Mixing
On numbers with small $varphi(n)/n$
$begingroup$
Let $Phi(n) = varphi(n)/n = prod_{p|n}(p-1)/p$ be the "normalized totient" of $n$.
Some facts:
$Phi(p) = (p-1)/p < 1$ for prime numbers with $lim_{prightarrow infty}Phi(p) = 1$
$Phi(n) = 1/2$ iff $n$ is a power of $2$
$Phi(n) < 1/2$ for all even $n$ that are not powers of $2$ and some odd $n$
if $Phi(n) > 1/2$ then $n$ is odd
I have some questions concerning numbers with $Phi(n) < 1/2$:
Are there numbers with arbitrary small $Phi(n)$? Or is there a lower bound $Phi_{text{min}} > 0$?
Are there odd numbers with arbitrary small $Phi(n)$?
How can this astonishing regularity been explained when displaying in a square spiral only those numbers with $Phi(n) < 1/3$ – a regular pattern of triples pointing right, down, left, up clockwise (with some irregularily distributed defects of course):
Note that the regular background pattern vanishes when choosing values other than 1/3, e.g. 0.3 (left) or 0.4 (right):
Since the cases $Phi(n) < 1/2$ and $Phi(n) < 1/3$ display regular patterns, one might suspect that also $Phi(n) < 1/5$ gives rise to some regularity. But the numbers envolved in creating that pattern are too big, so I cannot visualize it.
- Supposed one would visualize $Phi(n) < 1/5$ which regular pattern would emerge (if any)?
elementary-number-theory prime-numbers prime-factorization totient-function
asked Jan 11 at 11:57
Hans StrickerHans Stricker
6,24343988
$endgroup$
|
show 6 more comments
$begingroup$
Let $Phi(n) = varphi(n)/n = prod_{p|n}(p-1)/p$ be the "normalized totient" of $n$.
Some facts:
$Phi(p) = (p-1)/p < 1$ for prime numbers with $lim_{prightarrow infty}Phi(p) = 1$
$Phi(n) = 1/2$ iff $n$ is a power of $2$
$Phi(n) < 1/2$ for all even $n$ that are not powers of $2$ and some odd $n$
if $Phi(n) > 1/2$ then $n$ is odd
I have some questions concerning numbers with $Phi(n) < 1/2$:
Are there numbers with arbitrary small $Phi(n)$? Or is there a lower bound $Phi_{text{min}} > 0$?
Are there odd numbers with arbitrary small $Phi(n)$?
How can this astonishing regularity been explained when displaying in a square spiral only those numbers with $Phi(n) < 1/3$ – a regular pattern of triples pointing right, down, left, up clockwise (with some irregularily distributed defects of course):
Note that the regular background pattern vanishes when choosing values other than 1/3, e.g. 0.3 (left) or 0.4 (right):
Since the cases $Phi(n) < 1/2$ and $Phi(n) < 1/3$ display regular patterns, one might suspect that also $Phi(n) < 1/5$ gives rise to some regularity. But the numbers envolved in creating that pattern are too big, so I cannot visualize it.
- Supposed one would visualize $Phi(n) < 1/5$ which regular pattern would emerge (if any)?
elementary-number-theory prime-numbers prime-factorization totient-function
asked Jan 11 at 11:57
Hans StrickerHans Stricker
6,24343988
$endgroup$
1
$begingroup$
Perhaps you should change the symbol for that normalization, as it is easy, I think, to confuse with Euler's function. Perhaps something like $;Phi(n);$ or perhaps even $;Psi(n);$ .
$endgroup$
– DonAntonio
Jan 11 at 12:04
1
$begingroup$
Done, thanks for the hint.
$endgroup$
– Hans Stricker
Jan 11 at 12:10
$begingroup$
Even for odd $n$, the value can get arbitary small since the product $$prod_{p prime} frac{p-1}{p}$$ diverges to $0$
$endgroup$
– Peter
Jan 11 at 12:14
$begingroup$
What might help for a deeper analyze is that we can replace $n$ by its radical (the product of the primes dividing $n$)
$endgroup$
– Peter
Jan 11 at 12:18
1
$begingroup$
Sorry, you are right. We have $$prod_{p prime, ple x} frac{p-1}{p}approx frac{e^{-gamma}}{ln(x)}$$ where $gamma$ is the Euler-Mascheroni-constant.
$endgroup$
– Peter
Jan 11 at 12:52
|
show 6 more comments
$begingroup$
Let $Phi(n) = varphi(n)/n = prod_{p|n}(p-1)/p$ be the "normalized totient" of $n$.
Some facts:
$Phi(p) = (p-1)/p < 1$ for prime numbers with $lim_{prightarrow infty}Phi(p) = 1$
$Phi(n) = 1/2$ iff $n$ is a power of $2$
$Phi(n) < 1/2$ for all even $n$ that are not powers of $2$ and some odd $n$
if $Phi(n) > 1/2$ then $n$ is odd
I have some questions concerning numbers with $Phi(n) < 1/2$:
Are there numbers with arbitrary small $Phi(n)$? Or is there a lower bound $Phi_{text{min}} > 0$?
Are there odd numbers with arbitrary small $Phi(n)$?
How can this astonishing regularity been explained when displaying in a square spiral only those numbers with $Phi(n) < 1/3$ – a regular pattern of triples pointing right, down, left, up clockwise (with some irregularily distributed defects of course):
Note that the regular background pattern vanishes when choosing values other than 1/3, e.g. 0.3 (left) or 0.4 (right):
Since the cases $Phi(n) < 1/2$ and $Phi(n) < 1/3$ display regular patterns, one might suspect that also $Phi(n) < 1/5$ gives rise to some regularity. But the numbers envolved in creating that pattern are too big, so I cannot visualize it.
- Supposed one would visualize $Phi(n) < 1/5$ which regular pattern would emerge (if any)?
elementary-number-theory prime-numbers prime-factorization totient-function
asked Jan 11 at 11:57
Hans StrickerHans Stricker
6,24343988
$endgroup$
Let $Phi(n) = varphi(n)/n = prod_{p|n}(p-1)/p$ be the "normalized totient" of $n$.
Some facts:
$Phi(p) = (p-1)/p < 1$ for prime numbers with $lim_{prightarrow infty}Phi(p) = 1$
$Phi(n) = 1/2$ iff $n$ is a power of $2$
$Phi(n) < 1/2$ for all even $n$ that are not powers of $2$ and some odd $n$
if $Phi(n) > 1/2$ then $n$ is odd
I have some questions concerning numbers with $Phi(n) < 1/2$:
Are there numbers with arbitrary small $Phi(n)$? Or is there a lower bound $Phi_{text{min}} > 0$?
Are there odd numbers with arbitrary small $Phi(n)$?
How can this astonishing regularity been explained when displaying in a square spiral only those numbers with $Phi(n) < 1/3$ – a regular pattern of triples pointing right, down, left, up clockwise (with some irregularily distributed defects of course):
Note that the regular background pattern vanishes when choosing values other than 1/3, e.g. 0.3 (left) or 0.4 (right):
Since the cases $Phi(n) < 1/2$ and $Phi(n) < 1/3$ display regular patterns, one might suspect that also $Phi(n) < 1/5$ gives rise to some regularity. But the numbers envolved in creating that pattern are too big, so I cannot visualize it.
- Supposed one would visualize $Phi(n) < 1/5$ which regular pattern would emerge (if any)?
elementary-number-theory prime-numbers prime-factorization totient-function
elementary-number-theory prime-numbers prime-factorization totient-function
asked Jan 11 at 11:57
Hans StrickerHans Stricker
6,24343988
asked Jan 11 at 11:57
Hans StrickerHans Stricker
6,24343988
edited Jan 11 at 15:32
Hans Stricker
asked Jan 11 at 11:57
Hans StrickerHans Stricker
6,24343988
asked Jan 11 at 11:57
Hans StrickerHans Stricker
6,24343988
asked Jan 11 at 11:57
Hans StrickerHans Stricker
6,24343988
6,24343988
1
$begingroup$
Perhaps you should change the symbol for that normalization, as it is easy, I think, to confuse with Euler's function. Perhaps something like $;Phi(n);$ or perhaps even $;Psi(n);$ .
$endgroup$
– DonAntonio
Jan 11 at 12:04
1
$begingroup$
Done, thanks for the hint.
$endgroup$
– Hans Stricker
Jan 11 at 12:10
$begingroup$
Even for odd $n$, the value can get arbitary small since the product $$prod_{p prime} frac{p-1}{p}$$ diverges to $0$
$endgroup$
– Peter
Jan 11 at 12:14
$begingroup$
What might help for a deeper analyze is that we can replace $n$ by its radical (the product of the primes dividing $n$)
$endgroup$
– Peter
Jan 11 at 12:18
1
$begingroup$
Sorry, you are right. We have $$prod_{p prime, ple x} frac{p-1}{p}approx frac{e^{-gamma}}{ln(x)}$$ where $gamma$ is the Euler-Mascheroni-constant.
$endgroup$
– Peter
Jan 11 at 12:52
|
show 6 more comments
1
$begingroup$
Perhaps you should change the symbol for that normalization, as it is easy, I think, to confuse with Euler's function. Perhaps something like $;Phi(n);$ or perhaps even $;Psi(n);$ .
$endgroup$
– DonAntonio
Jan 11 at 12:04
1
$begingroup$
Done, thanks for the hint.
$endgroup$
– Hans Stricker
Jan 11 at 12:10
$begingroup$
Even for odd $n$, the value can get arbitary small since the product $$prod_{p prime} frac{p-1}{p}$$ diverges to $0$
$endgroup$
– Peter
Jan 11 at 12:14
$begingroup$
What might help for a deeper analyze is that we can replace $n$ by its radical (the product of the primes dividing $n$)
$endgroup$
– Peter
Jan 11 at 12:18
1
$begingroup$
Sorry, you are right. We have $$prod_{p prime, ple x} frac{p-1}{p}approx frac{e^{-gamma}}{ln(x)}$$ where $gamma$ is the Euler-Mascheroni-constant.
$endgroup$
– Peter
Jan 11 at 12:52
1
1
$begingroup$
Perhaps you should change the symbol for that normalization, as it is easy, I think, to confuse with Euler's function. Perhaps something like $;Phi(n);$ or perhaps even $;Psi(n);$ .
$endgroup$
– DonAntonio
Jan 11 at 12:04
$begingroup$
Perhaps you should change the symbol for that normalization, as it is easy, I think, to confuse with Euler's function. Perhaps something like $;Phi(n);$ or perhaps even $;Psi(n);$ .
$endgroup$
– DonAntonio
Jan 11 at 12:04
1
1
$begingroup$
Done, thanks for the hint.
$endgroup$
– Hans Stricker
Jan 11 at 12:10
$begingroup$
Done, thanks for the hint.
$endgroup$
– Hans Stricker
Jan 11 at 12:10
$begingroup$
Even for odd $n$, the value can get arbitary small since the product $$prod_{p prime} frac{p-1}{p}$$ diverges to $0$
$endgroup$
– Peter
Jan 11 at 12:14
$begingroup$
Even for odd $n$, the value can get arbitary small since the product $$prod_{p prime} frac{p-1}{p}$$ diverges to $0$
$endgroup$
– Peter
Jan 11 at 12:14
$begingroup$
What might help for a deeper analyze is that we can replace $n$ by its radical (the product of the primes dividing $n$)
$endgroup$
– Peter
Jan 11 at 12:18
$begingroup$
What might help for a deeper analyze is that we can replace $n$ by its radical (the product of the primes dividing $n$)
$endgroup$
– Peter
Jan 11 at 12:18
1
1
$begingroup$
Sorry, you are right. We have $$prod_{p prime, ple x} frac{p-1}{p}approx frac{e^{-gamma}}{ln(x)}$$ where $gamma$ is the Euler-Mascheroni-constant.
$endgroup$
– Peter
Jan 11 at 12:52
$begingroup$
Sorry, you are right. We have $$prod_{p prime, ple x} frac{p-1}{p}approx frac{e^{-gamma}}{ln(x)}$$ where $gamma$ is the Euler-Mascheroni-constant.
$endgroup$
– Peter
Jan 11 at 12:52
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The answer seems simple: Having a closer look at the numbers in the spiral reveals that most of them are - not surprisingly - multiples of $6 = 2cdot 3$:
And the spiral forces the multiples of $6$ to arrange in triples (except along the lower right diagonal):
But not all multiples of $6$ have $Phi(n) < 1/3$, e.g. $n = 2^kcdot 3$, and not all $n$ with $Phi(n) < 1/3$ are multiples of $6$, the smallest one being $770 = 2cdot 5cdot 7 cdot 11$.
To answer partly the last of my questions: This is how the integers divisible by 8 and 10 are distributed (the second picture giving the blueprint for the case $Phi(n) < 1/5$):
Note that in the right picture the "arrows" (5-tuples) go counter-clockwise.
answered Jan 11 at 14:27
Hans StrickerHans Stricker
6,24343988
$endgroup$
add a comment |
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$begingroup$
The answer seems simple: Having a closer look at the numbers in the spiral reveals that most of them are - not surprisingly - multiples of $6 = 2cdot 3$:
And the spiral forces the multiples of $6$ to arrange in triples (except along the lower right diagonal):
But not all multiples of $6$ have $Phi(n) < 1/3$, e.g. $n = 2^kcdot 3$, and not all $n$ with $Phi(n) < 1/3$ are multiples of $6$, the smallest one being $770 = 2cdot 5cdot 7 cdot 11$.
To answer partly the last of my questions: This is how the integers divisible by 8 and 10 are distributed (the second picture giving the blueprint for the case $Phi(n) < 1/5$):
Note that in the right picture the "arrows" (5-tuples) go counter-clockwise.
answered Jan 11 at 14:27
Hans StrickerHans Stricker
6,24343988
$endgroup$
add a comment |
$begingroup$
The answer seems simple: Having a closer look at the numbers in the spiral reveals that most of them are - not surprisingly - multiples of $6 = 2cdot 3$:
And the spiral forces the multiples of $6$ to arrange in triples (except along the lower right diagonal):
But not all multiples of $6$ have $Phi(n) < 1/3$, e.g. $n = 2^kcdot 3$, and not all $n$ with $Phi(n) < 1/3$ are multiples of $6$, the smallest one being $770 = 2cdot 5cdot 7 cdot 11$.
To answer partly the last of my questions: This is how the integers divisible by 8 and 10 are distributed (the second picture giving the blueprint for the case $Phi(n) < 1/5$):
Note that in the right picture the "arrows" (5-tuples) go counter-clockwise.
answered Jan 11 at 14:27
Hans StrickerHans Stricker
6,24343988
$endgroup$
add a comment |
$begingroup$
The answer seems simple: Having a closer look at the numbers in the spiral reveals that most of them are - not surprisingly - multiples of $6 = 2cdot 3$:
And the spiral forces the multiples of $6$ to arrange in triples (except along the lower right diagonal):
But not all multiples of $6$ have $Phi(n) < 1/3$, e.g. $n = 2^kcdot 3$, and not all $n$ with $Phi(n) < 1/3$ are multiples of $6$, the smallest one being $770 = 2cdot 5cdot 7 cdot 11$.
To answer partly the last of my questions: This is how the integers divisible by 8 and 10 are distributed (the second picture giving the blueprint for the case $Phi(n) < 1/5$):
Note that in the right picture the "arrows" (5-tuples) go counter-clockwise.
answered Jan 11 at 14:27
Hans StrickerHans Stricker
6,24343988
$endgroup$
The answer seems simple: Having a closer look at the numbers in the spiral reveals that most of them are - not surprisingly - multiples of $6 = 2cdot 3$:
And the spiral forces the multiples of $6$ to arrange in triples (except along the lower right diagonal):
But not all multiples of $6$ have $Phi(n) < 1/3$, e.g. $n = 2^kcdot 3$, and not all $n$ with $Phi(n) < 1/3$ are multiples of $6$, the smallest one being $770 = 2cdot 5cdot 7 cdot 11$.
To answer partly the last of my questions: This is how the integers divisible by 8 and 10 are distributed (the second picture giving the blueprint for the case $Phi(n) < 1/5$):
Note that in the right picture the "arrows" (5-tuples) go counter-clockwise.
answered Jan 11 at 14:27
Hans StrickerHans Stricker
6,24343988
edited Jan 11 at 15:24
answered Jan 11 at 14:27
Hans StrickerHans Stricker
6,24343988
answered Jan 11 at 14:27
Hans StrickerHans Stricker
6,24343988
answered Jan 11 at 14:27
Hans StrickerHans Stricker
6,24343988
6,24343988
add a comment |
add a comment |
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$begingroup$
Perhaps you should change the symbol for that normalization, as it is easy, I think, to confuse with Euler's function. Perhaps something like $;Phi(n);$ or perhaps even $;Psi(n);$ .
$endgroup$
– DonAntonio
Jan 11 at 12:04
1
$begingroup$
Done, thanks for the hint.
$endgroup$
– Hans Stricker
Jan 11 at 12:10
$begingroup$
Even for odd $n$, the value can get arbitary small since the product $$prod_{p prime} frac{p-1}{p}$$ diverges to $0$
$endgroup$
– Peter
Jan 11 at 12:14
$begingroup$
What might help for a deeper analyze is that we can replace $n$ by its radical (the product of the primes dividing $n$)
$endgroup$
– Peter
Jan 11 at 12:18
1
$begingroup$
Sorry, you are right. We have $$prod_{p prime, ple x} frac{p-1}{p}approx frac{e^{-gamma}}{ln(x)}$$ where $gamma$ is the Euler-Mascheroni-constant.
$endgroup$
– Peter
Jan 11 at 12:52