$d$-dimensional Einstein equations

I am reading this paper https://arxiv.org/abs/2002.02577. On page 13, it is written that the $$d$$-dimensional Einstein equations are

$$G_{\mu\nu}+\frac{(d-1)(d-2)}{6}\Lambda g_{\mu\nu}=8\pi T_{\mu\nu}.\tag{2.6}$$

Of course, for $$d=4$$ we take what we expect. But I don't see how the the factor $$(d-1)(d-2)/6$$ shows up if the action is

$$S=\int (\frac{1}{2\kappa}(R-2\Lambda)+L_{M})\sqrt{-g} d^{d}x.$$

If I variate with respect to the metric, I just take $$G_{\mu\nu}+\Lambda g_{\mu\nu}=8\pi T_{\mu\nu}$$ for any value of $$d$$. What's my mistake?

• This is a Gauss-Bonnet theory, the action is not the Einstein-Hilbert action. The action in the Einstein-Gauss-Bonnet theory does lead to an equation like this as I remember. – Y2H May 29 at 9:45
• I am sorry, I should explain more. I am interested in Einstein gravity in higher dimensions now. $L_{M}$ is not the Gauss-Bonnet term here, it just describes any matter fields. So actually the problem is the term that inculdes the cosmological constant. – john May 29 at 10:01

You didn't make a math mistake. The only mistake is assuming that equation (2.6) was derived starting from the action shown in the question. To get the factor $$(d-1)(d-2)/6$$, it must be inserted by hand into the action prior to starting the derivation.
The motive for including that factor is not stated in the cited paper as far as I noticed, but whatever the motive might be, introducing a $$d$$-dependent factor into that term in the action is a legal thing to do, since the coefficient $$\Lambda$$ is arbitrary anyway.