Totient Function and phi(n)==phi(2n) for odd n

[timo1023]

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):

\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}

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

\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}

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.

[DeaconJohn]

That's interesting. I'll try to get something to get back to you with.

[DeaconJohn]

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):

\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}

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

\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}

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.

This is no more than a comment. It is not an answer. Hence best phrased as a question.

You ask "How can the phi function know to compensate for the number of prime numbers between n and 2n?" But, does the data indicate that it really does that as opposed to compensating for the number of co-prime numbers between n and 2n?

Nothing more than a (hopefully) seminal question, but, I really would like it if you could give me your best shot at an answer.

DJ

[timo1023]

Upon further investigation, it seems as though you are right. Here is what I came up with (please forgive my lack of precision in stating x is a positive integer, etc.):


\phi (2n)=\left{|}\{x|x\leq 2n-1\}-\{x|x \textrm{ is even}\} - \{i\cdot x | x \textrm{ odd}, x|n, i \textrm{ odd}\}\right{|}


So, for n=15, we have:


\phi (30) = |\{1,\dots, 29\}-\{\textrm{evens}\}-\{3,9,15,21,27,5,25\}|


\phi (30) = | 29-14-7 | = 8


As we already know, \phi (30)=\phi (15)=8, and thus this method works. So it would seem that it has nothing to do with the number of primes between 1 and 2n.

Please give me your thoughts.