There are few posts asking if factoring $N-1$ can help factor $N$. In those posts the focus was on the factors of $N-1$ and $N$ which can never be the same. The conclusion therefore was that factoring $N-1$ cannot help factor $N$.
In this post the focus is shifted to the numbers $N$ and $N-1$ neighbors. It turned out that when the two numbers have a neighbor of a special form with the same numerical value it becomes possible to use factoring $N-1$ to factor $N$.
We consider numbers of the form $N=pq$ with $p,q$ primes of the form $p=6k_{1}+1$ and $q=6k_{2}+1$. Then $N$ can be written as $N=36k_{1}k_{2} + 6(k_{1} + k_{2}) + 1$.
It turned out that many numbers share the same value of $36k_{1}k_{2}$ though they have of course different factors. We will show below how we go from factoring $N-1=90$ to factoring $N=91$.
$91=7*13=(6*1+1)(6*2+1)= 6*6*2 + 6*(1+2) +1 = 6*12 +3*6 +1= 72 + 19$
$90=9*10= (8 + 1)(9 + 1)= 8*9 + 18 = 72 + 18 = 6*12 + 18$
Since $91$ and $90$ have the same value of $36k_{1}k_{2}=72$, it is therefore possible to use the same value of $72$ to get that of $91$ by writing $91= (6+1)*(12+1)$ simply because $8*9=6*12=72$ since $72$ can be written as $72=(p-1)(q-1)=6*12$
Basically it was possible to use factoring of $N-1$ to factor $N$ by using the common value of $72$ which is $72=(p-1)(q-1)$ for $91=pq$ and $72=(f-1)(h-1)$ for $90=fg$. But there are more numbers having the same value of $72$ because $72$ can be written as:
$72=1*72=2*36=3*24=4*18=6*12=8*9$ giving the corresponding numbers $M=146,111,100,95,91,90$. However $90$ was the most useful since we were looking for numbers of the form $N-1$ to see if they can help factor $N$. But $95=91+4$ would have done the job too but we couldn't know in advance which number would share the value of $36k_{1}k_{2}$ of $N=91$.
For example $133=7*19$ share the same value of $36k_{1}k_{2}=108=6*18=9*12$ as $N-3=130=10*13$.
Large numbers of the form $N=pq$ with $p,q$ primes would have a large value of $36k_{1}k_{2}$ whose decomposition will produce many more numbers that could in theory be possible candidates to play the role of $(N-1),(N-2),(N-3)...$. What we don't want is replacing the difficulty of factoring $N$ with the difficulty of finding a number $(N-x)$ or $(N+x)$ that can be easily factored. These numbers are not of the form $L=pq$ like the example of $90=2*3^2*5$ or $130=2*5*13$ so the hope of them being easier to factor is there.
The main question is: How easy is-it to find and factor one of those candidates when they are large, that is when we are dealing with numbers with hundreds of digits? We may still spend time searching for a good candidate by varying $x$ in $(N-x)$ or $(N+x)$.