Let $V$ be a graded vector space and for $j\geq 1$ let $m_j:V^{\otimes j}\to V$ be maps of degree $2-j$ such that for al $v\geq 0$
$$\sum_{v=j+q-1, j=r+1+t}(-1)^{rq+t}m_j(1^{\otimes r}\otimes m_q\otimes 1^{\otimes t})=0,$$
where $1$ is the identity map on $V$. According to the appendix of this paper, defining the maps $\tilde{m}_j=(-1)^{\frac{j(j-1)}{2}}m_j$, one obtains the relation
$$\sum_{v=j+q-1, j=r+1+t}(-1)^{vq+(r+1)(q-1)}\tilde{m}_j(1^{\otimes r}\otimes \tilde{m}_q\otimes 1^{\otimes t})=0.$$
In the paper the author writes $k$ instead of $r+1$, but that is the value of $k$ in the context. Replacing the maps $m_j$ with $(-1)^{\frac{j(j-1)}{2}}\tilde{m}_j$ in the first equation, the exponent I get on the sign is
$$rq+t+\frac{(r+1+t)(r+t)}{2}+\frac{q(q-1)}{2}$$
I don't know how to go from this to the exponent on the second equation, I've tried different combinations using the equation $v=j+q-1$ but I'm not getting the correct sign. What can I do?
Some ideas
To make $qv$ appear I can use the equation $v-j=q-1$ from the subindex of the sum, so the above expression becomes
$$rq+t+\frac{(r+1+t)(r+t)}{2}+\frac{q(v-j)}{2}=rq+t+\frac{(r+1+t)(r+t)}{2}+\frac{q(v-r-1-t)}{2}$$
I can also replace $rq$ by $(v-t-q)q$ to yield
$$(v-t-q)q+t+\frac{(r+1+t)(r+t)}{2}+\frac{q(v-r-1-t)}{2}$$
This already gives me $vq$. I also need $(r+1)(q-1)$ so I would replace $r+t$ with $v-q$ on the first fraction
$$(v-t-q)q+t+\frac{(r+1+t)(v-q)}{2}+\frac{q(v-r-1-t)}{2}$$
Now I take a the common factor $(r+1+t)$ so
$$(v-t-q)q+t+\frac{(r+1+t)(v-q-q)+qv}{2}$$
Looking and the fraction, it can be reduced to
$$\frac{(r+1+t)(v-2q+2-2)+qv}{2}=\frac{(r+1)(-2q+2)+(r+1)(v-2)+t(v-2q)+qv}{2}$$
Since this is an exponent on a sign, I only care about the value modulo $2$, so I get
$$(r+1)(q-1)+\frac{(r+1)(v-2)+t(v-2q)+qv}{2}$$
So all in all I have mod $2$
$$vq+(r+1)(q-1)+q(q+t)+t+\frac{(r+1)(v-2)+t(v-2q)+qv}{2}$$
I've managed to obtain $(r+1)(q-1)$ and $vq$, but I get lost trying to see that the rest vanish, especially trying to eliminate denominators.