Ruth-Aaron numbers: Difference between revisions

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@@ Line 165: / Line 165: @@
 {{libheader|Wren-seq}}
 {{libheader|Wren-fmt}}
-Have only looked for the first Ruth-Aaron triple based on factors (around 2.2 seconds overall) as, judging by the Raku experience, looking for the first triple based on divisors is going to be very slow indeed using Wren.
+To find the first thirty Ruth-Aaron pairs and the first triple based on factors takes around 2.2 seconds.
+However, with nearly 90 million trios of numbers to slog through, it takes around 68 minutes to find the first triple based on divisors.
 <lang ecmascript>import "./math" for Int, Nums
 import "./seq" for Lst
@@ Line 183: / Line 185: @@
 var countD = 0
 var countT = 0
-while (countF < 30 || countD < 30 || countT < 1) {
+while (countT < 1 || countD < 30 || countF < 30) {
     factors1 = factors2
     factors2 = factors3
@@ Line 194: / Line 196: @@
         countF = countF + 1
     }
-    if (countT < 1 && sum1 == sum2 && sum2 == sum3) {
+    if (sum1 == sum2 && sum2 == sum3) {
         resT.add(n)
         countT = countT + 1
@@ Line 216: / Line 218: @@
 System.print(resD.join(" "))
 System.print("\nFirst Ruth-Aaron triple (factors):")
+System.print(resT[0])
+resT = [] // divisors only
+n = 2
+factors1 = []
+factors2 = [2]
+factors3 = [3]
+sum1 = 0
+sum2 = 2
+sum3 = 3
+countT = 0
+while (countT < 1) {
+    factors1 = factors2
+    factors2 = factors3
+    factors3 = Int.primeFactors(n+2)
+    Lst.prune(factors3)
+    sum1 = sum2
+    sum2 = sum3
+    sum3 = Nums.sum(factors3)
+    if (sum1 == sum2 && sum2 == sum3) {
+        resT.add(n)
+        countT = countT + 1
+    }
+    n = n + 1
+}
+System.print("\nFirst Ruth-Aaron triple (divisors):")
 System.print(resT[0])</lang>
@@ Line 228: / Line 257: @@
 First Ruth-Aaron triple (factors):
+First Ruth-Aaron triple (divisors):
+89460294
 </pre>