# On nontotient Fibonacci numbers

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.

• Why in particular $F_{6t}$? Why not all $F_n$? Commented Aug 2, 2020 at 23:41
• Because only $6t$ indices can be totient except beginning of sequence. Commented Aug 3, 2020 at 5:32

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].