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 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 but I thought I would ask here.
Best Answer
Glaisher's I-numbers are described in J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., 31 (1899), 216-235.