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Does OEIS sequence A252502 contain all even numbers which are totients and all odd numbers $n$ with $n-1$ totients?

数论 Math StackExchange 0 票 1 回答 57 浏览 提问者: Richard Chen 2026-07-30 12:33
number-theory

问题内容

For even number $n$, if $n$ is not a totient (i.e. not in the range of Euler totient function), then $n$ is not in OEIS sequence A252502, but if $n$ is a totient, must be $n$ in A252502?

For odd number $n$, if $n-1$ is not a totient (i.e. not in the range of Euler totient function), then $n$ is not in OEIS sequence A252502 (except $n=1$), but if $n-1$ is a totient, must be $n$ in A252502?

If this conjecture is true, then the famous conjecture that there is no $n$ such that there is only one $x$ whose Euler totient function is $n$ is also true.

回答 (1)

Oscar Lanzi 0 票 2026-07-30 14:08 原文

For numbers that are twice a prime, consider $2×3=6$ as an example. Since this is twice an odd prime and also one less than the odd prime $7$, it is the Euler totient of $14$. Then we have a cyclotomic polynomial $x^6-x^5+x^4-x^3+x^2-x+1$ of order $14$, so constructed that for $x=10$ the value lies strictly between $9×10^5$ and $10^6$ hence having six digits. Thus $6$ enters the sequence, and similarly for any number that is both twice a prime and one less than a larger prime.