On shifted symmetric power sums The functions $p^*_k(x)=\sum_{i=1}^N ((x_i-i)^k-(-i)^k)$ are analogues of power sum symmetric functions, called shifted symmetric by Okounkov and Olshanski. Define $p^*_{(k_1,k_2,...)}=p^*_{k_1}p^*_{k_2}...$.
We know that the sum $\frac{1}{n!}\sum_{\mu\vdash n}C_\mu \chi_\lambda(\mu)p_\mu(x)=s_\lambda(x)$ returns Schur functions (where $C_\mu$ is the size of a conjugacy class in the permutation group and $\chi$ are the irreducible characters of this group).
My question is: what is known about $f_\lambda(x):=\frac{1}{n!}\sum_{\mu\vdash n}C_\mu \chi_\lambda(\mu)p_\mu^*(x)$? Is this related to shifted Schur functions? What properties does this function share with the usual Schur functions?
 A: This is only a partial answer, but perhaps it will help. While the symmetric functions $\Lambda$ are graded by polynomial degree, the shifted symmetric functions $\Lambda^*$ are only filtered. It is fairly easy to see from the definition (and proved in Okounkov-Olshanski's paper Shifted Schur Functions) that in fact the associated graded algebra of $\Lambda^*$ with respect to this filtration is isomorphic to $\Lambda$,
$$\text{gr}(\Lambda^*) \cong \Lambda.$$
In particular, in terms of polynomial degree the shifted Schur function $s_\lambda^*$ and shifted power sum function $p_{\lambda}^*$ can be written as
$s_\lambda^* = s_\lambda +$ l.o.t.,
$p_\lambda^* = p_\lambda +$ l.o.t.
So if we only look at the top degree terms of your $f_\lambda$ we get exactly $s_\lambda$, but in general $f_\lambda \neq s_\lambda^*$. I don't know if anything is known about the lower degree terms of $f_\lambda$.
I should mention that Okounkov-Olshanski introduce another shifted analogue to the power sum functions which they denote by $p^\#_\lambda$. These do have the property that,
$$s_\lambda^* = \frac{1}{n!}\sum_{\mu \vdash m} C_\mu\chi^{\lambda}(\mu) p^\#_\mu.$$
$p^\#_\mu$ also has the wonderful property (at the expense of having a nice explicit formula) that if $\mu \vdash k$ and we evaluate $p^\#_\mu$ on a partition $\lambda = (\lambda_1,\lambda_2, \dots, \lambda_r) \vdash n$ by setting $x_1 = \lambda_1, x_2 = \lambda_2, \dots, x_r = \lambda_r$, and all other $x_i = 0$, then we get
$$p_\mu^\#(\lambda) = \begin{cases}
\frac{(n \downharpoonright k)}{\dim L^\lambda} \chi^\lambda(\mu) & k \leq n,\\
0 & \text{otherwise}
\end{cases}$$
where $(n \downharpoonright k)$ is the falling factorial and $\dim L^\lambda$ is the dimension of the simple $S(n)$-representation associated to $\lambda$.
A: The super Schur function $s_\lambda(x/y)$ is obtained by applying to
$s_\lambda(x,y)$ (a Schur function in two sets of variables
$x=(x_1,x_2,\dots)$ and $y=(y_1,y_2,\dots)$) the algebra homomorphism
$\omega_y\colon \Lambda(x,y)\to\Lambda(x)\otimes \Lambda(y)$, where
$\Lambda(z)$ is the ring of symmetric functions in the variables $z$,
defined by 
  $$ \omega_yp_n(x,y)=p_n(x)+(-1)^{n-1}p_n(y). $$
Equivalently,
   $$ s_\lambda(x/y) =
   \sum_{\mu\subseteq\lambda}s_\mu(x)s_{\lambda'/\mu'}(y). $$
Thus $f_\lambda(x)=s_\lambda(x_1-1,x_2-2,\dots/1,2,3,\dots)$.
