TITLE: Asymptotics of A283190
QUESTION [9 upvotes]: OEIS sequence A283190 gives the  number of different values $n \mod k$ for $1 \le k \le \lfloor n/2 \rfloor$.  Yes, I know this
is taking $\mod k$ as a function rather than an equivalence relation:  $n \mod k$ is interpreted as the value $y \in [0,1,\ldots,k-1]$ with $n \equiv y \mod k$.  Thus $A283190(14) = 3$ with $14 \equiv 0 \mod k$ for $k = 1,2,7$, and
$14\equiv 2$ for $k=3, 4, 6$, and finally $14\equiv 4$ for $k=5$. 
Equivalently, $A283190(n)$ is $1$ plus the number of $y \in [1,2,\ldots, \lfloor n/2\rfloor]$ such that 
 the least prime divisor of $n-y$ is less than $(n-y)/y$. 
Empirically, it appears that $A283190(n)/n$ converges to a constant, approximately $0.2296$.  Since we know something about the statistical distribution of least prime divisors, I'm guessing this may be provable. 
What is the actual value of the constant?

REPLY [2 votes]: When $k\in(n/3,n/2]$, so $n=2k+r$, the remainders include half the numbers from $0$ to $n/3$.
When $k\in(n/4,n/3]$, so $n=3k+r$, the remainders include a third of the numbers from $0$ to $n/4$.
No new remainders happen for $k\in(n/5,n/4]$ as $4k+r=2(2k)+r$;
a fifth of the numbers from $0$ to $n/6$;...  
So the remainders include:
* $(1-\frac12)$ the numbers from $[n/4,n/3)$;
* $(1-\frac12\frac23)$ the numbers from $[n/6,n/4)$;
* $(1-\frac12\frac23\frac45)$ the numbers from $[n/8,n/6)$;
* $(1-\frac12\frac23\frac45\frac67)$ the numbers from $[n/12,n/8)$  
I get two ungainly forms for the fraction:
$$\frac12\times\frac13+\frac14(\frac12\times\frac13)+\frac16(\frac12\frac23\frac15)+\frac18(\frac12\frac23\frac45\frac17)+\frac1{12}(\frac12\frac23\frac45\frac67\frac1{11})+...\\
=\sum_{p\in P}\frac1{p(p+1)}\prod_{q<p\\q\in P}\frac{q-1}q$$
where $p$ and $q$ are taken over primes.