First of all, let $p$ and $q$ be some prime numbers with $q/2<p<q$. Then $pq\mid\binom{n}{pq}$ if

$$n\equiv a \pmod {p^2}$$

and

$$n\equiv b \pmod {q^2}$$

with some $a,b>0$ and $a<p(q-p)$, $b<pq$. To prove this, apply [Lucas's theorem][wiki] on binomial coefficients and also note that $pq$ is a two-digit number $\overline{p0}$ in base $q$ and a three-digit number $\overline{1(q-p)0}$ in base $p$. Now lets assume $n$ is large enough.

It is a well-known fact that the interval $(x,x+x^{2/3})$ contains at least one prime for $x$ sufficiently large. Now let us choose $q$ to lie in the interval $(\sqrt{n}-n^{3/8},\sqrt{n}-n^{0.35})$ (which is clearly possible for $n$ large enough) and $p$ to lie in the interval $(q/2,\sqrt{n}/2)$.

Then we have the following:

$$q=\sqrt{n}-A$$

with $0<A<n^{3/8}$. Therefore,

$$q^2=n-2A\sqrt{n}+A^2,$$

so $0<n-q^2<2n^{7/8}=o(n)=o(pq)$, which is required. And we also have

$$p=\sqrt{n}/2-B$$

with $0<B<n^{0.35}$, therefore

$$0<n-4p^2=O(n^{0.85})=o(n)=o((q-p)p),$$

which is required. So, for $n$ large enough one can find some $k$ around $n/2$ that satisfies your condition, so $f(n)>1$ (and in fact even grows with $n$). [wiki]:https://en.wikipedia.org/wiki/Lucas%27s_theorem

First of all, let $p$ and $q$ be some prime numbers with $q/2<p<q$. Then $pq\mid\binom{n}{pq}$ if

$$n\equiv a \pmod {p^2}$$

and

$$n\equiv b \pmod {q^2}$$

with some $a,b>0$ and $a<p(q-p)$, $b<pq$. To prove this, apply Lucas's theorem on binomial coefficients and also note that $pq$ is a two-digit number $\overline{p0}$ in base $q$ and a three-digit number $\overline{1(q-p)0}$ in base $p$. Now lets assume $n$ is large enough.

It is a well-known fact that the interval $(x,x+x^{2/3})$ contains at least one prime for $x$ sufficiently large. Now let us choose $q$ to lie in the interval $(\sqrt{n}-n^{3/8},\sqrt{n}-n^{0.35})$ (which is clearly possible for $n$ large enough) and $p$ to lie in the interval $(q/2,\sqrt{n}/2)$.

Then we have the following:

$$q=\sqrt{n}-A$$

with $0<A<n^{3/8}$. Therefore,

$$q^2=n-2A\sqrt{n}+A^2,$$

so $0<n-q^2<2n^{7/8}=o(n)=o(pq)$, which is required. And we also have

$$p=\sqrt{n}/2-B$$

with $0<B<n^{0.35}$, therefore

$$0<n-4p^2=O(n^{0.85})=o(n)=o((q-p)p),$$

which is required. So, for $n$ large enough one can find some $k$ around $n/2$ that satisfies your condition, so $f(n)>1$ (and in fact even grows with $n$).

First of all, let $p$ and $q$ be some prime numbers with $q/2<p<q$. Then $pq\mid\binom{n}{pq}$ if

$$n\equiv a \pmod {p^2}$$

and

$$n\equiv b \pmod {q^2}$$

with some $a,b>0$ and $a<p(q-p)$, $b<pq$. To prove this, apply [Lucas's theorem][wiki] on binomial coefficients and also note that $pq$ is a two-digit number $\overline{p0}$ in base $q$ and a three-digit number $\overline{1(q-p)0}$ in base $p$. Now lets assume $n$ is large enough.

It is a well-known fact that the interval $(x,x+x^{2/3})$ contains at least one prime for $x$ sufficiently large. Now let us choose $q$ to lie in the interval $(\sqrt{n}-n^{3/8},\sqrt{n}-n^{0.35})$ (which is clearly possible for $n$ large enough) and $p$ to lie in the interval $(q/2,\sqrt{n}/2)$.

Then we have the following:

$$q=\sqrt{n}-A$$

with $0<A<n^{3/8}$. Therefore,

$$q^2=n-2A\sqrt{n}+A^2,$$

so $0<n-q^2<2n^{7/8}=o(n)=o(pq)$, which is required. And we also have

$$p=\sqrt{n}/2-B$$

with $0<B<n^{0.35}$, therefore

$$0<n-4p^2=O(n^{0.85})=o(n)=o((q-p)*p),$$

which is required. So, for $n$ large enough one can find some $k$ around $n/2$ that satisfies your condition, so $f(n)>1$ (and in fact even grows with $n$). [wiki]:https://en.wikipedia.org/wiki/Lucas%27s_theorem

First of all, let $p$ and $q$ be some prime numbers with $q/2<p<q$. Then $pq\mid\binom{n}{pq}$ if

$$n\equiv a \pmod {p^2}$$

and

$$n\equiv b \pmod {q^2}$$

with some $a,b>0$ and $a<p(q-p)$, $b<pq$. To prove this, apply [Lucas's theorem][wiki] on binomial coefficients and also note that $pq$ is a two-digit number $\overline{p0}$ in base $q$ and a three-digit number $\overline{1(q-p)0}$ in base $p$. Now lets assume $n$ is large enough.

It is a well-known fact that the interval $(x,x+x^{2/3})$ contains at least one prime for $x$ sufficiently large. Now let us choose $q$ to lie in the interval $(\sqrt{n}-n^{3/8},\sqrt{n}-n^{0.35})$ (which is clearly possible for $n$ large enough) and $p$ to lie in the interval $(q/2,\sqrt{n}/2)$.

Then we have the following:

$$q=\sqrt{n}-A$$

with $0<A<n^{3/8}$. Therefore,

$$q^2=n-2A\sqrt{n}+A^2,$$

so $0<n-q^2<2n^{7/8}=o(n)=o(pq)$, which is required. And we also have

$$p=\sqrt{n}/2-B$$

with $0<B<n^{0.35}$, therefore

$$0<n-4p^2=O(n^{0.85})=o(n)=o((q-p)p),$$

which is required. So, for $n$ large enough one can find some $k$ around $n/2$ that satisfies your condition, so $f(n)>1$ (and in fact even grows with $n$). [wiki]:https://en.wikipedia.org/wiki/Lucas%27s_theorem