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.
Here is some interesting info on throwies from evilmadscientist.com
[Connecting a 1.7 V LED directly to a CR2032]
Wait-- 107 mA?!--‽
Yes, this is reproducible. (That is to say, we wasted used up another battery just because we didn't believe it either.) But holy cow anyway.
"And they said this was safe?" There are a couple of legitimate concerns here. Lithium coin cells aren't designed to source nearly that much power-- and aren't lithium batteries a fire hazard? And why does my LED-- rated for 25 mA continuous current survive this? I've certainly seen enough LEDs destroyed by overcurrent, and this one was over 25 mA for ten minutes solid. But, and perhaps against my better judgement, I do believe that this actually is safe in practice. With all of the throwies and similar things out there -- don't forget the keychain flashlights -- they just don't seem to be exploding or catching on fire. (Breaking, falling apart, running out of photons, yes-- but those modes of failure are usually not as dangerous.)
[...]
So.... Do you need a resistor? No, not really. As we said, it seems to be reasonably safe without one. Should you use a resistor? Yeah you should, if you want a red, yellow, or orange LED to last more than a day or so. So you can solder a resistor in place, but that somewhat defeats the purpose of easy to assemble throwies.
So to make it not a total pain-in-the rear to add resistors, here's a way to do it without soldering: just twist it.
Source: http://www.evilmadscientist.com/article.php/throw
Best Answer
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) Correct. (b) Incorrect.
In your first photo the resistor is in series with the LED. This is correct.
In your second photo your resistor is inserted into two points on the same track. The track is bypassing the resistor with its own much lower resistance (almost zero in comparison) so there is no current limiting to the LED. It should be clear from Figure 1b that you have applied 9 V directly across the LED.
The red and blue lines on the edge of the board indicate that those strips are continuous.
simulate this circuit
Figure 2. (b) is more efficient than (a).
You can extend your battery life by wiring your LEDs in strings of three. The same current can then light up three LEDs rather than one so your battery will last three times longer.