First we start with a list of prime numbers, then write the cumulative sums for the prime numbers (these images were taken with my laptop camera off a whiteboard): My method for computing P(n,k) is pretty simple. How big is P(2011,k) compared to, the number of primes less than or equal 2011? Let us write P(n,k) as the number of primes less than or equal to n that can be written as a sum of at least k consecutive primes. Naturally I wonder, how rare are these numbers - numbers that are prime but also a sum of a bunch of consecutive primes? This seems like a problem easily solved with some programming. Furthermore (as I already said), 2011 is prime. Not only that, but according to numerous math bloggers and twitterers on the internet, 2011 is the sum of 11 consecutive primes - that is, 2011 = 157 + 163 + 167 + 173 + 179 + 181 + 191 + 193 + 197 + 199 + 211.
0 Comments
Leave a Reply. |