Formula for genus of a Fano variety

Question

This is a simple question, just looking for a reference for a formula.

As far I understand the genus of a prime Fano $n$-fold is defined to be the genus of a complete intersection of $n-1$ smooth divisors in the system $|-K_{X}|$ (see https://www.math.ens.fr/~debarre/ExposePoitiers2013.pdf). For $n=3$ this number equals $$\frac{(-K_{X})^3}{2} +1.$$

Question: What is the general formula for the genus of Fano variety?

Unfortunately I was not able to find this in the literature.

Edit: I thank Pop for the formula for the quantity given above. However, this does not agree with the use of term genus used in the literature (that was an incorrect guess on my part). By https://www.math.ens.fr/~debarre/ExposePoitiers2013.pdf page 7 it seems that the most widely used definition of genus is $$\frac{(-K_{X})^n}{2 \cdot \iota(X)^n } +1 ,$$ where $\iota(X)$ is the index of $X$ i.e. the maximal divisibility of $c_{1}(X)$ in $H^{2}(X,\mathbb{Z})$.

I checked a few examples and this agrees with the way that other authors use the term genus in their papers.

Question 2: Why is this the accepted notion of genus, is there a geometric reason for this?

One could think that this is the genus of the intersection of $(n-1)$ smooth divisors in the system $$|\frac{-K_{X}}{\iota(X)}|$$, but this is not the case. For example if the index is $1$ then the formula does not give the formula of Pop. Is the definition related to $K3$ surfaces?

Edit 2: Just to make it clear that the question is resolved I write a brief explanation.

As Tom Ducat pointed out, the formula I referenced made a strong assumption on the index. Indeed the genus of a Fano variety is the genus of a curve given by intersecting divisors in the class $A = -K_{X}/\iota(X)$, where $\iota(X)$ is the index of $X$.Then by the computation of Pop's answer one can recover the formula $$g= 1+\frac{n-\iota(X)-1}{2}A^n$$, which agrees with Fujita's paper and the examples from the literature.

Answer 1

It's not clear from your question if you really want a reference (as per the first line) or the formula itself (as per what is written after "Question"). But the computation of the genus is not hard so let me write it here in case that answers the question.

If $C$ is such a smooth curve in $X$, then the normal bundle is $N_{C/X} = (-K_X)_{|C}^{\oplus n-1}$. So adjunction says that

\begin{align*} \operatorname{deg} K_C &= \operatorname{deg} (K_X)_{|C} + \operatorname{deg} \wedge^{n-1} N_{C/X} \\ &= \operatorname{deg} (K_X)_{|C} + (n-1) \operatorname{deg} (-K_X)_{|C} \\ &= (n-2) \operatorname{deg} (-K_X)_{|C}\\ &= (n-2)(-K_X)^n \end{align*} Since $\operatorname{deg} K_C = 2g(C)-2$ you get

\begin{align*} g(C) = \frac12 (n-2)(-K_X)^n + 1 \end{align*}