The least amount of heat required would be to let the temperature drop after use to as low as possible but obviously starting heating long enough before the next soak so that you will get to the desired temperature in time. The reason this is the most energy efficient is that heat loss is proportional to the temperature difference between the spa water and the outside air temperature. So the longer period of time with a lower spa water temperature, the less heat loss so that much less heat needed to be replaced. If you were to constantly maintain a high spa temperature, you would be losing heat more quickly and therefore replacing (heating) it more as well.
There are diminishing returns for this if the temperature drop is smaller, the outside air temperature warmer, and the time between uses shorter. Here are two examples to give you a rough idea of the differences. For simplicity I assume that the rate of heat loss and therefore temperature drop (since water volume is constant) is exactly proportional to the temperature difference between the spa water and the air temperature and I assume a constant proportionality.
dT/dt = -k*(T - Tair)
T = Tair + (T0 - Tair)*exp(-k*t)
Let's assume that k = 0.5 so that with Tair = 40ºF and T0 = 104ºF, then after one day so t=1 we have
T = 40ºF + (104ºF - 40ºF)*exp(-0.5) = 78.8ºF
If we had the spa heating set to 85ºF, then the temperature would get to that temperature at
t = -ln((T - Tair)/(T0 - Tair))/k = -ln((85-40)/(104-40))/0.5 = 0.70 days
so you would be maintaining an 85ºF temperature for 1-0.70 = 0.30 days with a heat loss dT/dt = -0.5*(85-40) = 22.5 so this is a total energy proportional to 0.30*22.5 = 6.75 (I'm using degrees as a proportional unit of energy).
You then need to heat up the spa from 85ºF to 104ºF which in these same proportional units (where heat is proportional to temperature deltas) is 19 where we assume near instant heating for simplicity. So that's a total proportional heat of 6.75+19 = 25.75
If instead you were to maintain a 104ºF temperature for the entire day, then this is a total energy proportional to 1*(104-40) = 64 so around 2.5 times more energy needed.
If the outside temperature were only 70ºF, then we have
t = -ln((T - Tair)/(T0 - Tair))/k = -ln((85-70)/(104-70))/0.5 = 1.6 days so the spa temperature wouldn't drop from 104 to 85 over just one day. Instead, the temperature would drop to
T = 70ºF + (104ºF - 70ºF)*exp(-0.5) = 90.6ºF
so we just have heating to 104ºF so 13.4 proportional units of heat.
If instead you maintain 104ºF temperature, then the total energy is proportional to 0.5*(104-70) = 17 so around 1.3 times as much energy is needed.