How can I prove this conjecture about a p-adic valuation? Define $$f(n)=5^n+6^n+10^n$$ for a positive integer $n$. I want to find out $v_7(f(n))$ , in other words the exponent belonging to $7$ in the prime factorization of $f(n)$. It seems that we simply have $$v_7(f(n))=v_7(n)+1$$ in the case $7\mid f(n)$ which is equivalent to $n$ not being divisible by $3$.

Is this correct for every positive integer $n$ not disible by $3$ ? If yes , how can it be proven ?

An analogue claim also holds for the prime factor $3$ , in this case using the $3$-valuation of $2^n+1$ can be used to prove it but here we have $3$ summands. I also tried to use the recurrence relation $$f(n)=21f(n-1)-140f(n-2)+300f(n-3)$$ but I could not prove my conjecture this way. It holds upto $n=10^6$.
 A: I wish I had more time to think about this, so I can only give a partial answer. Hopefully it helps to steer in a right direction.
Just to help clarify my perspective on the problem, I'll describe how I'm thinking about it first. If we fix a sixth root of unity $\zeta \in \mathbb{Q}_7$ we have for fixed $m$,
$$g(m):=\lim_{k \to \infty}f(7^km)=\zeta^m+\zeta^{3m}+\zeta^{5m}$$
This only depends on $m \mod 6$, and we have $g(0)=3$, $g(3)=-3$ and $g(m)=0$ otherwise.
Now this is basically how I think of $f(n)$ as $n \to 0$ in $\mathbb{N}^+$ (with the $7$-adic metric of course). Each power of $7$ is getting us closer to this limit of the Teichmuller characters.
More concretely, if $a$ is order $t$ modulo $p$, $a^t \equiv 1 \mod p$, then we can successively lift to $\omega \in \mathbb{Q}_p$ with $\omega$ having order $t$, $\omega^t=1$ by the following procedure of iteratively raising to the $p$th power:
$$a^{p^s} = \omega \mod p^{s+1}$$
In terms of the $p$-adic valuation this means,
$$v_p(a^{p^s}-\omega) \ge s+1$$
Hopefully it's not too much of a jump to then say that when $m$ is not divisible by $3$ we have $g(m)=0$ and so,
$$v_7(f(7^km)) = v_7(f(7^km)-g(m)) \ge k + 1$$
Which establishes half of the equality,
$$v_7(f(n)) \ge v_7(n) + 1$$
If I've been too terse or something doesn't work, I'll try to clarify/rectify it if possible this weekend.
