Hmm...oddly you get to the right answer, but the wrong algebra.
I am assuming you are planning to use a full wave rectifier (not bridge) and solid state diodes rather than a vacuum rectifier. If that is correct, then the math is more like this (approximation):
500VCT is 250/0/-250 vac => 250vac * 1.43 - 1.2v = 356vdc. I am pulling the diode drop off the top of my head. I can't recall if this is right, but as you can see it has very little impact. If vacuum tubes are to be used the diode drop is much larger, like 25v for damper diodes to 50+V for regular VT rectifiers.
P.S. this also assumes that this is a cap input filter where the first cap is relatively large. For choke input filters, the multiplying factor is roughly 0.9 instead of 1.43. You get less voltage but better regulations and ripple waveform with a choke input filter.
note: nevermind...I see that broskie made the formula simpler for folks. 500vct/sqrt(2) = 500vct/2*sqrt(2). So yeah, your math is correct, just not the way I think about it.