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Greg Martin
Greg Martin
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A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ uniformly and independently, the abobeabove probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ chosen at random we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?

A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ the abobe probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ chosen at random we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?

A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ uniformly and independently, the above probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ chosen at random we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?

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Alex G
Alex G
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A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ the abobe probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ chosen at random we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?

A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ the abobe probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?

A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ the abobe probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ chosen at random we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?

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Alex G
Alex G
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Probability of all combinations of k numbers among n being coprime

A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ the abobe probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?