Let $p \equiv 1 \pmod{3}$ be a prime and denote $H\_{n,m} = \sum\_{k = 1}^n 1/k^m$ as the $n,m$-th generalized harmonic number. I'm interested in computing $H\_{(p-1)/3,\, 2}$ and $H\_{(p-1)/6,\,2}$ modulo $p$. From [this](https://www.jstor.org/stable/1968791) paper I know
\begin{align\*}
H\_{(p-1)/6,2} \equiv -\frac{B\_{2p-3}(1/6)}{2p-3}\pmod p
\end{align\*}
and
\begin{align\*}
H\_{(p-1)/3,2} \equiv -\frac{B\_{2p-3}(1/3)}{2p-3}\pmod p
\end{align\*}
where $B\_n(x) = \sum\_{k = 0}^n {n \choose k}x^{n-k}B\_k$ is the $n$-th Bernoulli polynomial.


It is well known, I think, that
\begin{align\*}
B\_n(1/6) = (1 - 3^{1-n})(1 - 2^{1-n})\frac{B\_n}{2\cdot 3^{n-1}}
\end{align\*}
and
\begin{align\*}
B\_n(1/3)=(1 - 3^{1-n})\frac{B\_n}{2\cdot 3^{n-1}}
\end{align\*}
**but only for even $n$**. From this we can eventually express $H\_{(p-1)/3} = H\_{(p-1)/3, 1}$ and $H\_{(p-1)/6} = H\_{(p-1)/6,1}$ modulo $p$ in terms of the Fermat quotients $q\_p(2)$ and $q\_p(3)$ where $q\_p(a) = (a^{p-1} - 1)/p$ for $a$ co-prime to $p$.


It is mentioned in the paper above that there is some kind of expression for for $B\_n(1/3)$ and $B\_n(1/6)$ when $n$ is odd in terms of *"$I$ numbers"*, but I can't find the source, or anything else that is relevant.


Is it possible to write $H\_{(p-1)/3,2}$ and $H\_{(p-1)/6,2}$ in terms of Fermat quotients modulo $p$ analogous to $H\_{(p-1)/3}$ and $H\_{(p-1)/6}$? What are these "$I$ numbers?"?


I've asked a similar question [here](https://math.stackexchange.com/q/4474341/414513) but I thought I would ask here.



A question about generalized harmonic numbers modulo $p$
Glaisher's I-numbers are described in J. W. L. Glaisher, [On a set of coefficients analogous to the Eulerian numbers](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/plms/s1-31.1.216), Proc. London Math. Soc., 31 (1899), 216-235.


