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Eric Naslund
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This question is directly related to when $d(n)=d(n+1)$ where $d(n)$ denotes the divisor function.

Solutions to $d(n)=d(n+1)$:

In 1952, Erdos and Mirsky conjectured that $d(n)=d(n+1)$ has infinitely many solutions. In 1984, Heath Brown proved this result, and gave a lower bound on the counting function. Let Let $\widetilde{D}(x)$ denote the number of $n\leq x$ satisfying $d(n)=d(n+1)$. Heath Brown showed that $$\widetilde{D}(x)$\gg \frac{x}{(\log x)^7}.$$

In 1987 Erdős, Pomerance and Sárközy gave the upper bound $$\widetilde{D}(x)\ll \frac{x}{(\log \log x)^\frac{1}{2}}.$$

Later that year, Hildebrand improved Heath Browns Result that $$\widetilde{D}(x)\gg \frac{x}{(\log \log x)^3},$$ showing that the correct magnitude involves a doubly logarithmic factor.

Consecutive integers with identical prime signature:

Note that Erdős, Pomerance and Sárközy result immediately implies that your counting function is $$\ll \frac{x}{(\log \log x)^\frac{1}{2}},$$ and so it is not linear even though the graph resembles a line. This is because $\log \log x$ grows extremely slowly.

It seems likely that one could use Hildebrands lower bound to prove that the set of consecutive integers with identical prime signature is infinite. This is because $d(n)=d(n+1)$ "often" implies that $n$ and $n+1$ have the same signature.

Some References: (Chronological Ordering)

Eric Naslund
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  • 106