How can resonance be added to a fractional order Butterworth low pass
Given the formula for second order sections:

$$
s_k = \omega_c e^{\frac{j(2k+n-1)\pi}{2n}}\qquad k = 1,2,3,\ldots, n \qquad s = i\omega
$$

$$
H(s)=\frac{G_0}{\prod_{k=1}^n (s-s_k)/\omega_c}
$$

Use k = 1 for the resonance section, add $Q$ in the normal way i.e. calculate $H_{k=1}^2 = H_{k=1}(s) \overline{H_{k=1}(s)}$ where $s = i\omega$ and the denominator will be of the form $w^4 + cw^2 + 1$ where $c$ is a constant calculated from $s_k$. We add $Q$ to the first order term (note that this is a quadratic in $\omega^2$). Using any positive real $n$ in $s_k$ gives valid values for $c$.

Let $R^2$ be a modified $H_{k=1}^2$ with $Q$ added in to the first order term of the quadratic in $w^2$ then to add resonance to the original equation in the question multiply it by:

$$
\frac {R^2}{H_{k=1}^2}
$$

This effectively removes the $k = 1$ second order section and adds a resonant section back in. It turns out once you do this, $n$ can be any positive real value when used in $c$ but note this wouldn't work for fractional roots of the sections (you would have to use fractional calculus if you wanted the roots of the fractional equation).

Without resonance:

$$
c = 2 \cos^2(\frac{\pi (1+n) } {2n}) - 2
$$

With resonance:

$$
c = \frac{2}{Q^2}\cos^2(\frac{\pi (1+n) } {2n}) - 2
$$