Abundant, deficient and perfect number classifications: Difference between revisions

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 deficient:15043, perfect:4, abundant:4953
 </pre>
+=={{header|Picat}}==
+<lang Picat>go =>
+  Classes = new_map([deficient=0,perfect=0,abundant=0]),
+  foreach(N in 1..20_000)
+    C = classify(N),
+    Classes.put(C,Classes.get(C)+1)
+  end,
+  println(Classes),
+  nl.
+% Classify a number N
+classify(N) = Class =>
+ S = sum_divisors(N),
+ if S < N then
+   Class1 = deficient
+ elseif S = N then
+   Class1 = perfect
+ elseif S > N then
+   Class1 = abundant
+ end,
+ Class = Class1.
+% Alternative (slightly slower) approach.
+classify2(N,S) = C, S <  N => C = deficient.
+classify2(N,S) = C, S == N => C = perfect.
+classify2(N,S) = C, S >  N => C = abundant.
+% Sum of divisors
+sum_divisors(N) = Sum =>
+  sum_divisors(2,N,cond(N>1,1,0),Sum).
+% Part 0: base case
+sum_divisors(I,N,Sum0,Sum), I > floor(sqrt(N)) =>
+  Sum = Sum0.
+% Part 1: I is a divisor of N
+sum_divisors(I,N,Sum0,Sum), N mod I == 0 =>
+  Sum1 = Sum0 + I,
+  (I != N div I ->
+    Sum2 = Sum1 + N div I
+    ;
+    Sum2 = Sum1
+  ),
+  sum_divisors(I+1,N,Sum2,Sum).
+% Part 2: I is not a divisor of N.
+sum_divisors(I,N,Sum0,Sum) =>
+  sum_divisors(I+1,N,Sum0,Sum).
+</lang>
+Output:
+<pre>(map)[perfect = 4,deficient = 15043,abundant = 4953]</pre>
 =={{header|PicoLisp}}==