mathematical equation for parallel voltage dividers?
For one voltage divider there is the following rule:


A voltage divider with a constant voltage input behaves like:
A constant voltage source with a resistor \$R_p\$ in series while
\$R_p\$ has the value of \$R_1||R_2\$ (the resistance of resistors \$R_1\$ and \$R_2\$ in parallel)


Let's look at the right voltage divider (\$R_1\$ and \$R_2\$) first:

Without \$R_3\$ connected the voltage between the resistors is 1.65V (\$3.3V *\frac{100k\Omega}{100k\Omega+100k\Omega}\$). \$R_p\$ is the value that we had if \$R_1\$ and \$R_2\$ were in parallel: \$R_p = R_1||R_2 = 50k\Omega\$.

So we can replace the right voltage divider by a constant voltage source of 1.65V with a 50 kOhms resistor in series.

Now the left voltage divider consists of a voltage divider with three resistors: \$R_p\$, \$R_3\$ and \$R_4\$ and an input voltage of 1.65V.

Calculating the voltage over \$R_3\$ and \$R_4\$ in this replacement we get: \$1.65V*\frac{R_3+R_4}{R_p+R_3+R_4} \approx 1.43V\$. This is nearly the value you measured.

Now we can calculate the output voltage of the left voltage divider in the case that there is no current flowing through the output:

\$V_{out} = 1.65V*\frac{R_4}{R_p+R_3+R_4} \approx 0.446V\$

... and we can calculate the "\$R_p\$" value of the left voltage divider; let's call this value \$R_q\$ because \$R_p\$ is already the value for the right voltage divider:

\$R_q = (R_p+R_3)||R_4 \approx 73k\Omega\$

This means the complex voltage divider shown in your schematic behaves like a constant voltage source with a voltage of 0.446 Volts with a resistor of 73 kOhms in series.

(However if you have no current flow at the \$V_{out}\$ pin the \$R_q\$ value is not of interest.)

Note:

\$R_1||R_2 = \frac 1 {\frac 1 {R_1} + \frac 1 {R_2}}\$