How many digits can you create with only one seven segment display?

I wrote a python program to check all possibilities and I got:

Actual Program:

There are

Reason:

NOTE: Too much for spoiler tags, so I only did the last.

EDIT: After all that work, I find out I have the same answer as wythagoras.

Seven lights

If all seven lights are on, there is one possible digit.

Six lights

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

Five lights

There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

 ---     ---  
        |   | 
 ---     ---  
|   |         
 ---     ---       

Therefore, there are 19 digits with 5 lights.

Four lights

There are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in the "o" section and still be disconnected. This is a total of 8 ways to disconnect. Also, if you turn on the two lights in opposite corners they will be disconnected. There are 2 ways to do this.

Also, you can orphan a vertical piece as follows;

 ---         ---
|   |       |
 ---    -->  
|   |       |   |
 ---

There are 4 ways to do this.

Lastly, you can have two vertical bars disconnected by turning off all three horizontal lights.

Thus, the total number of ways to have four lights is 35-8-2-4-1=20.

Three lights

With three lights, you can have a "C" and "U" and "n" and backwards "C", both on the top and on the bottom for a total of 4x2=8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right (kind of a zig zag), or its reverse, for a total of 2 more. Grand total is 16.

Two lights

You can have 4 "L" shaped pieces around every corner on the top and another 4 on the bottom. Also, you can have 2 vertical digits; one on the left and one on the right. Total is 10.

One light

There are 7 digits with 1 light trivially.

So, the total is:

I am going to assume that you need to be able to see the indicator "in the dark". That means that there are several shapes that "look the same" with just an offset difference; and such shapes would not be distinguishable. I do assume you will be able to tell the difference between a "short vertical" and a "long vertical". If that is not the case you will have one fewer "digit".

I can't see a way to avoid enumerating all possibilities:

Of the single-element shapes, there are just two possibilities: horizontal, or vertical.

Of the two-element shapes, there are five possibilities: both vertical (on left or right), and four L shapes.

Of the three-element shapes, there are once again four L's; also there are two T's and four "little" C's, and two "lightning bolts".

Of the four-element shapes, there are four ways to do an F, two ways to do a C, four ways to do an "incomplete P" and four ways to do a "lower case h". There are also four ways to do a "shepherd's crook" and a single "small square".

The five-element "digit" can be in several possible configurations: for three horizontal and two vertical, there are four allowed positions (looking like E, 3, 5 and 2). For two horizontal and three vertical, connectivity is always ensured, meaning there are 3x4=12 possible configurations; and with one horizontal and four vertical, there are 3 configurations.

The six-element digit is always connected: there are 7 of these (any one of the segments could be off) and they are all distinct.

Finally there is the seven digit element.

The total is

Here is a map with all of them - it shows that there are 15 "missing characters" which are actually duplicates (just shifted horizontally, vertically, or both). Add that to my total, and you get the same answer as some of the other posts.

Note - this is lower than other answers because of the "in the dark" element. If that is not included, some of the patterns identified (especially for the lower element counts) can be repeated in multiple locations on the indicator.

@wythagoras and @Trenin have already provided correct answers to this puzzle, but for the sake of completeness, here is an explicit table of all possible digits, sorted by the number of segments turned on.

One can easily see that there are

possible configurations in total.

The answer is

Because of the

The easiest way to tackle this problem is to realise there are 7 segments and

It might help to look at a simpler version of the problem. For example, imagine if we had only a 3-segment display. The first segment could be off/on, and for each of those two settings the second segment could be off/on, which gives 2 × 2 = 4 variants. For these 4 variants, the final segment could be off/on, which gives 8 variants, or 2³.