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I'm trying to find insights about the following puzzle, to see if I can find it on the OEIS (and add it if it's not already there):

Suppose I give you a triangular array of light bulbs with side length $n$:

o o o o o o o o o o o o o o o o o o o o o 1 2 ... n

I'm going to turn on three lightbulbs that form an "upright" equilateral triangle as in the following example:

o o x o o o o o o o o x o o x o o o o o o

Before I turn on the lights, your job is to remove as many lightbulbs as possible from the array—without losing the ability to deduce the triangle of bulbs that has been turned on. To be clear, if a lightbulb has been removed, it is not lit up when its position is turned on.

For example, if you removed the following bulbs (marked by . ) you would only see the following two lights turn on (marked by x ), which is enough uniquely deduce the third (unlit) position:

. . . o . x . . o . . o o o o . => o o o . o o o o . o x o o . <- the third unlit position o . . . o o o . . . o o

Let $a(n)$ be the maximum number of bulbs that can be removed without introducing any ambiguities.

With a naive algorithm, I have checked values up to a triangle with side length 7, as seen below:

. . . o . . o o . o . . . . . o . o o . . . . . o o o o o . o o . o . . . . . o o o o . o o o o o . o . o . o o . . . o o . o o o o . . o o o . . . o o o . o . o o o a(2) = 3 a(3) = 4 a(4) = 5 a(5) = 7 a(6) = 9 a(7) = 11

Searching for this sequence on the OEIS turns up dozens of results.

As an upper bound for this sequence, we need the different configurations of 3, 2, 1, or 0 lights to be able to represent all of the $\binom{n + 1}{3}$ possible triangles. That is:

$$\binom{n + 1}{3} \leq \binom{b(n) - a(n)}{3} + \binom{b(n) - a(n)}{2} + \binom{b(n) - a(n)}{1} + \binom{b(n) - a(n)}{0}$$ where $b(n) = \frac{1}{2}n(n+1)$.