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| visits | member for | 1 year, 5 months |
| seen | 19 hours ago | |
| stats | profile views | 23 |
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Oct 30 |
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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.". |
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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. |
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Nov 27 |
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Integer exponentiation algorithm for the special case $3^n$ @Simone: I believe so even though I don't have the referenced print physically available. |
