Project Euler 23 implementation
I played around with this for a good half hour without any luck. I was consistently getting ~57 seconds on my crappy M3 laptop.

As a last ditch effort though, I changed

(-&gt;&gt; abundants
     (filter #(&lt; % num))
     set)


to 

(-&gt;&gt; abundants
     (filterv #(&lt; % num)) ; Note the v
     set)


And got it down to 25 seconds! Since you seem to be using a faster computer than me, you'll likely get an even lower time.

Why does this cause such a dramatic change? ¯\_(ツ)_/¯. filterv is strict while filter is lazy, but I have no idea why that makes such a difference here. I would have expected it to be slower since the strict filtering would require a full iteration on its own, then the adding to the set would require another full iteration. Apparently something else is going on. I'll update this answer if I think of what may be happening.



I cleaned up the code a bit too, although none of the other changes made any performance impact. These are mostly just personal stylistic and practice recommendations. See the comments:

(def lower-limit 12) ; Defining these so they aren't magic,
(def upper-limit 28123) ; especially since upper-limit is used in two places

(defn sum-div [num]
  (-&gt;&gt; (range 2 (inc (int (Math/ceil (Math/sqrt num)))))
       (filter #(= (mod num %) 0))
       (map (fn [x]
              (let [d (/ num x)]
                (if (= x d)
                  x
                  [x d]))))
       (flatten) ; I personally prefer wrapping for consistency
       (distinct)
       (apply +) ; apply is generally regarded as more idiomatic for summing
       (inc))) ; Use inc instead of (+ 1)

(def abundants
  (-&gt;&gt; (range lower-limit upper-limit)
       (map #(vector % (sum-div %)))
       (filter #(&gt; (second %) (first %)))
       (map first)
       (set)))

(defn is-sum-of-two-abundants?
  [num]
  (let [sub-ab (-&gt;&gt; abundants
                    (filterv #(&lt; % num))
                    (set))]
    (some (fn [ab]
            (sub-ab (- num ab)))
          sub-ab)))

(time
  (-&gt;&gt; (range (inc upper-limit)) ; Use upper limit
       (remove is-sum-of-two-abundants?) ; remove is literally equal to (filter (complement ...))
       (apply +)))

"Elapsed time: 26714.867469 msecs"
=&gt; 4179871