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This limit is 0.

The reason is that the central binomials are almost always divisible by 3, 5 etc (and by every specific prime). To be more precise, we have

Observation. for every fixed $p$ we have $\bigl|n\leqslant x: p\nmid{2n\choose n}\bigr|=o(x)$.

Proof. Recall that by Kummer's theorem $p$ does not divide ${n+k\choose k}$ if and only if the base $p$ expansions $n=\sum c_i p^i$ and $k=\sum b_i p^i$, $0\leqslant b_i,c_i\leqslant p-1$, satisfy $b_i+c_i\leqslant p-1$ for all $i$ (i.e., there are no carries if we sum up $n$ and $k$ in base $p$). For ${2n\choose n}$ this means that all base $p$ digits of $n$ must be less than $p/2$, there are at most $\lceil p/2\rceil^{1+\log_p x}=o(x)$ such numbers not exceeding $x$.

Thus, for every fixed $M$ the number of $n\leqslant x$ for which there exists prime $p<M$ not dividing ${2n\choose n}$ is $o(x)$. Fix $M$ and call such $n$ $M$-strange.

Counting numbers of the form $p+{2n\choose n}$ not exceeding $x$, we note that there exist $O(x/\log x)$ choices of $p$ and $o(\log x)$ choices of $M$-strange $n$. Totally this gives $o(x)$ numbers of the form $p+{2n\choose n}$ with $M$-strange $n$. For not $M$-strange $n$, the number $y=p+{2n\choose n}\leqslant x$ either satisfies $p\leqslant M$ (there are $O(\log x)$ such numbers), or $y$ does not have a prime divisor $q<M$, since $q$ divides ${2n\choose n}$ and does not divide $p$ (there are at most $x\cdot \prod_{p<M}(1-1/p)+o(x)$ such numbers $y$). Since the product $\prod_p (1-1/p)$ equals 0, we get that there are only $o(x)$ numbers less than $x$ of the form $p+{2n\choose n}$.

The limit (not only liminf) $s_1$ equals 0, i. e., $|S\cap [1,x]|=o(x)$.

The reason is that the central binomials are almost always divisible by 3, 5 etc (and by every specific prime). To be more precise, we have

Observation. for every fixed $p$ we have $\bigl|n\leqslant x: p\nmid{2n\choose n}\bigr|=o(x)$.

Proof. Recall that by Kummer's theorem $p$ does not divide ${n+k\choose k}$ if and only if the base $p$ expansions $n=\sum c_i p^i$ and $k=\sum b_i p^i$, $0\leqslant b_i,c_i\leqslant p-1$, satisfy $b_i+c_i\leqslant p-1$ for all $i$ (i.e., there are no carries if we sum up $n$ and $k$ in base $p$). For ${2n\choose n}$ this means that all base $p$ digits of $n$ must be less than $p/2$, there are at most $\lceil p/2\rceil^{1+\log_p x}=o(x)$ such numbers not exceeding $x$.

Thus, for every fixed $M$ the number of $n\leqslant x$ for which there exists prime $p<M$ not dividing ${2n\choose n}$ is $o(x)$. Fix $M$ and call such $n$ $M$-strange.

Counting numbers of the form $p+{2n\choose n}$ not exceeding $x$, we note that there exist $O(x/\log x)$ choices of $p$ and $o(\log x)$ choices of $M$-strange $n$. Totally this gives $o(x)$ numbers of the form $p+{2n\choose n}$ with $M$-strange $n$. For not $M$-strange $n$, the number $y=p+{2n\choose n}\leqslant x$ either satisfies $p\leqslant M$ (there are $O(\log x)$ such numbers), or $y$ does not have a prime divisor $q<M$, since $q$ divides ${2n\choose n}$ and does not divide $p$ (there are at most $x\cdot \prod_{p<M}(1-1/p)+o(x)$ such numbers $y$). Since the product $\prod_p (1-1/p)$ equals 0, we get that there are only $o(x)$ numbers less than $x$ of the form $p+{2n\choose n}$.

This limit is 0.

The reason is that the central binomials are almost always divisible by 3, 5 etc (and by every specific prime). To be more precise, we have

Observation. for every fixed $p$ we have $|n\leqslant x: p\not|{2n\choose n}|=o(x)$.

Proof. Recall that by Kummer's theorem $p$ does not divide ${n+k\choose k}$ if and only if the base $p$ expansions $n=\sum c_i p^i$ and $k=\sum b_i p^i$, $0\leqslant b_i,c_i\leqslant p-1$, satisfy $b_i+c_i\leqslant p-1$ for all $i$ (i.e., there are no carries if we sum up $n$ and $k$ in base $p$). For ${2n\choose n}$ this means that all base $p$ digits of $n$ must be less than $p/2$, there are at most $\lceil p/2\rceil^{1+\log_p x}=o(x)$ such numbers not exceeding $x$.

Thus, for every fixed $M$ the number of $n\leqslant x$ for which there exists prime $p<M$ not dividing ${2n\choose n}$ is $o(x)$. Fix $M$ and call such $n$ $M$-strange.

Counting numbers of the form $p+{2n\choose n}$ not exceeding $x$, we note that there exist $O(x/\log x)$ choices of $p$ and $o(\log x)$ choices of $M$-strange $n$. Totally this gives $o(x)$ numbers of the form $p+{2n\choose n}$ with $M$-strange $n$. For not $M$-strange $n$, the number $y=p+{2n\choose n}\leqslant x$ either satisfies $p\leqslant M$ (there are $O(\log x)$ such numbers), or $y$ does not have a prime divisor $q<M$, since $q$ divides ${2n\choose n}$ and does not divide $p$ (there are at most $x\cdot \prod_{p<M}(1-1/p)+o(x)$ such numbers $y$). Since the product $\prod_p (1-1/p)$ equals 0, we get that there are only $o(x)$ numbers less than $x$ of the form $p+{2n\choose n}$.

This limit is 0.

The reason is that the central binomials are almost always divisible by 3, 5 etc (and by every specific prime). To be more precise, we have

Observation. for every fixed $p$ we have $\bigl|n\leqslant x: p\nmid{2n\choose n}\bigr|=o(x)$.

Proof. Recall that by Kummer's theorem $p$ does not divide ${n+k\choose k}$ if and only if the base $p$ expansions $n=\sum c_i p^i$ and $k=\sum b_i p^i$, $0\leqslant b_i,c_i\leqslant p-1$, satisfy $b_i+c_i\leqslant p-1$ for all $i$ (i.e., there are no carries if we sum up $n$ and $k$ in base $p$). For ${2n\choose n}$ this means that all base $p$ digits of $n$ must be less than $p/2$, there are at most $\lceil p/2\rceil^{1+\log_p x}=o(x)$ such numbers not exceeding $x$.

Thus, for every fixed $M$ the number of $n\leqslant x$ for which there exists prime $p<M$ not dividing ${2n\choose n}$ is $o(x)$. Fix $M$ and call such $n$ $M$-strange.

Counting numbers of the form $p+{2n\choose n}$ not exceeding $x$, we note that there exist $O(x/\log x)$ choices of $p$ and $o(\log x)$ choices of $M$-strange $n$. Totally this gives $o(x)$ numbers of the form $p+{2n\choose n}$ with $M$-strange $n$. For not $M$-strange $n$, the number $y=p+{2n\choose n}\leqslant x$ either satisfies $p\leqslant M$ (there are $O(\log x)$ such numbers), or $y$ does not have a prime divisor $q<M$, since $q$ divides ${2n\choose n}$ and does not divide $p$ (there are at most $x\cdot \prod_{p<M}(1-1/p)+o(x)$ such numbers $y$). Since the product $\prod_p (1-1/p)$ equals 0, we get that there are only $o(x)$ numbers less than $x$ of the form $p+{2n\choose n}$.

the limit is 0, because central binomials are almost always divisible by 3, 5 etc (and by every specific prime), while for large enough $M$ at least 99 percent of integers have a prime divisor not exceeding $M$

This limit is 0.

The reason is that the central binomials are almost always divisible by 3, 5 etc (and by every specific prime). To be more precise, we have

Observation. for every fixed $p$ we have $|n\leqslant x: p\not|{2n\choose n}|=o(x)$.

Proof. Recall that by Kummer's theorem $p$ does not divide ${n+k\choose k}$ if and only if the base $p$ expansions $n=\sum c_i p^i$ and $k=\sum b_i p^i$, $0\leqslant b_i,c_i\leqslant p-1$, satisfy $b_i+c_i\leqslant p-1$ for all $i$ (i.e., there are no carries if we sum up $n$ and $k$ in base $p$). For ${2n\choose n}$ this means that all base $p$ digits of $n$ must be less than $p/2$, there are at most $\lceil p/2\rceil^{1+\log_p x}=o(x)$ such numbers not exceeding $x$.

Thus, for every fixed $M$ the number of $n\leqslant x$ for which there exists prime $p<M$ not dividing ${2n\choose n}$ is $o(x)$. Fix $M$ and call such $n$ $M$-strange.

Counting numbers of the form $p+{2n\choose n}$ not exceeding $x$, we note that there exist $O(x/\log x)$ choices of $p$ and $o(\log x)$ choices of $M$-strange $n$. Totally this gives $o(x)$ numbers of the form $p+{2n\choose n}$ with $M$-strange $n$. For not $M$-strange $n$, the number $y=p+{2n\choose n}\leqslant x$ either satisfies $p\leqslant M$ (there are $O(\log x)$ such numbers), or $y$ does not have a prime divisor $q<M$, since $q$ divides ${2n\choose n}$ and does not divide $p$ (there are at most $x\cdot \prod_{p<M}(1-1/p)+o(x)$ such numbers $y$). Since the product $\prod_p (1-1/p)$ equals 0, we get that there are only $o(x)$ numbers less than $x$ of the form $p+{2n\choose n}$.