Deducing the central charge of the Ising model from the free energy My question is inspired by Di Francesco, Mathieu, and Senechal's Conformal Field Theory problem 3.5.
Namely, the problem gives the fact that the free energy per particle of the 2D classical Ising model on an infinite strip of width $L$ is $f/L = f_0 - \frac{\pi}{6L^2}c + O(1/L^3)$ where $c$ is the central charge of the model, and asks one to plot and fit the free energy with the parameters $f_0$ and $c$. The aim is to deduce $c=1/2$. The free energy follows straightforwardly from the largest eigenvalue of the $2^L$ by $2^L$ transfer matrix of the model.
My numerical calculations for the 2D Ising model yield the following plot for $f/L$ against L, the free energy per particle:

I have added a red line signifying Onsager's result for the infinite lattice, which is approximately $-0.929695$. My numerics seem to get the right $f_0$, so I'm tempted I haven't done anything wrong.

However, I have no confidence in a numerical fit to the recommended ansatz of $f/L = f_0 - \frac{\pi}{6L^2}c + O(1/L^3)$, since the function's $L$ behavior seems to be dominated by a term proportional to $(-1)^L$ and also decaying with $L$.
Strangely, fitting only the even points to the ansatz above gets a $c$ close to the correct points of $1/2$ (and odd points would predict a $c$ of close to $-1$), so I'm wondering if perhaps the textbook's ansatz was meant only for even $L$!
That is, I find the absolute values of the difference between $f/L$ and the Onsager value are the following, and are suspiciously well fit by the functions in the legend:

Should there be such striking parity-of-$L$ (even vs. odd) dependence in the free energy per particle? If so, is there a better ansatz based on conformal field theory that will be able to deduce the central charge?
 A: As pointed out by Christophe in the comments, I was accidentally finding the free energy of the antiferromagnetic ising model on the cylinder. This can be mapped to the ferromagnetic ising model by flipping spins on a checkerboard sublattice, but with boundary conditions depending on the parity of the width of the cylinder. That is, for even $L$, the boundary conditions are periodic, while for odd $L$, the boundary conditions are antiperiodic - this is because the antiferromagnet is frustrated in the case of odd $L$.
This is important, because the ansatz I gave in $$f/L = f_0 - \frac{\pi}{6L^2}c + O(1/L^3)$$
is actually not the most general ansatz. Instead, as pointed out in an answer by Christophe to another question of mine (particularly in the citation to Izmailian and Yeh) the more general ansatz is
$$f(L)=f(\infty)+{f_{\rm surf.}\over L}+{p\pi\over L^2}
  \left({c\over 24}-\Delta\right)
  +{\cal O}\left({1\over L^3}\right)$$
where $p$ is $4$ for periodic and antiperiodic boundary conditions, and $1$ for open boundary conditions including fixed and free boundaries. The surface free energy $f_{\rm surf.}$ is zero for closed boundaries (periodic or antiperiodic). Thus, the important missing contribution from my ansatz is just the factor of $\Delta$, which is the highest conformal weight. Note that $\Delta$ is zero for periodic boundary conditions, but can be nonzero depending for other boundary conditions.
In particular, for the Ising model with $c=1/2$, the Izmailian and Yeh reference above notes that antiperiodic boundary conditions give $\Delta=\frac{1}{16}$.
This means that for the Ising model, I should expect
$$f/L = f_0 - \frac{\pi}{12L^2} + O(1/L^3)$$ for periodic boundary conditions but
$$f/L = f_0 + \frac{\pi}{6L^2} + O(1/L^3)$$ for antiperiodic boundary conditions.
These two equations, along with the fact noted above that my antiferromagnetic ising model with periodic boundary conditions was mappable to the ferromagnetic ising model with boundary condition depending on the parity of $L$, perfectly explain my two figures in my question. In particular, it explains the unexpected factor of $-2$ in the $\frac{1}{L^2}$ piece of $f$ for odd $L$ as stemming from the nonzero $\Delta$ in antiperiodic boundary conditions.
