Are the Bernoulli numbers $B_{2n}$ given by an elementary function of $n$? I found an old question asking for a proof that the factorial function is nonelementary, and the  Claim 2 section of the answer there (by Vincenzo Oliva) doesn't quite make sense to me: https://math.stackexchange.com/a/1394130/314780.
My issue is when they claim that the $2n$-th Bernoulli number $B_{2n}$ is an elementary function of $n$. Their justification is that it is defined as the constant term of the $2n$-th Bernoulli polynomial $B_{2n}(m)$, and polynomials are elementary. But the fact that $B_2(m), B_4(m), B_6(m), ...$ are each individually elementary functions of $m$ does not imply that the function $B_{2n}(0)$ is an elementary function of $n$. Is there a way to fix this proof (or am I just misunderstanding something)? 
 A: You ask if there exists an elementary function $f(2n)$ that evaluates $B_{2n}$. I suspect any such function is  not elementary. Please correct me if I am wrong. 
Note the following identity:
$$B_n = (-1)^nn \zeta(1-n)$$
$(-1)^n$ and $n$ are elementary functions of $n$. However, the zeta function is not an elementary function. The product of an elementary function and a non-elementary function is non-elementary, thus the RHS of the equation should be non-elementary. Hence any function $f(n) = B_n = (-1)^nn \zeta(1-n)$ should be non-elementary. 
Your question specifically asks about the even Bernoulli numbers (recall all odd Bernoulli numbers are zero). Let us show the existence of an elementary function for the even Bernoulli numbers would lead to a contradiction. Suppose that there is an elementary function $g(n) = B_{2n}$.  Then we could write the following function:
$$f(n) = \frac{g(n)+(-1)^ng(n)}{2}$$
When $n$ is even, we get $g(n) = B_{2n}$, and when odd, we get zero (i.e. the odd Bernoulli numbers). Hence $f(n)$ calculates $B_n$ and is composed of elementary functions, and thereby is elementary. However this contradicts our earlier line of reasoning. 
