Add sumDivisors and peterson implementation

sumDivisors/Program.fs

@@ -0,0 +1,43 @@
+open System
+
+let factorsWithMolteplicity n =
+  let rec loop c p =
+    if c < (p * p) then [c]
+    elif c % p = 0 then p :: (loop (c/p) p)
+    else loop c (p + 1)
+  loop n 2
+  |> List.countBy id
+
+let isPerfect n =
+  let sumDiv k =
+    k
+    |> factorsWithMolteplicity
+    |> List.map (fun (p,m) -> ((pown p (m+1))-1)/(p-1))
+    |> List.fold (*) 1
+  sumDiv n = 2*n
+
+// brute-force approach
+let divisors n =
+    [1..n/2]
+    |> Seq.filter (fun f -> n % f = 0)
+let sumProperDivisors n =
+    n
+    |> divisors
+    |> Seq.sum
+
+// perfect
+let isPerfect2 n = (sumProperDivisors n = n)
+
+let timer = new System.Diagnostics.Stopwatch()
+timer.Start()
+printfn "%A" (isPerfect 8128)
+timer.Stop()
+printfn "%f" timer.Elapsed.TotalMilliseconds
+printfn "Elapsed Time: %i" timer.ElapsedMilliseconds
+
+let timer2 = new System.Diagnostics.Stopwatch()
+timer2.Start()
+printfn "%A" (isPerfect2 8128)
+timer2.Stop()
+printfn "%f" timer2.Elapsed.TotalMilliseconds
+printfn "Elapsed Time: %i" timer2.ElapsedMilliseconds

sumDivisors/sumDivisors.m

@@ -0,0 +1,24 @@
+N=1e7;
+
+for n=2:N
+	% A contains the unique prime factors of n; M their multiplicities
+	[A,M]=factor(n);
+	l=length(A);
+	s=1;
+	for i=1:l
+		% formula for the sum of divisors of p^k
+		c(i)=((A(i)^(M(i)+1))-1)/(A(i)-1);
+		s*=c(i);
+	end
+	if 2*n==s
+		printf("%d is Perfect!\n",n)
+		m=n;
+		k=0;
+		while mod(m,2)==0
+			m=m/2;
+			k++;
+		end
+		k++;
+		printf("The corresponding Mersenne prime is %d\n",((2^k)-1))
+	end
+end