There is a very simple formulation for the character of irreducible representations of $S\_n$ evaluated on an n-cycle, i.e. that it is 0 on all non-hook partitions, and $(-1)^m$ on hooks. Is there an analogous computation for irreducible characters of $B\_n$, the hyperoctahedral group, evaluated on signed 2n-cycles? That is, on the conjugacy class indexed by the bipartition $([n],\emptyset)$?



Evaluation of irreducible representations of the hyperoctahedral group at bipartition $(\lambda,\mu)=([n],\emptyset)$
In general, if $(\lambda,\mu)$ is a bipartition of $n$, then
 $$ \prod\_i(p\_{\lambda\_i}(x)+p\_{\lambda\_i}(y))\cdot\prod\_j
 (p\_{\mu\_j}(x)-p\_{\mu\_j}(y)) = \sum\_{(\alpha,\beta)}
 \chi^{\alpha,\beta}(\lambda,\mu)s\_\alpha(x)s\_\beta(y), $$
where $(\alpha,\beta)$ ranges over all bipartitions of $n$ and $\chi^{\alpha,\beta}$ is the irreducible character of $B\_n$ indexed by $(\alpha,\beta)$. Setting $(\lambda,\mu)=
(n,\emptyset)$ gives
 $$ p\_n(x)+p\_n(y) = \sum\_{(\alpha,\beta)}
 \chi^{\alpha,\beta}(n,\emptyset)s\_\alpha(x)s\_\beta(y). $$
But (as alluded to in the question)
 $$ p\_n = \sum\_{i=0}^{n-1} (-1)^i s\_{n-i,1^i}, $$
so
 $$ p\_n(x)+p\_n(y)= \sum\_{i=0}^{n-1} (-1)^i s\_{n-i,1^i}(x) +
 \sum\_{i=0}^{n-1} (-1)^i s\_{n-i,1^i}(y). $$


