Is the identity $$\binom{p^k}{jp^{k-2}}\equiv \binom{p^2}{j} \;\mathrm{mod}\,p^3$$ true for all $k\geq 2$ and $j=1,\dots,p^2$ when $p=3$?

By Wolstenholme's theorem, we know it is true for all primes $p\geq 5$. Also, by Babbage's theorem, we know it is true for all primes if we replace modulo $p^3$ by $p^2$. However, I cannot find any proof or counterexample to this specific case with $p=3$. I have tried to apply Granville's result but I have not succeeded. Is there one? Is it actually true?

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    $\begingroup$ Equation (12.69.3) in Darij Grinberg, Victor Reiner, Hopf Algebras in Combinatorics, arXiv:1409.8356v5, version with solutions (ancillary file) says that every prime $p$, every positive integer $n$, every integer $q$ and every rational $r$ satisfy $\dbinom{qn/p}{rn/p} \equiv \dbinom{qn}{rn} \mod p^{v_p\left(n\right)}$, where $p^{v_p\left(n\right)}$ is the highest power of $p$ that divides $n$. Here, we understand a binomial coefficient $\dbinom{a}{k}$ to be $0$ whenever $k$ is not ... $\endgroup$ – darij grinberg Mar 11 at 19:47
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    $\begingroup$ ... a nonnegative integer. (This congruence is folklore; I certainly claim no ownership.) This congruence lets us divide both "numerator" and "denominator" of the binomial coefficient on your left hand side by $p$ until the coefficient becomes $\dbinom{p^2}{j}$. Each time you get a congruence modulo $p^{v_p\left(p^i\right)} = p^i$ for some $i \geq 3$. Thus you end up with a congruence modulo $p^3$. This holds for any prime $p$, even for $p = 2$. $\endgroup$ – darij grinberg Mar 11 at 19:49
  • $\begingroup$ Oh great, I am so happy it holds! Thanks a lot! $\endgroup$ – bd99 Mar 12 at 21:05
  • $\begingroup$ I have a question concerning the combinatoric proof you give of (12.69.3) in the book your mention. There, you say that orbits of the action by the permutation $i\mapsto i+1$ on subsets of the second class have size divisible by $p^{v_p(n)}$. Why is that the case? $\endgroup$ – bd99 Mar 13 at 19:18
  • $\begingroup$ Look at how the permutation $i \mapsto i + qn/p^{v_p\left(n\right)}$ acts on them. Its $p^{v_p\left(n\right)}$-th power acts as the identity, so each orbit has size dividing $p^{v_p\left(n\right)}$. But if an orbit's size was a proper divisor of $p^{v_p\left(n\right)}$, then the subsets in this orbit would be first class (since a proper divisor of $p^{v_p\left(n\right)}$ is always a divisor of $p^{v_p\left(n\right)-1}$), which contradicts the fact that they are first class. Now, the orbits of the permutation $i \mapsto i+1$ are unions of orbits ... $\endgroup$ – darij grinberg Mar 13 at 20:17

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