Effective Realization of GCD of middle binomials?

Question

So, it is well-known that $$ \gcd \left(\binom{m}{k}\mid 1\leq k\lt m \right) = e^{\Lambda(m)}$$ which can incidentally be sparsified for prime $p$ $$ \gcd \left(\binom{p^{r+1}}{1},\binom{p^{r+1}}{p^r}\right) = p $$ and composite $m$ $$ \gcd \left(m,\binom{m}{p^k} \mid p^k \| m\right) = 1 $$

While Understanding that it'll be tricky to get out something as intermittent as von-Mangoldt without putting in something fiddly (like, say, Möbius ...)

1. Are there clean/effective realizations of these $\gcd$s as particular integer combinations?

   1.1. in particular, is there one that, in some natural sense, treats the prime and composite cases the same?

2. Have you a favourite such?

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For example, it is far from a proof or algorithm, but several small cases can be worked-out, with various tricks, from the fact that for $\omega$ a primitive 3rd root of unity, $1+\omega$ is a square root of $\omega$, e.g.: $$ \omega^2 = (1+\omega)^4 = 1 + \binom{4}{1} \omega + \binom{4}{2} \omega^2 + \binom{4}{3} \omega^3 + \omega^4 $$ $$ 1 = \omega + \binom{4}{1} \omega^2 + \binom{4}{2} + \binom{4}{1} \omega + \omega^2 $$ $$ 2 = \binom{4}{2} - \binom{4}{1} $$ I anticipate, however, that these various tricks probably won't work for $\binom{25}{n}$.

Answer 1

It can be done in the following way (this construction is taken from the paper Coefficient rings of formal groups, it was also a part of the answer to the question Is there a better proof of this fact in number theory/formal group theory?).

If $n=p^k$ then $\binom{n}{p^{k-1}}\equiv p\pmod{p^2},$ so we can easely find $\lambda_{p^{k-1}}$ such that $\lambda_{p^{k-1}}\binom{p^k}{p^{k-1}}\equiv p\pmod {p^{k}}$. So for some $\lambda_{1}$ $$\lambda_{p^{k-1}}\binom{p^k}{p^{k-1}}+\lambda_{1}\binom{p^k}{1}=p.$$

Now let $n=p_1^{k_1}\ldots p_s^{k_s}$, where $s>1$. Then by Kummer's theorem $\mathrm{ord}_{p_i}\binom{n}{p_i^{k_i}}=0$ and $\mathrm{ord}_{p_j}\binom{n}{p_i^{k_i}}\ge k_j$ ($j\ne i$). Taking $\lambda_{p_i^{k_i}}\equiv \binom{n}{p_i^{k_i}}^{-1}\pmod{p_i^{k_i}}$ we'll have

$$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}\equiv 1\pmod n.$$ So for some $\lambda_1$ $$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}+\lambda_{1}\binom{n}{1}=1.$$