How can I simplify or write the following definition in standard mathematical format?

Question

$\gamma(r,n,d)$ is an integer defined as follows: for all $d|n$, $0<=r<=n-1$
call $u1$ the list $\mu(s) (d/s)$ over all divisors $s|d$,
for n=6, and d=1,2,3,6 respectively we get {1}, {2,-1}, {3,-1}, {6,-3,-2,1} (summing to $\phi(d)$).
call $u2$ the intersection of the divisors of n with the divisors of $r$
(if fact, for n=6, we want this intersection to be with the divisors of 1 .. 6 but will be counting $r=$ 0 .. 5 using $1+{(k-1)\bmod 6}$).
for n=6, r=0..5 the six tuples of $u2$ are
{1,2,3,6}, {1}, {1,2}, {1,3}, {1,2}, {1}
then $\gamma(r,n,d)$ equals the sum of the (signed) d-th element in $u1$ whose absolute value occurs in the (r+1)-th part of $u2$.
I could not find a more concise description or notation, and would be very surprised if $\gamma(r,n,d)$ didn't show up in other combinatorial contexts.

Example 1:
choose $n=6$ and $r=3$
$u1=$ {1}, {2,-1}, {3,-1}, {6,-3,-2,1} for $d$ ={1,2,3,6} respectively.
$u2=$ {1,3} from $\{1,3\} \cap \{1,2,3,6\}$
so $\gamma(3,6,1) = 1$; $\gamma(3,6,2) = -1$; $\gamma(3,6,3) = 3-1 =2$; $\gamma(3,6,6) = -3+1= -2$

Motivation :
I encountered it in the following context:
the principal specialisation of the Schur function $s_\lambda(1,q,q^2,..,q^{n-1}) , \lambda \vdash n$ is given by
$q^{b(\lambda)} \prod_{u \in \lambda},[n+c(u)]/[h(u)]$
with $b(\lambda) = \sum{_i} (i-1) \lambda_i $ and $[x] = (1-q^x)$ and with $h(u)$ the hooklength of cell u, and similarly $c(u)$ the content of cell u (see R. P. Stanley (Enum. Comb. Vol. 2 (1999) pg 375 Theorem 7.21.2).
The resulting polynomial in q has degree at most $n (n-1)$. If we choose q so that $q^k = q^{k\bmod n}$ like in $q=e^{i \pi k/n}$ then the $n(n-1)$ terms are collapsed into $n$.
These can be calculated directly. Choose any value $r$ in the range 0 to n-1. The coefficient of $q^r$ seems to be given by: $c_r(\lambda)= (1/n) \sum _{d|n} \beta_d(\lambda) \delta_d(\lambda) \gamma(r,n,d) \prod_{j=1}^d t (\lambda_j,n/d)$
where $\beta_d(\lambda)$ is $1$ iff the d-core of $\lambda$ is empty, else 0 ; $\delta_d(\lambda)$ is the sign of the $d$-quotients of $\lambda$ (see http://web.math.ku.dk/~olsson/manus/comb_rep_all.pdf (pg 22, example following remark 3.11).
$t(\lambda_j,n/d)$ is the count of SSYT of j-th d-quotient of $\lambda$ with max element $n/d$
(Stanley's hook content formula).

Example 2:
n=6; $\lambda = \{4,1,1\}$
all d-cores of $\lambda$ are empty, so all $\beta_d$ are $1$.
The signs of the $d$-quotients $\delta_n(\lambda)$ are {1,-1,1,1} for $d$={1,2,3,6} respectively.
The d-quotients are the following $\lambda_j$ :
d=1; {4,1,1}
d=2; {} , {2,1}
d=3; {2}, {}, {}
d=6; {}, {}, {}, {1}, {}, {}
and so the product over $t(\lambda_j,6/d)$ equals
d=1; $t(\{4,1,1\},6) = 840$
d=2; $t(\{2,1\},3) = 8$
d=3; $t(\{2\},2) = 3$
d=6; $t(\{1\},1) = 1$
where all factors $t(\{\},d)$ are equal to 1. Sofar we already have: $c_r(\{4,1,1\})= \frac{1}{6} (840 \gamma(r,6,1)- 8 \gamma(r,6,2) +3 \gamma(r,6,3) + \gamma(r,6,6))$
Using the values at $r=3$ of $\gamma(r,6,d)$ calculated above, we then find
$c_4(\{4,1,1\}) = (1/6)(840 -8(-1)+ (3)(2) -2) = 142$.
Check:
$s_{\{4,1,1\}}(1,q,q^2,..,q^5)$ under $q^j=q^{j \pmod 6}$ equals
$140+141q^1+138q^2+\mathbf {142q^3}+138q^4+141q^5$
I like the esthetics of the divisors appearing in $u1$ and $u2$ combining with the partition quotients in a natural way.