# For every sufficiently large $m$ there exists $k$ such that $m = k + \tau(k)$

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.

We are looking at, a specific composition. Included, is a multiplicative partition( number of divisors, in this case). Highly composite numbers are likely to have the highest number of multiplicative partitions due to highest divisor number up to a point. These may not be so dense on the naturals however. 12 divisors comes about in one of 4 ways:$$p^{11}\\p^5q\\p^3q^2\\p^2qr$$