Let's say that we have a Sierpiński triangle of width 1 and height 1
I want to find the function $y(x)$ for $0<x<0.5$ that gives the fraction of the Sierpiński triangle contained in $[0,x]$
If we take each of the 3 main triangles as 1/3 of the fractal, then at $x=0.5$ we have 50% of the fractal to the left, or the leftmost triangle, more half of the upper triangle: $y(x)=1/3+1/2*1/3=1/2$
At $x=0.25$ there is only half of the leftmost triangle, so the fraction is $y(x)=1/2*1/3=1/6$ It also can be calculated as the sum of each triangle to the left: $1/9*(1+1/3+1/3^2+1/3/^3+...)=1/6$
At $x=3/8$, by counting triangles, $y(x)=1/3$
There is a way to write a function of x? Only $0<x<0.5$ is needed
It looks like $y(x)=3y(x/2)$
By counting triangles, for
$x=(0, 1/16, 2/16, 3/16, 4/16, 5/16, 6/16, 7/16, 8/16)$
$y(x)=0/54,1/54, 3/54, 6/54, 9/54, 13/54, 18/54, 23/54, 27/54$
The last series is OEIS A282720, described as "Number of nonzero terms in first n rows of the base-2 generalized Pascal triangle P_2" Finding the Pascal triangle here is no surprising, becasue the Pascal triangle is famous for generating a Serpinsky triangle with his odd numbers:
Every column of triangles is given by the OEIS A007306 series, which has the formula
$$
\sum_{k=0}^n \left[ \binom{n+k-1}{2k} \bmod 2 \right] = 1, 1, 2, 3, 3, 4, 5, 5, 4, 5, 7, 8,,,,
$$
I'm interested in finding some kind of integral of the triangle.