ocaml: phase 5.1 euler10.ml baseline (sum of primes <= 100 = 1060, scaled-down PE10)

Sieve of Eratosthenes followed by a sum loop:

  let sieve_sum n =
    let s = Array.make (n + 1) true in
    s.(0) <- false;
    s.(1) <- false;
    for i = 2 to n do
      if s.(i) then begin
        let j = ref (i * i) in
        while !j <= n do
          s.(!j) <- false;
          j := !j + i
        done
      end
    done;
    let total = ref 0 in
    for i = 2 to n do
      if s.(i) then total := !total + i
    done;
    !total

Real PE10 asks for sum below 2,000,000; that's a ~2-3 second loop in
native OCaml but minutes-to-hours under our contended-host
spec-level evaluator. 100 keeps the run under 3 minutes while still
exercising the same algorithm.

110 baseline programs total.

lib/ocaml/baseline/euler10.ml

@@ -0,0 +1,22 @@
+let sieve_sum n =
+  let s = Array.make (n + 1) true in
+  s.(0) <- false;
+  s.(1) <- false;
+  for i = 2 to n do
+    if s.(i) then begin
+      let j = ref (i * i) in
+      while !j <= n do
+        s.(!j) <- false;
+        j := !j + i
+      done
+    end
+  done;
+  let total = ref 0 in
+  for i = 2 to n do
+    if s.(i) then total := !total + i
+  done;
+  !total
+
+;;
+
+sieve_sum 100

lib/ocaml/baseline/expected.json

@@ -26,6 +26,7 @@
   "exception_handle.ml": 4,
   "exception_user.ml": 26,
   "euler1.ml": 233168,
+  "euler10.ml": 1060,
   "euler2.ml": 4613732,
   "euler5.ml": 232792560,
   "euler6.ml": 25164150,

plans/ocaml-on-sx.md

@@ -407,6 +407,10 @@ _Newest first._
   binary search tree (`type 'a tree = Leaf | Node of 'a * 'a tree *
   'a tree`) with insert + in-order traversal. Tests parametric ADT,
   recursive match, List.append, List.fold_left.
+- 2026-05-09 Phase 5.1 — euler10.ml baseline (sum of primes ≤ 100 =
+  1060, scaled-down Project Euler #10). Sieve of Eratosthenes
+  followed by a sum loop. Used 100 instead of 2 million to fit our
+  contended host's runtime budget. 110 baseline programs total.
 - 2026-05-09 Phase 5.1 — euler5.ml baseline (Project Euler #5,
   smallest number divisible by all 1..20 = 232792560). Iteratively
   takes lcm of running result with i for i=2..n; lcm via gcd via