I'd think the 56? resistor would be moot, it'd only drop a volt and a half or less.
But a bridge and filter will charge up close to 170V (120 ?2) under no-load conditions.
Actually, I might use the 170V peak voltage for the calculations anyway, so string up 50 LEDs instead of 36. Or then use a dropping resistor, or better, a capacitor, which will dissipate very little heat. For 36 LEDs at 170V, we'd want to drop 50V, so 2000? assuming 25mA. A 1.2µF capacitor will have a reactance of 2210? at 60Hz, and will dissipate almost no power, where a 2200? resistor will dissipate over a watt.
Another thing to consider, to run directly on AC, wire two strings in reverse parallel to use both halves of the cycle.