Since the Stirling numbers are the coefficients of a polynomial of degree $n$ which is already factored, it can be evaluated at the roots of unity in $O(n)$ O(n\log n)$multiplications. Then, by Fourier transform, the coefficients can be found in another$O(n\log n)$multiplications, of roughly$O( n)$bit numbers. This will find an entire row of the Stirling triangle in time$O(n^2 \log^k n),$or$O(n \log^k n)$time per Stirling number. The exponent$k$is something like$2+\epsilon.$REMARK The recurrence approach takes$O(n^2)$arithmetic operations, or$O(n^3)$bit operations to generate either one, or all of the Stirling numbers, so if the goal is to generate all of them up to a certain size, the simple approach is better. However, if one needs either a single number or a row, the approach I give is considerably faster. 4 Added remark Since the Stirling numbers are the coefficients of a polynomial of degree$n$which is already factored, it can be evaluated at the roots of unity in$O(n)$multiplications. Then, by Fourier transform, the coefficients can be found in another$O(n\log n)$multiplications, of roughly$O( n)$bit numbers. This will find an entire row of the Stirling triangle in time$O(n^2 \log^k n),$or$O(n \log^k n)$time per Stirling number. The exponent$k$is something like$2+\epsilon.$REMARK The recurrence approach takes$O(n^2)$arithmetic operations, or$O(n^3)$bit operations to generate either one, or all of the Stirling numbers, so if the goal is to generate all of them up to a certain size, the simple approach is better. However, if one needs either a single number or a row, the approach I give is considerably faster. 3 corrected silliness Since the Stirling numbers are the coefficients of a polynomial of degree$n$which is already factored, it can be evaluated at the roots of unity in$O(n)$multiplications. Then, by Fourier transform, the coefficients can be found in another$O(n\log n)$multiplications, of roughly$O(\log O( n)$bit numbers. This will find an entire row of the Stirling triangle in time$O(n O(n^2 \log^k n),$or polylog$O(n \log^k n)$time per Stirling number. The exponent$k$is something like$2+\epsilon.\$