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359 points sdsykes | 7 comments | | HN request time: 0.214s | source | bottom
1. stevefan1999 ◴[] No.41886077[source]
But why do we have to "discover" it when we know the formula would be 2^N - 1...? Are we trying to prove a corollary or what?
replies(2): >>41886104 #>>41886109 #
2. aaronmdjones ◴[] No.41886104[source]
Not all 2^N - 1 are prime. For example, N=18 makes 2^N - 1 = 262143, which can also be written as 3^3 * 7 * 19 * 73 (not prime).
replies(2): >>41886111 #>>41887510 #
3. Jabrov ◴[] No.41886109[source]
There’s tons of numbers of the form 2^N-1 which are not prime.

To discover the primes, we have to iterate through the numbers and test their primality. With numbers that are so big, it’s very compute intensive

4. stevefan1999 ◴[] No.41886111[source]
Oh, right, all Mersenne number is in the form of 2^N -1, but Mersenne prime is Mersenne number plus being prime
replies(1): >>41887050 #
5. mort96 ◴[] No.41887050{3}[source]
And we look for Mersenne primes, AFAIU, mainly because Mersenne numbers are more likely to be prime than other numbers, so it's easier to find big primes that way.
6. umanwizard ◴[] No.41887510[source]
A lot simpler example is 2^4-1=5*3
replies(1): >>41890565 #
7. lupire ◴[] No.41890565{3}[source]
2^11-1 (prime power) for a less trivial example