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 $\widetilde{D}(x)$ 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:

Let $\widetilde{\mathcal{P}}(x)$ denote the number of integers $n\leq x$ such that $n$ and $n+1$ have the same prime signature. Then $\widetilde{\mathcal{P}}(x)\leq \widetilde{D}(x)$, and so Erdős, Pomerance and Sárközy result immediately implies that $$\widetilde{\mathcal{P}}(x)\ll \frac{x}{(\log \log x)^\frac{1}{2}}.$$ This means that the counting function is not linear even though the graph resembles a straight line. ($\log \log x$ grows extremely slowly, and is nearly unnoticeable)

Since $d(n)=d(n+1)$ "often" implies that $n$ and $n+1$ have the same signature, it seems likely that one could use Hildebrands lower bound to prove that the set of consecutive integers with identical prime signature is infinite. Bounding the number of times we have $d(n)= d(n+1)$, yet difference signatures, seems like a fruitful approach.

Some References: (Chronological Ordering)

• Erdös, Mirsky 1952: The distribution of the values of $d(n)$.

• Heath-Brown 1984: The divisor function at consecutive integers.

• Erdős, Pomerance and Sárközy 1987: On locally repeated values of certain arithmetic functions. III.

• Hildebrand 1987: The divisor function at consecutive integers.

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 $\widetilde{D}(x)$ 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:

Let $\widetilde{\mathcal{P}}(x)$ denote the number of integers $n\leq x$ such that $n$ and $n+1$ have the same prime signature. Then $\widetilde{\mathcal{P}}(x)\leq \widetilde{D}(x)$, and so Erdős, Pomerance and Sárközy result immediately implies that $$\widetilde{\mathcal{P}}(x)\ll \frac{x}{(\log \log x)^\frac{1}{2}}.$$ This means that the counting function is not linear even though the graph resembles a straight line. ($\log \log x$ grows extremely slowly, and is nearly unnoticeable)

Since $d(n)=d(n+1)$ "often" implies that $n$ and $n+1$ have the same signature, it seems likely that one could use Hildebrands lower bound to prove that the set of consecutive integers with identical prime signature is infinite. Bounding the number of times we have $d(n)= d(n+1)$, yet difference signatures, seems like a fruitful approach.

Some References: (Chronological Ordering)

• Erdös, Mirsky 1952: The distribution of the values of $d(n)$.

• Heath-Brown 1984: The divisor function at consecutive integers.

• Erdős, Pomerance and Sárközy 1987: On locally repeated values of certain arithmetic functions. III.

• Hildebrand 1987: The divisor function at consecutive integers.

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 $d(n)$.

• Heath-Brown 1984: The divisor function at consecutive integers.

• Erdős, Pomerance and Sárközy 1987 On locally repeated values of certain arithmetic functions. III.

• Hildebrand 1987: The divisor function at consecutive integers.

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 $\widetilde{D}(x)$ 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:

Let $\widetilde{\mathcal{P}}(x)$ denote the number of integers $n\leq x$ such that $n$ and $n+1$ have the same prime signature. Then $\widetilde{\mathcal{P}}(x)\leq \widetilde{D}(x)$, and so Erdős, Pomerance and Sárközy result immediately implies that $$\widetilde{\mathcal{P}}(x)\ll \frac{x}{(\log \log x)^\frac{1}{2}}.$$ This means that the counting function is not linear even though the graph resembles a straight line. ($\log \log x$ grows extremely slowly, and is nearly unnoticeable)

Since $d(n)=d(n+1)$ "often" implies that $n$ and $n+1$ have the same signature, it seems likely that one could use Hildebrands lower bound to prove that the set of consecutive integers with identical prime signature is infinite. Bounding the number of times we have $d(n)= d(n+1)$, yet difference signatures, seems like a fruitful approach.

Some References: (Chronological Ordering)

• Erdös, Mirsky 1952: The distribution of the values of $d(n)$.

• Heath-Brown 1984: The divisor function at consecutive integers.

• Erdős, Pomerance and Sárközy 1987: On locally repeated values of certain arithmetic functions. III.

• Hildebrand 1987: The divisor function at consecutive integers.