On nontotient Fibonacci numbers

Question

This question is related to sequence of numbers $t$ such that $F_{6t}$ is a nontotient where $F_n$ represents the sequence of Fibonacci numbers for $n\geq 0$.

The online encyclopedia Wikipedia has the articles Fibonacci number and Nontotient, respectively.

There are very few terms in https://oeis.org/A335976 and some prime numbers appear as initial terms as expected.

Conjecture. There are infinitely many numbers $t$ such that $F_{6t}$ is a nontotient.

Question. Can someone prove or disprove above conjecture?

Initial terms of sequence of composite numbers $t_{c}$ such that $F_{6t_{c}}$ is a nontotient are also very welcome as helpful comment. Additionally, I couldn't find that question in literature yet, but if one can find references that have these or strongly related results, I will be very grateful for this response.

Thanks.

Answer 1

I've extended OEIS A335976 with many terms. The numerical data so far is in favor of the conjecture, although I think it may be hard to prove it rigorously.

Still, we can note a few major factors that limit the possibility for $F_{6t}$ be totient:

• the small value of $\nu_2(F_{6t})$;
• the small value of $\Omega(F_{6t})$;
• abundance of prime factors $\equiv 1\pmod3$.

Essentially $F_{6t}$ is totient iff we can partition the prime factors of $F_{6t}$ into $\nu_2(F_{6t})$ or smaller number of subsets (each including at least one prime $2$) such that the product in each subset is $\varphi(q)$ for some prime power $q$. Furthermore, if a subset has just one prime $\equiv 2\pmod3$ the corresponding $q$ (if exists) cannot be prime, which makes its existence less likely.

We have $\nu_2(F_{6t})=3$ for odd $t$ and $\nu_2(F_{6t})\geq 4$ for even $t$, and that explains why we mostly see odd terms in A335976. As for $\Omega(F_{6t})$ values, those from A335976 have these values in the interval $[7,24]$ with more than half concentrated in the "middle" subinterval [14,17].