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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.

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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:$$p^{11}\\p^5q\\p^3q^2\\p^2qr$$

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  • $\begingroup$ I have understood this observation before. But I don't know how to use it... $\endgroup$ – Mikhail Goltvanitsa Jun 10 at 18:05
  • $\begingroup$ show that most numbers m can be written as a highly composite number plus a natural number with a highly composite number of divisors. $\endgroup$ – Roddy MacPhee Jun 10 at 18:37

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