4 ms·
Ah yes log2(n!) is O(n.log2(n)) bits, not actually n.log2(n) bits. From Stirling's approximation* it seems it's more like n.log2(n)-n+O(log2(n)) bits. https://
by darkmighty 11y ago
Ah yes log2(n!) is O(n.log2(n)) bits, not actually n.log2(n) bits. From Stirling's approximation* it seems it's more like n.log2(n)-n+O(log2(n)) bits.
https://en.wikipedia.org/wiki/Stirling%27s_approximation https://en.wikipedia.org/wiki/Stirling%27s_approximation