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 Apr 28 awarded Scholar Apr 27 accepted Seeking proper terminology Apr 27 asked Seeking proper terminology Oct 30 comment Why are Mersenne primes easier to find? Any prime divisor of 2^p-1 must be of the form 2kp+1. Thus "in particular no prime less than 2p can be a divisor.". Mar 31 awarded Supporter Nov 29 awarded Teacher Nov 29 answered computing primes Nov 27 comment Integer exponentiation algorithm for the special case $3^n$ @HenningMakholm: That's correct. The referenced Knuth's algorithm exploits the binary representation of the exponent, so let me clarify - I need the binary result to work with it in an ordinary binary computer. Nov 27 comment Integer exponentiation algorithm for the special case $3^n$ @Simone: I believe so even though I don't have the referenced print physically available. Nov 27 awarded Student Nov 27 awarded Analytical Nov 27 asked Integer exponentiation algorithm for the special case $3^n$