If you put LEDs in series the forward voltage adds up, so 6.2v is more than 5v, which makes it almost sure the LEDs won't light up.
So you either use a led driver to boost the voltage while limiting the current to your desired value (for best efficiency, but probably not worth it in your case), or you connect the LEDs in parallel, each with it own resistor to limit current.
Figure the current you want in Amps (0.01A = 10mA) and the you take an average forward voltage (let's say 3.4v in your case, but keep in mind the forward voltage lowers as the temperature of the LED goes up) and now you can easily forward the resistor value from the elementary formula V = IxR
So V (input) - V (led) = I x R which means R = (Vin - Vled) / I ,,, in our case (5-3.4) / 0.01= 160 ohm which is not a standard value so you could go with a 150 ohm (and allow more than 10mA to go through led) or with 180 ohm but let slightly less current go through led at least initially.
With 180 ohm , you'd have I = (5v - 3.4v) / 180 = 0.0088 A or 8.8 mA but as the LED warms up the forward voltage could drop down to about 3.2v and then you'd have I = (5v-3.2v) / 180 = 0.01A or 10mA
One other thing worth knowing is the power wasted in the resistor, so that you'd know what resistors to use. Formula is P = IxIxR ... for our 180ohm and 0.01A numbers we have P = 0.01x0.01x180 = 0.018w or 18mW which means you can safely use cheap standard through hole 0.125w rated resistors , or even tiny cheap smd resistors (0603 and up) which typically are rated for 50mW and up. Generally, you don't want to dissipate more than around 75% of a resistor's rating
This is also worth knowing when you want to figure if it's worth moving from cheap resistors to a led driver, to get better battery life. In our example with 10mA leds and 3.2v forward voltage and 180 ohm resistors, we'd use 2 x 10mA x 3.2v = 64 mW on light, and 2 x 18mW = 36 mW on resistors as heat, so our circuit is only about 64% efficient.
A led driver IC could in theory raise this efficiency at around 90%, which means only around 5mW or so would be lost in the led driver circuit, which depending on what battery you choose, may extend the time the leds stays on.
Here's an example of such a chip, cheap and simple to use and still easy to solder by hand:
http://uk.farnell.com/microchip/mic2287cyd5-tr/pwm-wled-driver-1-2mhz-sot23-5/dp/2509928The datasheet has an example circuit in the first page, where they show how to use it to light up 3 white leds in series. The efficiency can go as high as 85%, compared to the 64% you get with resistors. Doesn't sound like much, but the energy savings can make a huge difference when the LEDs are more powerful, let's say 100mA or something like that.
It takes in up to 5.5v and can produce up to 34v with a switch current of 500mA , which means you can put in series as many leds as you wish as long as the sum of their forward voltages doesn't go over 34v.