Can I use Binary Exponentiation with this problem? 2^2^n mod k

Revision en1, by baowilliam, 2022-09-26 18:44:00

Hi, I'm doing an exercise on mods for large numbers, i don't know if there is a more efficient solution than using binary exponentiation? Hope to help you, thanks. (Sorry about my bad english :()

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  Rev. Lang. By When Δ Comment
en4 English baowilliam 2022-09-26 19:02:58 70 Tiny change: 'with (0\leq n\l' -> 'with $(0\leq n\l'
en3 English baowilliam 2022-09-26 18:48:01 10 Tiny change: 'caculate: 2^2^n mod k (1 ' -> 'caculate: (2^2^n + 1) mod k (1 '
en2 English baowilliam 2022-09-26 18:47:06 69
en1 English baowilliam 2022-09-26 18:44:00 257 Initial revision (published)