Background
I've been learning the basics of Bernoulli numbers and this is a reference to What is the simplest way to get Bernoulli numbers? and vadim123's succinct reply, where he stated:
The simplest way to calculate them, using very few fancy tools, is the following recursive definition:
$$B_n=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}$$
in other words
$$B_n=1-{n\choose 0}\frac{B_0}{n-0+1}-{n\choose 1}\frac{B_1}{n-1+1}-\cdots -{n\choose n-1}\frac{B_{n-1}}{n-(n-1)+1}$$Here, ${a\choose b}$ denotes a binomial coefficient. So, we begin with $B_0=1, B_1=\frac{1}{2}$, and we can calculate $B_2$ using the above recursive definition. That is, $B_2=1-{2\choose 0}\frac{B_0}{3}-{2\choose 1}\frac{B_1}{2}=1-\frac{1}{3}-2\frac{\frac{1}{2}}{2}=\frac{1}{6}$.
Now, with $B_2$ in hand, we can calculate $B_3$. And so on.
I've decided to find $B_4$, but I wanted to see if my process was correct.
My Attempt
Using the following recursive definition:
$$\begin{align*}
B_3&=1-{3\choose 0}\frac{B_0}{3-0+1}-{3\choose 1}\frac{B_1}{3-1+1}-{3\choose 2}\frac{B_2}{3-2+1}\\
&=1-\frac{1}{4}-3\frac{\frac{1}{2}}{3}-3\frac{\frac{1}{6}}{2}\\
&=0\\
\\
B_4&=1-{4\choose 0}\frac{B_0}{4-0+1}-{4\choose 1}\frac{B_1}{4-1+1}-{4\choose 2}\frac{B_2}{4-2+1}-{4\choose 3}\frac{B_3}{4-2+1}\\
&=1-\frac{1}{5}-4\frac{\frac{1}{2}}{4}-6\frac{\frac{1}{6}}{3}-{4\frac{0}{2}}\\
&=\boxed{-\frac{1}{30}}\\
\end{align*}$$
Would this process correct? Also, for higher Bernoulli numbers, say $B_{50}$, is there a better way to find them?
Edit @RobPratt, thank you kindly for correcting my errors! And, for providing the link.
Best Answer
Yes, this process is correct, but you had a couple errors that I just corrected for $B_3$. Note that $B_n=0$ for odd $n>1$, so you can skip the calculation. For efficient computation, see https://en.wikipedia.org/wiki/Bernoulli_number#Efficient_computation_of_Bernoulli_numbers