No wonder they're dim, they're only using half of the wave and there isn't enough voltage.
The most blue LEDs you can have in series is four, giving a total voltage drop of 14V, less than the peak voltage of 15.6V, after the bridge rectifier.
Another thing is, calculating the resistor to use isn't as simple as Ohm's law.
For a start, the duty cycle isn't 100% as the LEDs only conduct when their peak voltage is exceeded and the current varies throughout the sinewave too.
Here's how I would do it:
First we need the duty cycle but to get that we need the angle at which the LEDs start conducting.
[latex] \theta = arcsin \frac{Vdrop}{Vpeak}[/latex]
Vdrop is the total voltage drop of the LEDs plus the rectifier loss, in this case we'll assume there are fou LEDs giving a total drop of 14V, plus the rectifier losses which gives 15.4V
Vpeak is the peak power supply voltage which is equal to 12*1.414 = 17V.
Putting the aforementioned into the equation gives:
[latex] \theta = arcsin \frac{15.4}{17}=64.9^o[/latex]
Now let's calculate the duty cycle.
[latex]Duty = \frac{90-\theta}{90} = \frac{90-64.9}{90} = 27.89%[/latex]
Here's where my maths leaves me a bit, we need to calculate the RMS current but the problem is it varies during the part of the cycle when the LED is conducting.
Oh well I won't bother, since the peak current is a close enough approximation.
Now suppose the LED is being powered from 17V continiously; first we calculate the resistor value for that:
[latex]R = \frac{V-Vrdop}{Iforward}[/latex]
Where If is the Iforward is the LED forwad current; let's set this at 10mA. As the LED is only conducting for 27.89% of the time we can safely increase this to 1/27.89%.
0.01/0.2789 = 35.9mA.
[latex]R=\frac{17-15.4}{0.0359} = 45\Omega[/latex]