Electronic – LED circuit and resistor recommendations

Well, let's analyze the circuit. We know that the power required in a DC circuit is:

P = Vsrc * Iout

We know that

I = (Vout - Vled)/R

and the power delivered to the LEDs is all that matters, so we want to maximize

Pr = (Vout - Vled) * I = (Vout - Vled)^2/R
Pled = Vled * I = Vled * (Vout - Vled)/R

Clearly, we want to minimize Pr and maximize Pled. We can do this without decreasing the current by reducing R and making Vled close to Vsrc.

This is accomplished by putting the LEDs in series.

However, your battery (isn't the 6LR61 a 9V battery?) will go from some nominal voltage (ex 9V) to a lower voltage - 9Vs are spec'd to be dead at 4.8V. This means that a passive solution will go dim while there's still charge left in the battery. For your original schematic, that might mean that you'd end up below the minimum current to turn the LED on, or for the series version, the voltage might go below the diode forward voltage.

A simple way to extract more brightness with the same power is to pulse the LEDs - Human eyes percieve blinking light to be brighter than continuous light, even if the average power is the same. A 555 timer or other oscillator/switch combination will be able to do this, no microcontroller required. Try playing with the duty cycle and frequency of your LEDs to see where it looks the brightest - You may be surprised!

Also, a switching power supply can increase the efficiency of your regulation circuit to 80, 90, or even 95%. However, that will drive up the cost and complexity of the design, and may not be necessary.

If you put all these LEDs in series you'll indeed need 38V, plus a bit for the series resistor. But you can make a circuit with several branches. Put 2 LEDs in series and you'll need 7.6V. So the remaining 9V - 7.6V = 1.4V is the voltage over your resistor. If you want 20mA through your circuit You divide this 1.4V / 20mA = 70 ohm, so a standard 68 ohm will do nicely.
Now that's for 2 LEDs, you could go for more if you had a higher voltage. the way to calculate the resistor is the same. If you want 10 LEDs from 9V, in theory you could put the circuit 5 times in parallel. That would be a total current of 100mA though (5 x 20mA), and that's a bit much for a 9V battery.

edit
It's been suggested that you could replace the 5 resistors by a single one, and branch from beneath this. In an ideal world that would be true; the resistor value would then be 1.4V / 100mA = 14 ohm. But this isn't an ideal world, and there may be small differences in LED voltages. In that case the branch with the lowest voltage will draw most of the current (100mA!) while the other LEDs will hardly light at all.