TITLE: For every sufficiently large $m$ there exists $k$ such that $m = k + \tau(k)$
QUESTION [6 upvotes]: Let $\tau(k)$ , be the number of positive divisors of natural number $k$. Is it true, that there exists $n_0$ , such that for every $m\geq n_0$ there exists $k  \in\mathbb{N}$ such that:
$$
m = k + \tau(k)
$$
I have tried to use the following formula for $\tau$:
$$
\tau(p_1^{k_1}\ldots p_{s}^{k_s}) = (k_1 +1)\cdot\ldots \cdot(k_s +1),
$$
Where $p_1, \ldots, p_s$ are different prime numbers.
Intuitively, I think that the answer will be no. So, we can assume the contrary (that such $n_0$ exists) and try some infinite series of numbers (primes, factorials, primorials, etc.), which can't be written in the form $k + \tau(k)$ for every $k$. But my attempts weren't successful.
So, I will be grateful for hints and ideas.

REPLY [2 votes]: contest problem related to divisor function 
Look at Loppukilpailija's answer. I think he solved your problem negatively.