Mixing
![How is $[P text AND (Q text OR R)] text IFF [(P text AND Q) text OR (P text AND R)]$ valid?](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
$3^k cdot (n + frac12) - frac12$ powers of $2$
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Does the above sequence $3^k (n + 1/2) - 1/2$ always have a power of 2 in it for all n and some k?
Let $s_nk = 3^k (n + frac12) - frac12$, does $s_nk == 2^m$ for some m and k.
I'm thinking that because 3 is coprime with 2 that the sequence will eventually hit a power of 2 but it is a hunch and I can't seem to prove it.
number-theory
edited Jul 15 at 20:07
DarÃo A. Gutiérrez
2,41521129
asked Jul 15 at 20:03
AbstractDissonance
456214
add a comment |Â
up vote
0
down vote
favorite
Does the above sequence $3^k (n + 1/2) - 1/2$ always have a power of 2 in it for all n and some k?
Let $s_nk = 3^k (n + frac12) - frac12$, does $s_nk == 2^m$ for some m and k.
I'm thinking that because 3 is coprime with 2 that the sequence will eventually hit a power of 2 but it is a hunch and I can't seem to prove it.
number-theory
edited Jul 15 at 20:07
DarÃo A. Gutiérrez
2,41521129
asked Jul 15 at 20:03
AbstractDissonance
456214
Are you fixing $n$ and looking at $a_k=3^k(n+1/2)-1/2$ or fixing $k$ and looking at $a_n=3^k(n+1/2)-1/2$? I can't quite tell from the question.
– Carl Schildkraut
Jul 15 at 20:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does the above sequence $3^k (n + 1/2) - 1/2$ always have a power of 2 in it for all n and some k?
Let $s_nk = 3^k (n + frac12) - frac12$, does $s_nk == 2^m$ for some m and k.
I'm thinking that because 3 is coprime with 2 that the sequence will eventually hit a power of 2 but it is a hunch and I can't seem to prove it.
number-theory
edited Jul 15 at 20:07
DarÃo A. Gutiérrez
2,41521129
asked Jul 15 at 20:03
AbstractDissonance
456214
Does the above sequence $3^k (n + 1/2) - 1/2$ always have a power of 2 in it for all n and some k?
Let $s_nk = 3^k (n + frac12) - frac12$, does $s_nk == 2^m$ for some m and k.
I'm thinking that because 3 is coprime with 2 that the sequence will eventually hit a power of 2 but it is a hunch and I can't seem to prove it.
number-theory
edited Jul 15 at 20:07
DarÃo A. Gutiérrez
2,41521129
asked Jul 15 at 20:03
AbstractDissonance
456214
edited Jul 15 at 20:07
DarÃo A. Gutiérrez
2,41521129
edited Jul 15 at 20:07
DarÃo A. Gutiérrez
2,41521129
edited Jul 15 at 20:07
DarÃo A. Gutiérrez
2,41521129
2,41521129
asked Jul 15 at 20:03
AbstractDissonance
456214
asked Jul 15 at 20:03
AbstractDissonance
456214
asked Jul 15 at 20:03
AbstractDissonance
456214
456214
Are you fixing $n$ and looking at $a_k=3^k(n+1/2)-1/2$ or fixing $k$ and looking at $a_n=3^k(n+1/2)-1/2$? I can't quite tell from the question.
– Carl Schildkraut
Jul 15 at 20:12
add a comment |Â
Are you fixing $n$ and looking at $a_k=3^k(n+1/2)-1/2$ or fixing $k$ and looking at $a_n=3^k(n+1/2)-1/2$? I can't quite tell from the question.
– Carl Schildkraut
Jul 15 at 20:12
Are you fixing $n$ and looking at $a_k=3^k(n+1/2)-1/2$ or fixing $k$ and looking at $a_n=3^k(n+1/2)-1/2$? I can't quite tell from the question.
– Carl Schildkraut
Jul 15 at 20:12
Are you fixing $n$ and looking at $a_k=3^k(n+1/2)-1/2$ or fixing $k$ and looking at $a_n=3^k(n+1/2)-1/2$? I can't quite tell from the question.
– Carl Schildkraut
Jul 15 at 20:12
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Your equation reduces to
$$3^k(2n+1)=2^m+1$$
for some $k,n,m$.
If you're fixing $k$, take $m=3^k-1$. Then, by Lifting the Exponent Lemma (you can find a description and proof of the lemma along with notation here),
$$nu_3left(2^3^k-1+1right)=nu_3left(2+1right)+nu_3left(3^k-1right)=k,$$
so
$$3^kbig|2^3^k-1+1implies frac2^3^k-1+13^k=2n+1$$
for some $n$. So, fixing $k$, there always exists a pair $(m,n)$ for which $3^k(2n+1)=2^m+1$.
If you're fixing $n$, take $n=3$. We have that the left hand side is a multiple of $7$, while
$$2^m+1in2,3,5bmod 7,$$
so there are no integers $K$ for which
$$frac2^m+12n+1=K,$$
and certainly no powers of $3$.
answered Jul 15 at 20:32
Carl Schildkraut
8,28711238
I am looking for the statement to hold for each n and some k and m. Basically any arbitrary positive integer(an odd integer works) n, their exists some k and m such that the stated relationship holds. So, given n(not fixed, but for every positive integer), does the equation $3^k(2n+1) - 2^m - 1 = 0$ have an positive integral solution $(k,m)$? Using the language of lifting: Given some arbitrary $n$, their exists a $k$ s.t. $v_p(3^k(n + 1/2) - 1/2) = 0$ for all $p > 2$
– AbstractDissonance
Jul 15 at 23:32
@AbstractDissonance Thanks. My example of $n=3$ in the second half of the answer should show that, for some $n$ ($n=3$ but probably others as well), there do not exist any $k,m$.
– Carl Schildkraut
Jul 15 at 23:55
The ‘$v_3(2^3+1)$’ should be ‘$v_3(2+1)$’.
– Szeto
Jul 16 at 1:08
@Szeto You're right; I've fixed it.
– Carl Schildkraut
Jul 16 at 2:01
Carl, would you mind to look at my answer? My impression is that there is no $k$ that for all $n$ the evaluation of $s_nk$ has powers $2^m$ with $m>0$.
– Gottfried Helms
Jul 16 at 6:30
 |Â
show 3 more comments
up vote
0
down vote
$$ s_nk=3^k cdot (n+ frac12) - frac12 \
s_nk=3^k n + 3^k-1over 2 \
$$
- For even $k$ we need even $n$ to have some power of $2$ in $s_nk$
- For odd $k$ we need odd $n$ to have some power of $2$ in $s_nk$
answered Jul 16 at 2:06
Gottfried Helms
22.6k24194
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your equation reduces to
$$3^k(2n+1)=2^m+1$$
for some $k,n,m$.
If you're fixing $k$, take $m=3^k-1$. Then, by Lifting the Exponent Lemma (you can find a description and proof of the lemma along with notation here),
$$nu_3left(2^3^k-1+1right)=nu_3left(2+1right)+nu_3left(3^k-1right)=k,$$
so
$$3^kbig|2^3^k-1+1implies frac2^3^k-1+13^k=2n+1$$
for some $n$. So, fixing $k$, there always exists a pair $(m,n)$ for which $3^k(2n+1)=2^m+1$.
If you're fixing $n$, take $n=3$. We have that the left hand side is a multiple of $7$, while
$$2^m+1in2,3,5bmod 7,$$
so there are no integers $K$ for which
$$frac2^m+12n+1=K,$$
and certainly no powers of $3$.
answered Jul 15 at 20:32
Carl Schildkraut
8,28711238
I am looking for the statement to hold for each n and some k and m. Basically any arbitrary positive integer(an odd integer works) n, their exists some k and m such that the stated relationship holds. So, given n(not fixed, but for every positive integer), does the equation $3^k(2n+1) - 2^m - 1 = 0$ have an positive integral solution $(k,m)$? Using the language of lifting: Given some arbitrary $n$, their exists a $k$ s.t. $v_p(3^k(n + 1/2) - 1/2) = 0$ for all $p > 2$
– AbstractDissonance
Jul 15 at 23:32
@AbstractDissonance Thanks. My example of $n=3$ in the second half of the answer should show that, for some $n$ ($n=3$ but probably others as well), there do not exist any $k,m$.
– Carl Schildkraut
Jul 15 at 23:55
The ‘$v_3(2^3+1)$’ should be ‘$v_3(2+1)$’.
– Szeto
Jul 16 at 1:08
@Szeto You're right; I've fixed it.
– Carl Schildkraut
Jul 16 at 2:01
Carl, would you mind to look at my answer? My impression is that there is no $k$ that for all $n$ the evaluation of $s_nk$ has powers $2^m$ with $m>0$.
– Gottfried Helms
Jul 16 at 6:30
 |Â
show 3 more comments
up vote
1
down vote
accepted
Your equation reduces to
$$3^k(2n+1)=2^m+1$$
for some $k,n,m$.
If you're fixing $k$, take $m=3^k-1$. Then, by Lifting the Exponent Lemma (you can find a description and proof of the lemma along with notation here),
$$nu_3left(2^3^k-1+1right)=nu_3left(2+1right)+nu_3left(3^k-1right)=k,$$
so
$$3^kbig|2^3^k-1+1implies frac2^3^k-1+13^k=2n+1$$
for some $n$. So, fixing $k$, there always exists a pair $(m,n)$ for which $3^k(2n+1)=2^m+1$.
If you're fixing $n$, take $n=3$. We have that the left hand side is a multiple of $7$, while
$$2^m+1in2,3,5bmod 7,$$
so there are no integers $K$ for which
$$frac2^m+12n+1=K,$$
and certainly no powers of $3$.
answered Jul 15 at 20:32
Carl Schildkraut
8,28711238
I am looking for the statement to hold for each n and some k and m. Basically any arbitrary positive integer(an odd integer works) n, their exists some k and m such that the stated relationship holds. So, given n(not fixed, but for every positive integer), does the equation $3^k(2n+1) - 2^m - 1 = 0$ have an positive integral solution $(k,m)$? Using the language of lifting: Given some arbitrary $n$, their exists a $k$ s.t. $v_p(3^k(n + 1/2) - 1/2) = 0$ for all $p > 2$
– AbstractDissonance
Jul 15 at 23:32
@AbstractDissonance Thanks. My example of $n=3$ in the second half of the answer should show that, for some $n$ ($n=3$ but probably others as well), there do not exist any $k,m$.
– Carl Schildkraut
Jul 15 at 23:55
The ‘$v_3(2^3+1)$’ should be ‘$v_3(2+1)$’.
– Szeto
Jul 16 at 1:08
@Szeto You're right; I've fixed it.
– Carl Schildkraut
Jul 16 at 2:01
Carl, would you mind to look at my answer? My impression is that there is no $k$ that for all $n$ the evaluation of $s_nk$ has powers $2^m$ with $m>0$.
– Gottfried Helms
Jul 16 at 6:30
 |Â
show 3 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your equation reduces to
$$3^k(2n+1)=2^m+1$$
for some $k,n,m$.
If you're fixing $k$, take $m=3^k-1$. Then, by Lifting the Exponent Lemma (you can find a description and proof of the lemma along with notation here),
$$nu_3left(2^3^k-1+1right)=nu_3left(2+1right)+nu_3left(3^k-1right)=k,$$
so
$$3^kbig|2^3^k-1+1implies frac2^3^k-1+13^k=2n+1$$
for some $n$. So, fixing $k$, there always exists a pair $(m,n)$ for which $3^k(2n+1)=2^m+1$.
If you're fixing $n$, take $n=3$. We have that the left hand side is a multiple of $7$, while
$$2^m+1in2,3,5bmod 7,$$
so there are no integers $K$ for which
$$frac2^m+12n+1=K,$$
and certainly no powers of $3$.
answered Jul 15 at 20:32
Carl Schildkraut
8,28711238
Your equation reduces to
$$3^k(2n+1)=2^m+1$$
for some $k,n,m$.
If you're fixing $k$, take $m=3^k-1$. Then, by Lifting the Exponent Lemma (you can find a description and proof of the lemma along with notation here),
$$nu_3left(2^3^k-1+1right)=nu_3left(2+1right)+nu_3left(3^k-1right)=k,$$
so
$$3^kbig|2^3^k-1+1implies frac2^3^k-1+13^k=2n+1$$
for some $n$. So, fixing $k$, there always exists a pair $(m,n)$ for which $3^k(2n+1)=2^m+1$.
If you're fixing $n$, take $n=3$. We have that the left hand side is a multiple of $7$, while
$$2^m+1in2,3,5bmod 7,$$
so there are no integers $K$ for which
$$frac2^m+12n+1=K,$$
and certainly no powers of $3$.
answered Jul 15 at 20:32
Carl Schildkraut
8,28711238
edited Jul 16 at 2:01
answered Jul 15 at 20:32
Carl Schildkraut
8,28711238
answered Jul 15 at 20:32
Carl Schildkraut
8,28711238
answered Jul 15 at 20:32
Carl Schildkraut
8,28711238
8,28711238
I am looking for the statement to hold for each n and some k and m. Basically any arbitrary positive integer(an odd integer works) n, their exists some k and m such that the stated relationship holds. So, given n(not fixed, but for every positive integer), does the equation $3^k(2n+1) - 2^m - 1 = 0$ have an positive integral solution $(k,m)$? Using the language of lifting: Given some arbitrary $n$, their exists a $k$ s.t. $v_p(3^k(n + 1/2) - 1/2) = 0$ for all $p > 2$
– AbstractDissonance
Jul 15 at 23:32
@AbstractDissonance Thanks. My example of $n=3$ in the second half of the answer should show that, for some $n$ ($n=3$ but probably others as well), there do not exist any $k,m$.
– Carl Schildkraut
Jul 15 at 23:55
The ‘$v_3(2^3+1)$’ should be ‘$v_3(2+1)$’.
– Szeto
Jul 16 at 1:08
@Szeto You're right; I've fixed it.
– Carl Schildkraut
Jul 16 at 2:01
Carl, would you mind to look at my answer? My impression is that there is no $k$ that for all $n$ the evaluation of $s_nk$ has powers $2^m$ with $m>0$.
– Gottfried Helms
Jul 16 at 6:30
 |Â
show 3 more comments
I am looking for the statement to hold for each n and some k and m. Basically any arbitrary positive integer(an odd integer works) n, their exists some k and m such that the stated relationship holds. So, given n(not fixed, but for every positive integer), does the equation $3^k(2n+1) - 2^m - 1 = 0$ have an positive integral solution $(k,m)$? Using the language of lifting: Given some arbitrary $n$, their exists a $k$ s.t. $v_p(3^k(n + 1/2) - 1/2) = 0$ for all $p > 2$
– AbstractDissonance
Jul 15 at 23:32
@AbstractDissonance Thanks. My example of $n=3$ in the second half of the answer should show that, for some $n$ ($n=3$ but probably others as well), there do not exist any $k,m$.
– Carl Schildkraut
Jul 15 at 23:55
The ‘$v_3(2^3+1)$’ should be ‘$v_3(2+1)$’.
– Szeto
Jul 16 at 1:08
@Szeto You're right; I've fixed it.
– Carl Schildkraut
Jul 16 at 2:01
Carl, would you mind to look at my answer? My impression is that there is no $k$ that for all $n$ the evaluation of $s_nk$ has powers $2^m$ with $m>0$.
– Gottfried Helms
Jul 16 at 6:30
I am looking for the statement to hold for each n and some k and m. Basically any arbitrary positive integer(an odd integer works) n, their exists some k and m such that the stated relationship holds. So, given n(not fixed, but for every positive integer), does the equation $3^k(2n+1) - 2^m - 1 = 0$ have an positive integral solution $(k,m)$? Using the language of lifting: Given some arbitrary $n$, their exists a $k$ s.t. $v_p(3^k(n + 1/2) - 1/2) = 0$ for all $p > 2$
– AbstractDissonance
Jul 15 at 23:32
I am looking for the statement to hold for each n and some k and m. Basically any arbitrary positive integer(an odd integer works) n, their exists some k and m such that the stated relationship holds. So, given n(not fixed, but for every positive integer), does the equation $3^k(2n+1) - 2^m - 1 = 0$ have an positive integral solution $(k,m)$? Using the language of lifting: Given some arbitrary $n$, their exists a $k$ s.t. $v_p(3^k(n + 1/2) - 1/2) = 0$ for all $p > 2$
– AbstractDissonance
Jul 15 at 23:32
@AbstractDissonance Thanks. My example of $n=3$ in the second half of the answer should show that, for some $n$ ($n=3$ but probably others as well), there do not exist any $k,m$.
– Carl Schildkraut
Jul 15 at 23:55
@AbstractDissonance Thanks. My example of $n=3$ in the second half of the answer should show that, for some $n$ ($n=3$ but probably others as well), there do not exist any $k,m$.
– Carl Schildkraut
Jul 15 at 23:55
The ‘$v_3(2^3+1)$’ should be ‘$v_3(2+1)$’.
– Szeto
Jul 16 at 1:08
The ‘$v_3(2^3+1)$’ should be ‘$v_3(2+1)$’.
– Szeto
Jul 16 at 1:08
@Szeto You're right; I've fixed it.
– Carl Schildkraut
Jul 16 at 2:01
@Szeto You're right; I've fixed it.
– Carl Schildkraut
Jul 16 at 2:01
Carl, would you mind to look at my answer? My impression is that there is no $k$ that for all $n$ the evaluation of $s_nk$ has powers $2^m$ with $m>0$.
– Gottfried Helms
Jul 16 at 6:30
Carl, would you mind to look at my answer? My impression is that there is no $k$ that for all $n$ the evaluation of $s_nk$ has powers $2^m$ with $m>0$.
– Gottfried Helms
Jul 16 at 6:30
 |Â
show 3 more comments
up vote
0
down vote
$$ s_nk=3^k cdot (n+ frac12) - frac12 \
s_nk=3^k n + 3^k-1over 2 \
$$
- For even $k$ we need even $n$ to have some power of $2$ in $s_nk$
- For odd $k$ we need odd $n$ to have some power of $2$ in $s_nk$
answered Jul 16 at 2:06
Gottfried Helms
22.6k24194
add a comment |Â
up vote
0
down vote
$$ s_nk=3^k cdot (n+ frac12) - frac12 \
s_nk=3^k n + 3^k-1over 2 \
$$
- For even $k$ we need even $n$ to have some power of $2$ in $s_nk$
- For odd $k$ we need odd $n$ to have some power of $2$ in $s_nk$
answered Jul 16 at 2:06
Gottfried Helms
22.6k24194
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$ s_nk=3^k cdot (n+ frac12) - frac12 \
s_nk=3^k n + 3^k-1over 2 \
$$
- For even $k$ we need even $n$ to have some power of $2$ in $s_nk$
- For odd $k$ we need odd $n$ to have some power of $2$ in $s_nk$
answered Jul 16 at 2:06
Gottfried Helms
22.6k24194
$$ s_nk=3^k cdot (n+ frac12) - frac12 \
s_nk=3^k n + 3^k-1over 2 \
$$
- For even $k$ we need even $n$ to have some power of $2$ in $s_nk$
- For odd $k$ we need odd $n$ to have some power of $2$ in $s_nk$
answered Jul 16 at 2:06
Gottfried Helms
22.6k24194
answered Jul 16 at 2:06
Gottfried Helms
22.6k24194
answered Jul 16 at 2:06
Gottfried Helms
22.6k24194
answered Jul 16 at 2:06
Gottfried Helms
22.6k24194
22.6k24194
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852808%2f3k-cdot-n-frac12-frac12-powers-of-2%23new-answer', 'question_page');
);
Post as a guest
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
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
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
Are you fixing $n$ and looking at $a_k=3^k(n+1/2)-1/2$ or fixing $k$ and looking at $a_n=3^k(n+1/2)-1/2$? I can't quite tell from the question.
– Carl Schildkraut
Jul 15 at 20:12