How many pairs $(x,y)$ such that $x+y=n$ have an $x$ or $y$ divisible by 3 but $x$ and $y$ are not equal to...
$begingroup$
Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+y=n$ where $mathbb{O}$ 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: $S_{10} ={(3,7),(5,5),(7,3)}$ and $f(10) = 3$.
$S_{14} = {(3,11), (5,9), (7,7), (9,5),(11,3)}$ and $f(14) = 5$
$S_{32} = {(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)}$
and $f(32) = 14$
It turns out that $f(n) = ((n/2) - 2)$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x neq 3$ and $yneq 3$.
Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3.
They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).
There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?
My guess is that as $n to infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.
number-theory goldbachs-conjecture
asked Dec 13 '18 at 16:29
temp wattstemp watts
262
$endgroup$
add a comment |
$begingroup$
Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+y=n$ where $mathbb{O}$ 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: $S_{10} ={(3,7),(5,5),(7,3)}$ and $f(10) = 3$.
$S_{14} = {(3,11), (5,9), (7,7), (9,5),(11,3)}$ and $f(14) = 5$
$S_{32} = {(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)}$
and $f(32) = 14$
It turns out that $f(n) = ((n/2) - 2)$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x neq 3$ and $yneq 3$.
Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3.
They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).
There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?
My guess is that as $n to infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.
number-theory goldbachs-conjecture
asked Dec 13 '18 at 16:29
temp wattstemp watts
262
$endgroup$
add a comment |
$begingroup$
Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+y=n$ where $mathbb{O}$ 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: $S_{10} ={(3,7),(5,5),(7,3)}$ and $f(10) = 3$.
$S_{14} = {(3,11), (5,9), (7,7), (9,5),(11,3)}$ and $f(14) = 5$
$S_{32} = {(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)}$
and $f(32) = 14$
It turns out that $f(n) = ((n/2) - 2)$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x neq 3$ and $yneq 3$.
Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3.
They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).
There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?
My guess is that as $n to infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.
number-theory goldbachs-conjecture
asked Dec 13 '18 at 16:29
temp wattstemp watts
262
$endgroup$
Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+y=n$ where $mathbb{O}$ 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: $S_{10} ={(3,7),(5,5),(7,3)}$ and $f(10) = 3$.
$S_{14} = {(3,11), (5,9), (7,7), (9,5),(11,3)}$ and $f(14) = 5$
$S_{32} = {(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)}$
and $f(32) = 14$
It turns out that $f(n) = ((n/2) - 2)$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x neq 3$ and $yneq 3$.
Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3.
They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).
There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?
My guess is that as $n to infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.
number-theory goldbachs-conjecture
number-theory goldbachs-conjecture
asked Dec 13 '18 at 16:29
temp wattstemp watts
262
asked Dec 13 '18 at 16:29
temp wattstemp watts
262
asked Dec 13 '18 at 16:29
temp wattstemp watts
262
asked Dec 13 '18 at 16:29
temp wattstemp watts
262
asked Dec 13 '18 at 16:29
temp wattstemp watts
262
262
add a comment |
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1 Answer
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$begingroup$
The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.
So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$
The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$
Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:
$$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$
$$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$
answered Dec 23 '18 at 1:57
François HuppéFrançois Huppé
357111
$endgroup$
add a comment |
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$begingroup$
The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.
So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$
The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$
Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:
$$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$
$$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$
answered Dec 23 '18 at 1:57
François HuppéFrançois Huppé
357111
$endgroup$
add a comment |
$begingroup$
The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.
So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$
The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$
Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:
$$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$
$$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$
answered Dec 23 '18 at 1:57
François HuppéFrançois Huppé
357111
$endgroup$
add a comment |
$begingroup$
The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.
So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$
The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$
Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:
$$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$
$$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$
answered Dec 23 '18 at 1:57
François HuppéFrançois Huppé
357111
$endgroup$
The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.
So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$
The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:
$$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$
Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:
$$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$
$$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$
answered Dec 23 '18 at 1:57
François HuppéFrançois Huppé
357111
answered Dec 23 '18 at 1:57
François HuppéFrançois Huppé
357111
answered Dec 23 '18 at 1:57
François HuppéFrançois Huppé
357111
answered Dec 23 '18 at 1:57
François HuppéFrançois Huppé
357111
357111
add a comment |
add a comment |
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