Electronic – Why does this PUT oscillator max out at about 25 Hz

The basic operation can be seen more clearly if you first assume that C1 is shorted at time = 0. In that case you have a simple voltage divider created by R1 and R3, so increasing the value of R3 just increases the initial voltage at the anode connection. C1 then begins to charge and the gate trigger point (set by R2,R4) is reached much quicker.

Then because the charge time was very short only a small amount of charge accumulated on C1, so the discharge time (also the LED on time) is also very short.

Placing a diode in parallel with R3 may give you the result you expected (with cathode to C1). In that case R3 is nearly out of the circuit during C1 charging, but during C1 discharge R3 would extend the on time.

There are, I think, a number of issues. Try this:

Reduce C2 to .01 uF, and increase R47 to 10k. You don't list your pot value, but I'm assuming 1k. Increase this to 100k, and put a 1k in series with it. As you've got it now, the pot value gets way too low for the circuit to operate properly.

Then, you need to do a major change on your transistor output. Try something like

^{simulate this circuit – Schematic created using CircuitLab}

Note that there are 2 LED outputs. Divide your LEDs into 2 equal groups and drive each group with one output. This will keep the PNP current levels within spec (3906s are only rated for 100 mA).

Alternatively, get rid of the transistor and R48/49 completely. The output current for a 555 is rated at 200 mA, that's about what your LEDs will draw. R48 and R49 are so low that you're drawing near that anyways.

Finally, I'd suggest reducing your base resistors (R17 - R32) to about 1k to 2k, just to make sure the transistors are driven hard on.