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# perfect totient number

An integer $n$ is a perfect totient number if

$n=\sum_{{i=1}}^{{c+1}}\phi^{i}(n)$ |

, where $\phi^{i}(x)$ is the iterated totient function and $c$ is the integer such that $\phi^{c}(n)=2$.

A082897 in Sloane’s OEIS lists the first few perfect totient numbers: 3, 9, 15, 27, 39, 81, 111, 183, 243, 255, 327, 363, 471, 729, 2187, 2199, 3063, 4359, 4375, etc. It can be observed that many of these are multiples of 3 (in fact, 4375 is the smallest one that is not divisible by 3) and in fact all $3^{x}$ for $x>0$ are perfect totient numbers.

Furthermore, $3p$ for a prime $p>3$ is a perfect totient number if and only if $p=4n+1$, where $n$ itself is also a perfect totient number. Mohan and Suryanarayana showed why $3p$ can’t be a perfect totient number when $p\equiv 3\mod 4$. In regards to $3^{2}p$, Ianucci et al showed that if it is a perfect totient number then $p$ is a prime of one of three specific forms listen in their paper. It is not known if there are any perfect totient numbers of the form $3^{x}p$ for $x>3$.

# References

- 1 Perez Cacho, “On the sum of totients of successive orders,” Revista Matematica Hispano-Americana 5.3 (1939): 45 - 50
- 2 D. E. Ianucci, D. Moujie & G. L. Cohen, “On Perfect Totient Numbers” Journal of Integer Sequences, 6, 2003: 03.4.5
- 3 R. K. Guy, Unsolved Problems in Number Theory New York: Springer-Verlag 2004: B42

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## Comments

## Question about 4n + 1

Apologies in advance for the tenor of this question. In all likelihood it will appear to be an extremely gauche question of little importance distracting from topics considered more important, such as the curvature of space-time with an irrational number of dimensions. To me right now, it seems that the answer is obvious but I can't figure it out for some reason.

If an integer is of the form 4n + 1, doesn't that automatically rule out the possibility that it could be equiv 3 mod 4, that is, 4n - 1?

## Re: Question about 4n + 1

Yes, if n is also an integer, no otherwise. If n is 1/2 for example then 4n+1=3. But when n is an integer then 4n+1 = 1 modulo 4. Just follow the definition: (4n+1)-1=4n is divisible by 4. 1 is not 3 modulo 4. Proof: 3-1=2 is not divisible by 4 so 1 and 3 are not congruent modulo 4.

## Re: Question about 4n + 1

First, there is no need to apologize. There is nothing wrong

with your question. Had you been posting spam, that would

have been a different story. There is nothing inappropriate

in asking a mathematical question, no matter how elementary

or advanced it might be.

Second, the answer to your question is "yes". Assume we have an

integer x such that x = 4n + 1 for some n and also x = 4m - 1 for

some m. Then, we would have 4n + 1 = 4m - 1. Manipulating

algebra, this is equivlent to 2(n - m) = 1. This is impossible ---

twice an integer cannot equal 1. Hence the possibility is

ruled out.

## Re: Question about 4n + 1

I think I may have figured out the source of PrimeFan's confusion:

From my understanding of skimming the Ianucci paper Lisa mentions in her entry, I think what Mohan and Suryanarayana proved that there is some specific reason why if p is congruent to 3 mod 4 then 3p can't be a perfect totient number. There is some other reason why congruent to 0 or 2 it can't be a PTN. Of course I need to read the Mohan paper before saying this 100%.

## Re: Question about 4n + 1

You're exactly right, Anton, that's why I was confused. You're probably right about the 0 and 2 cases, too. Thanks for explaining that.