# $1$ as difference of composites with same number of prime factors

I noticed and found only first three cases:

We can write $1$ as difference of two composites that have one prime factor $$3^2-2^3=1$$

and as difference of two composites that have two prime factors $$3\cdot 5 - 7\cdot 2 = 1$$

and as difference of two composites that have three prime factors $$2^2 \cdot 3^2 \cdot 43-7 \cdot 13 \cdot 17=1$$

I believe that this holds for every $k \in \mathbb N$, that is, that for every $k \in \mathbb N$ there exist composites $a_k$ and $b_k$ that have exactly $k$ prime factors and are such that we have $a_k-b_k=1$.

Is my belief true? Is this known? What is known about all of this and similar problems? Can someone find solutions for some larger $k$´s?

Examples exist at least for $k=1,2,...11$ by this list .

• Also asked on MSE: math.stackexchange.com/questions/2700917/… – Shalom Mar 22 '18 at 8:58
• Why close? This seems to be pretty research level. – Fedor Petrov Mar 22 '18 at 9:49
• The smallest example for $k=3$ is $3\cdot7\cdot11-2\cdot5\cdot23=231-230=1$. – Gerry Myerson Mar 22 '18 at 12:03
• @GerryMyerson And the smallest examples, so far, for $k=1,2,3$, all three have $3$ as a factor of a bigger number, and $2$ as a factor of a smaller number. – Shalom Mar 22 '18 at 12:08
• The smallest example for $k=4$ is $7315-7314=1$ (and 3 is not a factor of 7315). – Gerry Myerson Mar 22 '18 at 12:11

"It is stil unkown if the equation $\omega(n+1)=\omega(n)$ has infinitely many solutions...It is known that $|\omega(n+1)-\omega(n)|$ takes a certain value infinitely often".
Here $\omega(x)$ denotes the number of prime factors of $x$.
• Do you know is $\omega(n)=\omega(n+k)$ open for every $k \in \mathbb N$, in the sense of infinity of solutions? – Shalom Mar 22 '18 at 10:19