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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.".
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.