 Solution

Choosing two slow legs in $h({\varphi }_{<}+{\varphi }_{>}{)}^6$ and contracting four fast legs gives $\left(\genfrac{}{}{0ex}{}{6}{2}\right)(3)h{I}_1^2{\varphi }_{<}^2$. Since the mass term is ${\mu }^2{\varphi }_{<}^2/2$,Choosing four slow legs and contracting the remaining pair gives $B=\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$. For the connected quartic-sextic cumulant, the factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{2}\right)2!=180$ and the cumulant has a minus sign, soAs a check, two quartic vertices give $-\frac{1}{2}{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}^{2}2!=-36$, matching the supplied coefficient.

Solved by gpt-5.6-sol high.