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How many unique integers to be obtained by $leftlceilfrac{F}{1:1:N}rightrceil$?
Assume that $F$ and $N$ are positive integers, and $N<F$. How many unique integers would I get if I apply the ceiling function to the division of $F$ divided by $1,2,ldots,N$, respectively. That is, I want a formula to (approximately) count
$mathrm{texttt{length}}(mathrm{texttt{unique}}left(mathrm{texttt{ceil}}(F./(1:1:N))right))$, which is expressed in Matlab language.
combinatorics number-theory elementary-number-theory combinations
asked Nov 20 '18 at 7:49
leeyee
656
add a comment |
Assume that $F$ and $N$ are positive integers, and $N<F$. How many unique integers would I get if I apply the ceiling function to the division of $F$ divided by $1,2,ldots,N$, respectively. That is, I want a formula to (approximately) count
$mathrm{texttt{length}}(mathrm{texttt{unique}}left(mathrm{texttt{ceil}}(F./(1:1:N))right))$, which is expressed in Matlab language.
combinatorics number-theory elementary-number-theory combinations
asked Nov 20 '18 at 7:49
leeyee
656
add a comment |
Assume that $F$ and $N$ are positive integers, and $N<F$. How many unique integers would I get if I apply the ceiling function to the division of $F$ divided by $1,2,ldots,N$, respectively. That is, I want a formula to (approximately) count
$mathrm{texttt{length}}(mathrm{texttt{unique}}left(mathrm{texttt{ceil}}(F./(1:1:N))right))$, which is expressed in Matlab language.
combinatorics number-theory elementary-number-theory combinations
asked Nov 20 '18 at 7:49
leeyee
656
Assume that $F$ and $N$ are positive integers, and $N<F$. How many unique integers would I get if I apply the ceiling function to the division of $F$ divided by $1,2,ldots,N$, respectively. That is, I want a formula to (approximately) count
$mathrm{texttt{length}}(mathrm{texttt{unique}}left(mathrm{texttt{ceil}}(F./(1:1:N))right))$, which is expressed in Matlab language.
combinatorics number-theory elementary-number-theory combinations
combinatorics number-theory elementary-number-theory combinations
asked Nov 20 '18 at 7:49
leeyee
656
asked Nov 20 '18 at 7:49
leeyee
656
asked Nov 20 '18 at 7:49
leeyee
656
asked Nov 20 '18 at 7:49
leeyee
656
asked Nov 20 '18 at 7:49
leeyee
656
656
add a comment |
add a comment |
1 Answer
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Let $g_f(n)=leftlceil frac{f}{n}rightrceil$. We may first look at the forward difference of $g_f$, which is $Delta[g_f](n)=g_f(n+1)-g_f(n)$. When $Delta[g_f](n)geq -1$ for all $ngeq n_0$, then $g_f(n)$ will take on all values from $g_f(n_0)$ to $g_f(f-1)=2$ when $ngeq n_0.$ Additionally, if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)= -1,$$ then for all $ngeq n_0$, $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)geq -1,$$ so $$Delta[g_f](n)geq -1.$$ We also note that if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)< -1,$$ then for all $n<n_0$, $$Delta[g_f](n)< -1,$$ so every value of $g_f(n)$ is unique for all $n<n_0$. When we solve for $n_0$, we get $n_0^2 +n_0-f=0$, so $$n_0=frac{-1+sqrt{1+4f}}{2}.$$ However, $$0leqleftlceilsqrt{f}rightrceil-leftlceilfrac{-1+sqrt{1+4f}}{2}rightrceilleq 1.$$ So if we were to use $n_1=sqrt{f}$ instead of $n_0$, the inequalities for $Delta[g_f](n)$ would still hold. Thus, the number of unique values of $g_f(n)$ where $1leq n<f$ is $$lceil n_1rceil-1+g_f(lceil n_1 rceil)-1=2leftlceil sqrt{f}rightrceil -2.$$ Now the number of unique values of $g_f(n)$, where $1leq n leq N < f$ is $$begin{cases}N & N<leftlceil sqrt{f}rightrceil\ 2leftlceil sqrt{f}rightrceil -leftlceilfrac{f}{N}rightrceil & Ngeq leftlceil sqrt{f}rightrceil. end{cases}$$ The first case comes from the fact that every value of $g_f(n)$ is unique for all $n<n_1$, and the second case comes from the fact that $g_f(n)$ will take on all values from $g_f(n_1)$ to $g_f(f-1)=2$ when $ngeq n_1,$ but we lose all the values from $g_f(N)-1$ and $2$ by the restriction $nleq N$.
answered Nov 20 '18 at 14:55
Jacob
1,092715
Shouldn't the final answer depend on both $F$ and $N$?
– Misha Lavrov
Nov 20 '18 at 15:55
Oh I misread it as counting all unique $g_f(n)$ where $1leq n < f$. I’ll see if I can fix it and resubmit my answer.
– Jacob
Nov 20 '18 at 16:09
@mishalavrov I’ve fixed it.
– Jacob
Nov 20 '18 at 16:42
add a comment |
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Let $g_f(n)=leftlceil frac{f}{n}rightrceil$. We may first look at the forward difference of $g_f$, which is $Delta[g_f](n)=g_f(n+1)-g_f(n)$. When $Delta[g_f](n)geq -1$ for all $ngeq n_0$, then $g_f(n)$ will take on all values from $g_f(n_0)$ to $g_f(f-1)=2$ when $ngeq n_0.$ Additionally, if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)= -1,$$ then for all $ngeq n_0$, $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)geq -1,$$ so $$Delta[g_f](n)geq -1.$$ We also note that if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)< -1,$$ then for all $n<n_0$, $$Delta[g_f](n)< -1,$$ so every value of $g_f(n)$ is unique for all $n<n_0$. When we solve for $n_0$, we get $n_0^2 +n_0-f=0$, so $$n_0=frac{-1+sqrt{1+4f}}{2}.$$ However, $$0leqleftlceilsqrt{f}rightrceil-leftlceilfrac{-1+sqrt{1+4f}}{2}rightrceilleq 1.$$ So if we were to use $n_1=sqrt{f}$ instead of $n_0$, the inequalities for $Delta[g_f](n)$ would still hold. Thus, the number of unique values of $g_f(n)$ where $1leq n<f$ is $$lceil n_1rceil-1+g_f(lceil n_1 rceil)-1=2leftlceil sqrt{f}rightrceil -2.$$ Now the number of unique values of $g_f(n)$, where $1leq n leq N < f$ is $$begin{cases}N & N<leftlceil sqrt{f}rightrceil\ 2leftlceil sqrt{f}rightrceil -leftlceilfrac{f}{N}rightrceil & Ngeq leftlceil sqrt{f}rightrceil. end{cases}$$ The first case comes from the fact that every value of $g_f(n)$ is unique for all $n<n_1$, and the second case comes from the fact that $g_f(n)$ will take on all values from $g_f(n_1)$ to $g_f(f-1)=2$ when $ngeq n_1,$ but we lose all the values from $g_f(N)-1$ and $2$ by the restriction $nleq N$.
answered Nov 20 '18 at 14:55
Jacob
1,092715
Shouldn't the final answer depend on both $F$ and $N$?
– Misha Lavrov
Nov 20 '18 at 15:55
Oh I misread it as counting all unique $g_f(n)$ where $1leq n < f$. I’ll see if I can fix it and resubmit my answer.
– Jacob
Nov 20 '18 at 16:09
@mishalavrov I’ve fixed it.
– Jacob
Nov 20 '18 at 16:42
add a comment |
Let $g_f(n)=leftlceil frac{f}{n}rightrceil$. We may first look at the forward difference of $g_f$, which is $Delta[g_f](n)=g_f(n+1)-g_f(n)$. When $Delta[g_f](n)geq -1$ for all $ngeq n_0$, then $g_f(n)$ will take on all values from $g_f(n_0)$ to $g_f(f-1)=2$ when $ngeq n_0.$ Additionally, if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)= -1,$$ then for all $ngeq n_0$, $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)geq -1,$$ so $$Delta[g_f](n)geq -1.$$ We also note that if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)< -1,$$ then for all $n<n_0$, $$Delta[g_f](n)< -1,$$ so every value of $g_f(n)$ is unique for all $n<n_0$. When we solve for $n_0$, we get $n_0^2 +n_0-f=0$, so $$n_0=frac{-1+sqrt{1+4f}}{2}.$$ However, $$0leqleftlceilsqrt{f}rightrceil-leftlceilfrac{-1+sqrt{1+4f}}{2}rightrceilleq 1.$$ So if we were to use $n_1=sqrt{f}$ instead of $n_0$, the inequalities for $Delta[g_f](n)$ would still hold. Thus, the number of unique values of $g_f(n)$ where $1leq n<f$ is $$lceil n_1rceil-1+g_f(lceil n_1 rceil)-1=2leftlceil sqrt{f}rightrceil -2.$$ Now the number of unique values of $g_f(n)$, where $1leq n leq N < f$ is $$begin{cases}N & N<leftlceil sqrt{f}rightrceil\ 2leftlceil sqrt{f}rightrceil -leftlceilfrac{f}{N}rightrceil & Ngeq leftlceil sqrt{f}rightrceil. end{cases}$$ The first case comes from the fact that every value of $g_f(n)$ is unique for all $n<n_1$, and the second case comes from the fact that $g_f(n)$ will take on all values from $g_f(n_1)$ to $g_f(f-1)=2$ when $ngeq n_1,$ but we lose all the values from $g_f(N)-1$ and $2$ by the restriction $nleq N$.
answered Nov 20 '18 at 14:55
Jacob
1,092715
Shouldn't the final answer depend on both $F$ and $N$?
– Misha Lavrov
Nov 20 '18 at 15:55
Oh I misread it as counting all unique $g_f(n)$ where $1leq n < f$. I’ll see if I can fix it and resubmit my answer.
– Jacob
Nov 20 '18 at 16:09
@mishalavrov I’ve fixed it.
– Jacob
Nov 20 '18 at 16:42
add a comment |
Let $g_f(n)=leftlceil frac{f}{n}rightrceil$. We may first look at the forward difference of $g_f$, which is $Delta[g_f](n)=g_f(n+1)-g_f(n)$. When $Delta[g_f](n)geq -1$ for all $ngeq n_0$, then $g_f(n)$ will take on all values from $g_f(n_0)$ to $g_f(f-1)=2$ when $ngeq n_0.$ Additionally, if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)= -1,$$ then for all $ngeq n_0$, $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)geq -1,$$ so $$Delta[g_f](n)geq -1.$$ We also note that if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)< -1,$$ then for all $n<n_0$, $$Delta[g_f](n)< -1,$$ so every value of $g_f(n)$ is unique for all $n<n_0$. When we solve for $n_0$, we get $n_0^2 +n_0-f=0$, so $$n_0=frac{-1+sqrt{1+4f}}{2}.$$ However, $$0leqleftlceilsqrt{f}rightrceil-leftlceilfrac{-1+sqrt{1+4f}}{2}rightrceilleq 1.$$ So if we were to use $n_1=sqrt{f}$ instead of $n_0$, the inequalities for $Delta[g_f](n)$ would still hold. Thus, the number of unique values of $g_f(n)$ where $1leq n<f$ is $$lceil n_1rceil-1+g_f(lceil n_1 rceil)-1=2leftlceil sqrt{f}rightrceil -2.$$ Now the number of unique values of $g_f(n)$, where $1leq n leq N < f$ is $$begin{cases}N & N<leftlceil sqrt{f}rightrceil\ 2leftlceil sqrt{f}rightrceil -leftlceilfrac{f}{N}rightrceil & Ngeq leftlceil sqrt{f}rightrceil. end{cases}$$ The first case comes from the fact that every value of $g_f(n)$ is unique for all $n<n_1$, and the second case comes from the fact that $g_f(n)$ will take on all values from $g_f(n_1)$ to $g_f(f-1)=2$ when $ngeq n_1,$ but we lose all the values from $g_f(N)-1$ and $2$ by the restriction $nleq N$.
answered Nov 20 '18 at 14:55
Jacob
1,092715
Let $g_f(n)=leftlceil frac{f}{n}rightrceil$. We may first look at the forward difference of $g_f$, which is $Delta[g_f](n)=g_f(n+1)-g_f(n)$. When $Delta[g_f](n)geq -1$ for all $ngeq n_0$, then $g_f(n)$ will take on all values from $g_f(n_0)$ to $g_f(f-1)=2$ when $ngeq n_0.$ Additionally, if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)= -1,$$ then for all $ngeq n_0$, $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)geq -1,$$ so $$Delta[g_f](n)geq -1.$$ We also note that if $$fleft(frac{1}{n_0+1}-frac{1}{n_0}right)< -1,$$ then for all $n<n_0$, $$Delta[g_f](n)< -1,$$ so every value of $g_f(n)$ is unique for all $n<n_0$. When we solve for $n_0$, we get $n_0^2 +n_0-f=0$, so $$n_0=frac{-1+sqrt{1+4f}}{2}.$$ However, $$0leqleftlceilsqrt{f}rightrceil-leftlceilfrac{-1+sqrt{1+4f}}{2}rightrceilleq 1.$$ So if we were to use $n_1=sqrt{f}$ instead of $n_0$, the inequalities for $Delta[g_f](n)$ would still hold. Thus, the number of unique values of $g_f(n)$ where $1leq n<f$ is $$lceil n_1rceil-1+g_f(lceil n_1 rceil)-1=2leftlceil sqrt{f}rightrceil -2.$$ Now the number of unique values of $g_f(n)$, where $1leq n leq N < f$ is $$begin{cases}N & N<leftlceil sqrt{f}rightrceil\ 2leftlceil sqrt{f}rightrceil -leftlceilfrac{f}{N}rightrceil & Ngeq leftlceil sqrt{f}rightrceil. end{cases}$$ The first case comes from the fact that every value of $g_f(n)$ is unique for all $n<n_1$, and the second case comes from the fact that $g_f(n)$ will take on all values from $g_f(n_1)$ to $g_f(f-1)=2$ when $ngeq n_1,$ but we lose all the values from $g_f(N)-1$ and $2$ by the restriction $nleq N$.
answered Nov 20 '18 at 14:55
Jacob
1,092715
edited Nov 20 '18 at 16:41
answered Nov 20 '18 at 14:55
Jacob
1,092715
answered Nov 20 '18 at 14:55
Jacob
1,092715
answered Nov 20 '18 at 14:55
Jacob
1,092715
1,092715
Shouldn't the final answer depend on both $F$ and $N$?
– Misha Lavrov
Nov 20 '18 at 15:55
Oh I misread it as counting all unique $g_f(n)$ where $1leq n < f$. I’ll see if I can fix it and resubmit my answer.
– Jacob
Nov 20 '18 at 16:09
@mishalavrov I’ve fixed it.
– Jacob
Nov 20 '18 at 16:42
add a comment |
Shouldn't the final answer depend on both $F$ and $N$?
– Misha Lavrov
Nov 20 '18 at 15:55
Oh I misread it as counting all unique $g_f(n)$ where $1leq n < f$. I’ll see if I can fix it and resubmit my answer.
– Jacob
Nov 20 '18 at 16:09
@mishalavrov I’ve fixed it.
– Jacob
Nov 20 '18 at 16:42
Shouldn't the final answer depend on both $F$ and $N$?
– Misha Lavrov
Nov 20 '18 at 15:55
Shouldn't the final answer depend on both $F$ and $N$?
– Misha Lavrov
Nov 20 '18 at 15:55
Oh I misread it as counting all unique $g_f(n)$ where $1leq n < f$. I’ll see if I can fix it and resubmit my answer.
– Jacob
Nov 20 '18 at 16:09
Oh I misread it as counting all unique $g_f(n)$ where $1leq n < f$. I’ll see if I can fix it and resubmit my answer.
– Jacob
Nov 20 '18 at 16:09
@mishalavrov I’ve fixed it.
– Jacob
Nov 20 '18 at 16:42
@mishalavrov I’ve fixed it.
– Jacob
Nov 20 '18 at 16:42
add a comment |
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