# proof that the sum of the iterated totient function is always odd

Given a positive integer $n$, it is always the case that

$$2|\u0338\sum _{i=1}^{c+1}{\varphi}^{i}(n),$$ |

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

Accepting as proven that $n>\varphi (n)$ and $2|\varphi (n)$ for $n>2$, it is clear that summing up the iterates of the totient function up to $c$ is summing up a series of even numbers^{} in descending order and that this sum is therefore itself even. Then, when we add the $c+1$ iterate, the sum turns odd.

As a bonus, this proves that no even number can be a perfect totient number.

Title | proof that the sum of the iterated totient function is always odd |
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Canonical name | ProofThatTheSumOfTheIteratedTotientFunctionIsAlwaysOdd |

Date of creation | 2013-03-22 16:34:26 |

Last modified on | 2013-03-22 16:34:26 |

Owner | PrimeFan (13766) |

Last modified by | PrimeFan (13766) |

Numerical id | 6 |

Author | PrimeFan (13766) |

Entry type | Proof |

Classification | msc 11A25 |