Let $X$ be the set of $(d,n)\in\mathbf{N}_{\ge 2}\times\mathbf{N}_{\ge 2}$ such that $B(d,n)$ is finite, and $X_d=\{n\in\mathbf{N}_{\ge 2}:(d,n)\in X\}$ the set of $n$ such that $B(d,n)$ is finite (by the infiniteness of $B(2,n)$ for large $n$, due to Adian and possibly others in the case of powers of 2). Then $X_d\supset X_{d+1}$ for all $d$, and $X_d$ is finite for all $d\ge 2$. Hence $X_d$ is eventually independent of $d$, and $X$ is thus a recursive subset of $\mathbf{N}^2$. Hence its complement, the set of $(d,n)$ such that $B(d,n)$ is infinite, is recursive.

Also, for given $k<\infty$, the set of $(d,n)$ such that $B(d,n)$ has cardinal $k$ is finite (obvious, since its cardinal is $\ge n^d\ge\max(n,2^d)$), hence recursive.

Of course this gives no information: it just shows that the question is not whether there is an algorithm, but whether we can describe a reasonable one (of course an efficient one is a bit hopeless since it would answer well-known open question, such as the question whether $B(2,5)$ is infinite).

Edit: Related fact: there is an (explicit, but ineffective) algorithm computing the cardinal of the

*restricted* Burnside group as a function of $(d,n)$: enumerate finite groups of exponent dividing $n$, and thus enumerate all homomorphism to those groups, and thus enumerate presentations of their kernels $K$. At some point, the kernel is aperiodic (i.e., has no nontrivial finite quotient), and we have to detect this: to do so, enumerate finite simple groups $S$ of exponent dividing $n$: there are finitely many and a list can be given (computably) in terms of $n$. We have to be careful because $K$ is only recursively presented, but the fact that there is no nontrivial homomorphism $K\to S$ is equivalent to the fact that some truncated presentation $K'$ of $K$ has no nontrivial homomorphism to $S$, so this will be eventually detected.

Now to go back to the initial question: I at least have: **if the Burnside groups have uniformly solvable word problem** (i.e. the input is $(d,n,w)$ with $w$ a group word in $d$ letters and the output is YES/NO according to whether $w\equiv 1$ in $B(d,n)$) **then here's an explicit algorithm for infiniteness of** $B(d,n)$: enumerate group words in $d$ and by the above word problem assumption, we can compute when they are distinct and thus compute an increasing sequence whose limit is the cardinal of $B(d,n)$. If $B(d,n)$ is infinite then this number will pass over the cardinal of the restricted $BR(d,n)$ which is also computable by the previous paragraph, so this will be detected; if $B(d,n)$ is finite this is detected also by standard means (enumerate all relations in $B(d,n)$ and stop if for some $k$ all elements of the $k$-ball belong to the $(k-1)$-ball.)

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