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## Main Question or Discussion Point

Hello,

Last year in number theory I discovered something with Euler's Totient Function that I couldn't explain. I asked my professor and he couldn't figure it out either. Here is what I discovered:

As a premise, it should be clear that phi(n)==phi(2n) where n is an odd, positive, integer. Using the formula for Euler's totient function, this is clear. However, looking at the numbers, it becomes more complex. For example, let n = 9. We know that phi(9)=phi(18)=6.

So we have the pos. integers from 1 to 9 ( n in red, coprime integers in blue, noncoprime integers in orange). Note that there are 4 primes in this range (2,3,5,7):

[itex]\definecolor{orange}{rgb}{1,0.5,0} {\color{blue}1}, {\color{blue}2}, {\color{orange}3}, {\color{blue}4}, {\color{blue}5}, {\color{orange}6}, {\color{blue}7}, {\color{blue}8}, {\color{red}9} [/itex]

Next, the integers from 1 to 18 for n=18. Note there are only 3 new primes from 9-18

[itex]\definecolor{orange}{rgb}{1,0.5,0} {\color{blue}1}, {\color{orange}2}, {\color{orange}3}, {\color{orange}4}, {\color{blue}5}, {\color{orange}6}, {\color{blue}7}, {\color{orange}8}, {\color{orange}9}, {\color{orange}10}, \color{blue}11}, \color{orange}12}, \color{blue}13}, \color{orange}14}, \color{orange}15}, \color{orange}16}, \color{blue}17}, \color{red}18}[/itex]

I'm having a hard time phrasing this, but I think what I mean to say is this: How can the phi function know to compensate for the number of prime numbers between n and 2n? Wouldn't this be related to the density of primes? Also, could this be used to prove that there exist infinite prime numbers?

I'm sorry If my wording is hard to understand. Once I gain some more understanding, I would like to write it up properly and show it to my math teacher or something.

Thanks in advance.

Last year in number theory I discovered something with Euler's Totient Function that I couldn't explain. I asked my professor and he couldn't figure it out either. Here is what I discovered:

As a premise, it should be clear that phi(n)==phi(2n) where n is an odd, positive, integer. Using the formula for Euler's totient function, this is clear. However, looking at the numbers, it becomes more complex. For example, let n = 9. We know that phi(9)=phi(18)=6.

So we have the pos. integers from 1 to 9 ( n in red, coprime integers in blue, noncoprime integers in orange). Note that there are 4 primes in this range (2,3,5,7):

[itex]\definecolor{orange}{rgb}{1,0.5,0} {\color{blue}1}, {\color{blue}2}, {\color{orange}3}, {\color{blue}4}, {\color{blue}5}, {\color{orange}6}, {\color{blue}7}, {\color{blue}8}, {\color{red}9} [/itex]

Next, the integers from 1 to 18 for n=18. Note there are only 3 new primes from 9-18

[itex]\definecolor{orange}{rgb}{1,0.5,0} {\color{blue}1}, {\color{orange}2}, {\color{orange}3}, {\color{orange}4}, {\color{blue}5}, {\color{orange}6}, {\color{blue}7}, {\color{orange}8}, {\color{orange}9}, {\color{orange}10}, \color{blue}11}, \color{orange}12}, \color{blue}13}, \color{orange}14}, \color{orange}15}, \color{orange}16}, \color{blue}17}, \color{red}18}[/itex]

I'm having a hard time phrasing this, but I think what I mean to say is this: How can the phi function know to compensate for the number of prime numbers between n and 2n? Wouldn't this be related to the density of primes? Also, could this be used to prove that there exist infinite prime numbers?

I'm sorry If my wording is hard to understand. Once I gain some more understanding, I would like to write it up properly and show it to my math teacher or something.

Thanks in advance.