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) - Erdös, Mirsky 1952: [The distribution of the values of $d(n)$.][1] - Heath-Brown 1984: [The divisor function at consecutive integers.][2] - Erdős, Pomerance and Sárközy 1987 [On locally repeated values of certain arithmetic functions. III.][3] - Hildebrand 1987: [The divisor function at consecutive integers.][4] [1]: http://www.ams.org/mathscinet/search/publdoc.html?r=1&pg1=CNO&s1=49932&loc=fromrevtext [2]: http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=6990892 [3]: http://www.jstor.org/stable/2046541?origin=crossref [4]: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1102690578