Pure mathematical solution in JS. For larger n's it can be faster than the loop version by roughly a factor of log₂ n.
function who_is_next(names,n){ let p=2**Math.floor(Math.log2(1+--n/names.length)) return names[Math.floor((n-(p-1)*names.length)/p)] }
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Pure mathematical solution in JS. For larger n's it can be faster than the loop version by roughly a factor of log₂ n.