1N4007 : In theory the reverse voltage would divide between the LEDs and not be too high for any of them
In reality i dont know how evenly it would divide (or even better : when the sring is outdoors and some moisture leaked in, and forms small leaks across several LEDs essentially transfering the reverse voltage to the others), and possibly there might happen that some LED will see too high voltage
You can not rely to this, LED's are extremely sensitive for the overvoltage.
Side note: LED's are the most ESD sensitive component used in the automotive industry today, that comparison include all sub-micron technology microcontrollers and similar devices...
Next question, if the LED begings to pass tiny current when it's max is reached and ballancing the sysem this way, and if this tiny current does not damage the LED itself
The LED start to degrade way sooner (for lower voltage) then there start to flow any current. It is a dielectric breakdown, where when exposed to too high voltage, charges accumulate on the spot of the maximum field intensity, pronouncing this effect even more, so over time the charges form a kind of string. At this time there is still no current yet, but as the local electrical field is increased, the charge string grw faster and faster. But when such string meet the other conductive part, this string cause a short circuit breakdown, what mean destruction. Only at this point start to flow some current, but that is already too late, such LED would not anymore emit any light.
Resistor : Without it the string is way more sensitive to line voltage fluctations - small change in line voltage would cause large change in LED current
Definitely for mains you need quite high voltage to drop on the resistor, mainly to have room for the overvoltage spikes.
For 230V the chain should have voltage drop maximum 170V, the rest should be on the resistor. But be aware, then with such chain the voltages and currents are by far not the same shape, so simple "230V = Vled + Vresistor" does not work at all, you have to work with the actual waveforms. But the best tool for these calculations is the regular table processor software (MS Excell, OpenOffice Calc,...).
Into one column (let say B; in column A you may put the time) you put the mains voltage, how it evolve with time for 1 or 1/2 (see below) sinewave period, in about 100 rows (B = Vmains*sqrt(2)*sin(2*Pi*Freq*A)). Keep the Vmains in a separate cell and use it as a parameter, to allow easy modifications, e.g. to check the sensitivity to mains fluctuations.
Into the 3'rd column (C) you put the equation calculating the resistor voltage drop as difference between the actual voltage on the mains (in given row) minus the expected LED voltage drop (you may keep it in separate cell as a parameter, to allow easy modifications).
Then in the next column (D) you put the equation for circuit current, keeping zero, when the mains voltage is below the LED drop (D = if(B>C;(B-C)/Res;0)). Take the "Res" from some separate cell as a parameter (to allow easy trials).
From the column D you then let the excell calculate:
DCcurrent = average(D)
LEDtotalPower = Vled * DCCurrent
MainsRMSCurrent = sqrt((stdev(D))^2 + DCcurrent^2); that is Excell math trick I use to calculate the rms value of an array of sample values
ResistorPowerDissipation = Res * MainsDCCurrent^2
Then tune the Res value so, you get required DC current for your LED string.
If you use only one halfwaves (via the common 1N4007), let the tool calculate the above from the whole sine period.
If you want to use both halfwaves, so e.g. greatz bridge or LED pairs connected antiparallel (that I would recommend, as it automatically clamp the LED reverse voltage to a safe value; you could use such pair in place of the single incandescent socket), you let the tool use only the first (positive) halfwave.
