There's a youtube video on the construction, but the chalk board is blurry, and I've seen 3 different versions, including Jacob Bernoulli's (shown in wikipedia/faulhaber's formula) which omits the zeroes. There are almost no articles on Faulhaber's triangle (just his formula) and it doesn't occur in my textbooks on number theory.
It resembles the binomial theorem terms but it also looks like infinite sums, and that's throwing me off. I see some patterns (diagonal zeroes, repeats of 1/n, repeats of 1/2, repeats of n/12) but the creation of new terms like the -1/30 and +1/42 eludes me.
$$ 1$$ $$ \frac{1}{2}+\frac{1}{2}$$ $$ \frac{2}{12}+\frac{1}{2}+\frac{1}{3}$$ $$ 0+\frac{3}{12}+\frac{1}{2}+\frac{1}{4}$$ $$ -\frac{1}{30}+0+\frac{4}{12}+\frac{1}{2}+\frac{1}{5}$$ $$ 0-\frac{1}{12}+0+\frac{5}{12}+\frac{1}{2}+\frac{1}{6}$$ $$ \frac{1}{42}+0-\frac{1}{6}+0+\frac{6}{12}+\frac{1}{2}+\frac{1}{7}$$ $$ 0+\frac{1}{12}+0-\frac{7}{24}+0+\frac{7}{12}+\frac{1}{2}+\frac{1}{8}$$
I get the impression there's some very simple arithmetic steps I'm missing to make it happen. One of the few texts (https://www.maa.org/sites/default/files/Torabi-Dashti-CMJ-2011.pdf) I found said "these are found by multiplying the number directly above and to the left by row i divided by column j (i.e., times i/j) and assuming my triangle were aligned left rather than centered in a table) but 1/6th for instance has 1/2 directly above it and nothing to the left of it, so how do you get 1/6th (which I have as 2/12 for pattern clarity) starting with 1/2 and nothing?
The triangle in your question numbered by rows $n$ from $0$ at the top and increasing downward, and by columns $k$ from $0$ at the left and increasing to the right we will denote by $\, T(n,k). \,$ Now $\, T(n,k)(n\!+\!1)!\,$ is the triangular OEIS sequence A080779 with more information. The $\, T(n,k) \,$ is the coefficient of $\, x^k \,$ in $\, \frac1x\int_0^x B_n(x+1)\, dx \,$ where $\, B_n(x) \,$ is the $n$-th Bernoulli polynomial.
One way to construct the triangle is this. Assuming that row $n-1$ is given, use the formula $\, T(n,k) = T(n-1, k-1) \, n/(k+1) \,$ to compute $\, T(n,1) \,$ up to $\, T(n,n) \,$ and then $\, T(n,0) = 1 - \sum_{k=1}^n T(n,k). \,$