:19
TheSimpliFireTheSimpliFire
12.4k62460
12.4k62460
$begingroup$
Thank you for your quick reply. Kudos for your excellent analysis in such a short time. Unfortunately I made a mistake in my question. Pairs like (25,27) I wish to ignore since 27 is divisible by 3. I apologize greatly for my mistake. You can call me an idiot if you want. Anyway, I fixed the question. If you can kindly solve the corrected question it would be greatly appreciated.
$endgroup$
– temp watts
Jan 9 at 21:11
$begingroup$
Feel free to modify the argument. It works very similarly! Please post your attempts in your main post as well so that I can see where you've made any mistakes :)
$endgroup$
– TheSimpliFire
Jan 9 at 21:38
add a comment |
$begingroup$
Thank you for your quick reply. Kudos for your excellent analysis in such a short time. Unfortunately I made a mistake in my question. Pairs like (25,27) I wish to ignore since 27 is divisible by 3. I apologize greatly for my mistake. You can call me an idiot if you want. Anyway, I fixed the question. If you can kindly solve the corrected question it would be greatly appreciated.
$endgroup$
– temp watts
Jan 9 at 21:11
$begingroup$
Feel free to modify the argument. It works very similarly! Please post your attempts in your main post as well so that I can see where you've made any mistakes :)
$endgroup$
– TheSimpliFire
Jan 9 at 21:38
$begingroup$
Thank you for your quick reply. Kudos for your excellent analysis in such a short time. Unfortunately I made a mistake in my question. Pairs like (25,27) I wish to ignore since 27 is divisible by 3. I apologize greatly for my mistake. You can call me an idiot if you want. Anyway, I fixed the question. If you can kindly solve the corrected question it would be greatly appreciated.
$endgroup$
– temp watts
Jan 9 at 21:11
$begingroup$
Thank you for your quick reply. Kudos for your excellent analysis in such a short time. Unfortunately I made a mistake in my question. Pairs like (25,27) I wish to ignore since 27 is divisible by 3. I apologize greatly for my mistake. You can call me an idiot if you want. Anyway, I fixed the question. If you can kindly solve the corrected question it would be greatly appreciated.
$endgroup$
– temp watts
Jan 9 at 21:11
$begingroup$
Feel free to modify the argument. It works very similarly! Please post your attempts in your main post as well so that I can see where you've made any mistakes :)
$endgroup$
– TheSimpliFire
Jan 9 at 21:38
$begingroup$
Feel free to modify the argument. It works very similarly! Please post your attempts in your main post as well so that I can see where you've made any mistakes :)
$endgroup$
– TheSimpliFire
Jan 9 at 21:38
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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Required, but never shown
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Post as a guest
Required, but never shown
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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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
