Mixing
Cardinality set of multiples
Given an arbitrarily large set of natural numbers greater than one,
S = {$p_0$, $p_1$, ... $p_n$}
product of S = $prod_{i=0}^n p_i$
define M as the set of all natural numbers that are multiples of any member of S, smaller or equal than the product and larger than zero.
Is there a formula for the cardinality of M given S?
E.g
S = {2,3}
M = {2,3,4,6}
cardinality of M = 4.
I can think of the formula for specific lengths but not of a general form.
multiplicative-function
asked Nov 22 '18 at 9:26
JFugger_jrJFugger_jr
133
add a comment |
Given an arbitrarily large set of natural numbers greater than one,
S = {$p_0$, $p_1$, ... $p_n$}
product of S = $prod_{i=0}^n p_i$
define M as the set of all natural numbers that are multiples of any member of S, smaller or equal than the product and larger than zero.
Is there a formula for the cardinality of M given S?
E.g
S = {2,3}
M = {2,3,4,6}
cardinality of M = 4.
I can think of the formula for specific lengths but not of a general form.
multiplicative-function
asked Nov 22 '18 at 9:26
JFugger_jrJFugger_jr
133
add a comment |
Given an arbitrarily large set of natural numbers greater than one,
S = {$p_0$, $p_1$, ... $p_n$}
product of S = $prod_{i=0}^n p_i$
define M as the set of all natural numbers that are multiples of any member of S, smaller or equal than the product and larger than zero.
Is there a formula for the cardinality of M given S?
E.g
S = {2,3}
M = {2,3,4,6}
cardinality of M = 4.
I can think of the formula for specific lengths but not of a general form.
multiplicative-function
asked Nov 22 '18 at 9:26
JFugger_jrJFugger_jr
133
Given an arbitrarily large set of natural numbers greater than one,
S = {$p_0$, $p_1$, ... $p_n$}
product of S = $prod_{i=0}^n p_i$
define M as the set of all natural numbers that are multiples of any member of S, smaller or equal than the product and larger than zero.
Is there a formula for the cardinality of M given S?
E.g
S = {2,3}
M = {2,3,4,6}
cardinality of M = 4.
I can think of the formula for specific lengths but not of a general form.
multiplicative-function
multiplicative-function
asked Nov 22 '18 at 9:26
JFugger_jrJFugger_jr
133
asked Nov 22 '18 at 9:26
JFugger_jrJFugger_jr
133
asked Nov 22 '18 at 9:26
JFugger_jrJFugger_jr
133
asked Nov 22 '18 at 9:26
JFugger_jrJFugger_jr
133
asked Nov 22 '18 at 9:26
JFugger_jrJFugger_jr
133
133
add a comment |
add a comment |
1 Answer
1
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oldest
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Let us consider all numbers from $1$ to $N=prod_k p_k$ identified with the elements of the ring $Bbb Z/Ncongprod_kBbb Z/p_k$ of integers modulo $N$. Then the elements divisible by one of the factors, including zero (i.e. $N$ modulo $N$), are the non-units. The number of the units is given by the Euler indicator
$$phi(N)=N;prod_kleft(1-frac 1{p_k}right) .$$
Now consider "the complement", to get the answer $N-phi(N)$. For example, for $N=6$ we get $6-phi(6)=6-2=4$.
answered Nov 22 '18 at 10:03
dan_fuleadan_fulea
6,3301312
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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Let us consider all numbers from $1$ to $N=prod_k p_k$ identified with the elements of the ring $Bbb Z/Ncongprod_kBbb Z/p_k$ of integers modulo $N$. Then the elements divisible by one of the factors, including zero (i.e. $N$ modulo $N$), are the non-units. The number of the units is given by the Euler indicator
$$phi(N)=N;prod_kleft(1-frac 1{p_k}right) .$$
Now consider "the complement", to get the answer $N-phi(N)$. For example, for $N=6$ we get $6-phi(6)=6-2=4$.
answered Nov 22 '18 at 10:03
dan_fuleadan_fulea
6,3301312
add a comment |
Let us consider all numbers from $1$ to $N=prod_k p_k$ identified with the elements of the ring $Bbb Z/Ncongprod_kBbb Z/p_k$ of integers modulo $N$. Then the elements divisible by one of the factors, including zero (i.e. $N$ modulo $N$), are the non-units. The number of the units is given by the Euler indicator
$$phi(N)=N;prod_kleft(1-frac 1{p_k}right) .$$
Now consider "the complement", to get the answer $N-phi(N)$. For example, for $N=6$ we get $6-phi(6)=6-2=4$.
answered Nov 22 '18 at 10:03
dan_fuleadan_fulea
6,3301312
add a comment |
Let us consider all numbers from $1$ to $N=prod_k p_k$ identified with the elements of the ring $Bbb Z/Ncongprod_kBbb Z/p_k$ of integers modulo $N$. Then the elements divisible by one of the factors, including zero (i.e. $N$ modulo $N$), are the non-units. The number of the units is given by the Euler indicator
$$phi(N)=N;prod_kleft(1-frac 1{p_k}right) .$$
Now consider "the complement", to get the answer $N-phi(N)$. For example, for $N=6$ we get $6-phi(6)=6-2=4$.
answered Nov 22 '18 at 10:03
dan_fuleadan_fulea
6,3301312
Let us consider all numbers from $1$ to $N=prod_k p_k$ identified with the elements of the ring $Bbb Z/Ncongprod_kBbb Z/p_k$ of integers modulo $N$. Then the elements divisible by one of the factors, including zero (i.e. $N$ modulo $N$), are the non-units. The number of the units is given by the Euler indicator
$$phi(N)=N;prod_kleft(1-frac 1{p_k}right) .$$
Now consider "the complement", to get the answer $N-phi(N)$. For example, for $N=6$ we get $6-phi(6)=6-2=4$.
answered Nov 22 '18 at 10:03
dan_fuleadan_fulea
6,3301312
answered Nov 22 '18 at 10:03
dan_fuleadan_fulea
6,3301312
answered Nov 22 '18 at 10:03
dan_fuleadan_fulea
6,3301312
answered Nov 22 '18 at 10:03
dan_fuleadan_fulea
6,3301312
6,3301312
add a comment |
add a comment |
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