# Consecutive Prime Sum — Project Euler (Problem 50)

Many problems in Project Euler relate to working with primes. I use primesieve-python to help solve such problems. It consists of Python bindings for the primesieve C++ library. Generates primes orders of magnitude faster than any pure Python code. Features:

• Generate a list of primes
• Count primes and prime k-tuplets
• Print primes and prime k-tuplets
• Find the nth prime
• Iterate over primes using little memory

Anyway, here’s Problem 50 from Project Euler:

Here’s how I did it:

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 # Question: Which prime, below one-million, can be written as the sum of the most consecutive primes from primesieve import * from math import * # Generate list of primes under a million primes_under_million = generate_primes(10**6) # Sum of consecutive primes is of order 0.5(n^2)(logn) # Calculate 'n' so that sum of consecutive primes is less than a million (and not necessarily prime) nsum = 1 n = 1 while nsum < 10**6: nsum = 0.5*(n**2)*(log(n, e)) n += 1 # Calculate index so that sum of first 'index' consecutive primes is under a million and also prime primes_subset = primes_under_million[:n] nsum = sum(primes_under_million[:n]) while nsum > 10**6: n -= 1 nsum = sum(primes_under_million[:n]) primes_sum = 0 index = 0 for i in range(len(primes_subset)): if i % 2 == 1: pass else: sumprimes = sum(primes_subset[:i]) if sumprimes > primes_sum and sumprimes < 10**6 and sumprimes in primes_under_million: primes_sum = sumprimes index = i # Print out sum of consecutive primes till 'index', index, n # print primes_sum, index, n # Check consecutive primes within a range (index to n) such that their number is greater than index and maximum j = index + 1 start = 0 while j <= n: while (j-start) >= (n-index): sumprimes = sum(primes_subset[start:j]) if sumprimes > primes_sum and sumprimes in primes_under_million: primes_sum = sumprimes start += 1 j += 1 start = 0 print primes_sum
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