Mixing
Proportion of odd digits in powers of $2$ base ten
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For $n ge 1,$ if $2^n$ in base ten has $d$ digits, $k$ of which are odd, define $f(n)=k/d.$ I'm wondering whether $f(n)$ can get arbitrarily close to $1,$ or on the other hand does $f(n)$ have a least upper bound which is less than $1.$ [If the latter can we find that lub?]
We have $f(9)=2/3,$ since $2^9=512.$ we can't have $f(n)=1$ since the units digit is even. I'd be interested in some specific cases where $f(n)$ is "fairly near" $1$ [whatever that means].
elementary-number-theory
asked Jan 31 at 18:11
coffeemathcoffeemath
2,9171415
$endgroup$
|
show 1 more comment
$begingroup$
For $n ge 1,$ if $2^n$ in base ten has $d$ digits, $k$ of which are odd, define $f(n)=k/d.$ I'm wondering whether $f(n)$ can get arbitrarily close to $1,$ or on the other hand does $f(n)$ have a least upper bound which is less than $1.$ [If the latter can we find that lub?]
We have $f(9)=2/3,$ since $2^9=512.$ we can't have $f(n)=1$ since the units digit is even. I'd be interested in some specific cases where $f(n)$ is "fairly near" $1$ [whatever that means].
elementary-number-theory
asked Jan 31 at 18:11
coffeemathcoffeemath
2,9171415
$endgroup$
4
$begingroup$
Worth noting: problems like this tend not to be solvable. See, for instance, the comparatively simpler Conjecture of Erdos regarding the ternary expressions of powers of $2$.
$endgroup$
– lulu
Jan 31 at 18:32
3
$begingroup$
I have computed $f(n)$ for $1le nle10^5$. Maximum is $f(76)=0.695652$. The mean is $0.499857$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:40
$begingroup$
The number of odd digits in $2^n$ is the number of digits $ge5$ in $2^{n-1}$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:42
1
$begingroup$
@JuliánAguirre a look at the larger numbers make it feels like when $n rightarrow infty$ it goes to $0$. look for $f(n)$ in $100<n<10^5$, and than $500<n<10^5$ and than $2000<n<10^5$.. Do you feel the same? EDIT: I use python on online compiler cause my computer sucks, I can look only up to n=1000..
$endgroup$
– Shaq
Jan 31 at 18:57
$begingroup$
The values cluster around $0.5$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:58
|
show 1 more comment
$begingroup$
For $n ge 1,$ if $2^n$ in base ten has $d$ digits, $k$ of which are odd, define $f(n)=k/d.$ I'm wondering whether $f(n)$ can get arbitrarily close to $1,$ or on the other hand does $f(n)$ have a least upper bound which is less than $1.$ [If the latter can we find that lub?]
We have $f(9)=2/3,$ since $2^9=512.$ we can't have $f(n)=1$ since the units digit is even. I'd be interested in some specific cases where $f(n)$ is "fairly near" $1$ [whatever that means].
elementary-number-theory
asked Jan 31 at 18:11
coffeemathcoffeemath
2,9171415
$endgroup$
For $n ge 1,$ if $2^n$ in base ten has $d$ digits, $k$ of which are odd, define $f(n)=k/d.$ I'm wondering whether $f(n)$ can get arbitrarily close to $1,$ or on the other hand does $f(n)$ have a least upper bound which is less than $1.$ [If the latter can we find that lub?]
We have $f(9)=2/3,$ since $2^9=512.$ we can't have $f(n)=1$ since the units digit is even. I'd be interested in some specific cases where $f(n)$ is "fairly near" $1$ [whatever that means].
elementary-number-theory
elementary-number-theory
asked Jan 31 at 18:11
coffeemathcoffeemath
2,9171415
asked Jan 31 at 18:11
coffeemathcoffeemath
2,9171415
edited Jan 31 at 18:29
coffeemath
asked Jan 31 at 18:11
coffeemathcoffeemath
2,9171415
asked Jan 31 at 18:11
coffeemathcoffeemath
2,9171415
asked Jan 31 at 18:11
coffeemathcoffeemath
2,9171415
2,9171415
4
$begingroup$
Worth noting: problems like this tend not to be solvable. See, for instance, the comparatively simpler Conjecture of Erdos regarding the ternary expressions of powers of $2$.
$endgroup$
– lulu
Jan 31 at 18:32
3
$begingroup$
I have computed $f(n)$ for $1le nle10^5$. Maximum is $f(76)=0.695652$. The mean is $0.499857$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:40
$begingroup$
The number of odd digits in $2^n$ is the number of digits $ge5$ in $2^{n-1}$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:42
1
$begingroup$
@JuliánAguirre a look at the larger numbers make it feels like when $n rightarrow infty$ it goes to $0$. look for $f(n)$ in $100<n<10^5$, and than $500<n<10^5$ and than $2000<n<10^5$.. Do you feel the same? EDIT: I use python on online compiler cause my computer sucks, I can look only up to n=1000..
$endgroup$
– Shaq
Jan 31 at 18:57
$begingroup$
The values cluster around $0.5$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:58
|
show 1 more comment
4
$begingroup$
Worth noting: problems like this tend not to be solvable. See, for instance, the comparatively simpler Conjecture of Erdos regarding the ternary expressions of powers of $2$.
$endgroup$
– lulu
Jan 31 at 18:32
3
$begingroup$
I have computed $f(n)$ for $1le nle10^5$. Maximum is $f(76)=0.695652$. The mean is $0.499857$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:40
$begingroup$
The number of odd digits in $2^n$ is the number of digits $ge5$ in $2^{n-1}$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:42
1
$begingroup$
@JuliánAguirre a look at the larger numbers make it feels like when $n rightarrow infty$ it goes to $0$. look for $f(n)$ in $100<n<10^5$, and than $500<n<10^5$ and than $2000<n<10^5$.. Do you feel the same? EDIT: I use python on online compiler cause my computer sucks, I can look only up to n=1000..
$endgroup$
– Shaq
Jan 31 at 18:57
$begingroup$
The values cluster around $0.5$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:58
4
4
$begingroup$
Worth noting: problems like this tend not to be solvable. See, for instance, the comparatively simpler Conjecture of Erdos regarding the ternary expressions of powers of $2$.
$endgroup$
– lulu
Jan 31 at 18:32
$begingroup$
Worth noting: problems like this tend not to be solvable. See, for instance, the comparatively simpler Conjecture of Erdos regarding the ternary expressions of powers of $2$.
$endgroup$
– lulu
Jan 31 at 18:32
3
3
$begingroup$
I have computed $f(n)$ for $1le nle10^5$. Maximum is $f(76)=0.695652$. The mean is $0.499857$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:40
$begingroup$
I have computed $f(n)$ for $1le nle10^5$. Maximum is $f(76)=0.695652$. The mean is $0.499857$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:40
$begingroup$
The number of odd digits in $2^n$ is the number of digits $ge5$ in $2^{n-1}$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:42
$begingroup$
The number of odd digits in $2^n$ is the number of digits $ge5$ in $2^{n-1}$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:42
1
1
$begingroup$
@JuliánAguirre a look at the larger numbers make it feels like when $n rightarrow infty$ it goes to $0$. look for $f(n)$ in $100<n<10^5$, and than $500<n<10^5$ and than $2000<n<10^5$.. Do you feel the same? EDIT: I use python on online compiler cause my computer sucks, I can look only up to n=1000..
$endgroup$
– Shaq
Jan 31 at 18:57
$begingroup$
@JuliánAguirre a look at the larger numbers make it feels like when $n rightarrow infty$ it goes to $0$. look for $f(n)$ in $100<n<10^5$, and than $500<n<10^5$ and than $2000<n<10^5$.. Do you feel the same? EDIT: I use python on online compiler cause my computer sucks, I can look only up to n=1000..
$endgroup$
– Shaq
Jan 31 at 18:57
$begingroup$
The values cluster around $0.5$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:58
$begingroup$
The values cluster around $0.5$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:58
|
show 1 more comment
0
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4
$begingroup$
Worth noting: problems like this tend not to be solvable. See, for instance, the comparatively simpler Conjecture of Erdos regarding the ternary expressions of powers of $2$.
$endgroup$
– lulu
Jan 31 at 18:32
3
$begingroup$
I have computed $f(n)$ for $1le nle10^5$. Maximum is $f(76)=0.695652$. The mean is $0.499857$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:40
$begingroup$
The number of odd digits in $2^n$ is the number of digits $ge5$ in $2^{n-1}$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:42
1
$begingroup$
@JuliánAguirre a look at the larger numbers make it feels like when $n rightarrow infty$ it goes to $0$. look for $f(n)$ in $100<n<10^5$, and than $500<n<10^5$ and than $2000<n<10^5$.. Do you feel the same? EDIT: I use python on online compiler cause my computer sucks, I can look only up to n=1000..
$endgroup$
– Shaq
Jan 31 at 18:57
$begingroup$
The values cluster around $0.5$.
$endgroup$
– Julián Aguirre
Jan 31 at 18:58