From df64e1015dee43eefce9b591b7bf6a62d57e8822 Mon Sep 17 00:00:00 2001
From: Philippe Zwietering <philippe.zwietering@gmail.com>
Date: Sat, 6 Mar 2021 01:32:47 +0100
Subject: [PATCH] Quite a slow solution for 47, but it works

---
 47/main.py | 68 ++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 68 insertions(+)
 create mode 100644 47/main.py

diff --git a/47/main.py b/47/main.py
new file mode 100644
index 0000000..9bb1800
--- /dev/null
+++ b/47/main.py
@@ -0,0 +1,68 @@
+'''
+The first two consecutive numbers to have two distinct prime factors are:
+
+14 = 2 × 7
+15 = 3 × 5
+
+The first three consecutive numbers to have three distinct prime factors are:
+
+644 = 2² × 7 × 23
+645 = 3 × 5 × 43
+646 = 2 × 17 × 19.
+
+Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?
+'''
+
+import math
+import numpy as np
+
+def sieve(n):
+    assert n > 1
+
+    ns = [True] * n
+
+    for i in range(2, math.ceil(np.sqrt(n))):
+        if ns[i]:
+            j = pow(i, 2)
+
+            while j < n:
+                ns[j] = False
+                j = j + i
+
+    return [i for i,val in enumerate(ns) if val][2:]
+
+def primeFactors(n, primes):
+    res = []
+
+    for p in primes:
+        if n % p == 0:
+            res.append(p)
+            while n % p == 0:
+                n /= p
+    
+    return res
+
+def main():
+    print("Hello this is Patrick")
+
+    end = 1000000
+    primeList = sieve(end)
+    primeSet = set(primeList)
+    consecs = 0
+
+    for n in range(100000, end):
+        if n not in primeSet:
+            if len(primeFactors(n, primeList)) == 4:
+                consecs += 1
+            else:
+                consecs = 0
+        else:
+            consecs = 0
+        
+        if consecs == 4:
+            for i in range(n-3, n+1):
+                print(i, primeFactors(i, primeList))
+            break
+
+if __name__ == "__main__":
+    main()
\ No newline at end of file