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| (defn pow [x n] | |
| (reduce * (repeat n x))) | |
| (pow 55 2) ; 3025 | |
| (defn square [x] | |
| (pow x 2)) | |
| (square 55) ; 3025 | |
| (def sum (comp (partial reduce +) map)) | |
| (sum identity (range 0 11)) ; 55 | |
| (sum square (range 0 11)) ; 385 | |
| (defn sum-square-difference [n] | |
| (let [nums (range 0 (+ 1 n))] | |
| (- (square (sum identity nums)) | |
| (sum square nums)))) | |
| (sum-square-difference 10) ; 2640 | |
| (sum-square-difference 100) ; 25164150 |
Why didn't you simply solve it in closed form? D(n) = T(n)^2-P(n) where T(n) is the n'th triangular number = n(n+1)/2 and P(n) is the n'th square pyramidal number = (2n^3+3n^2+n)/6. This simplifies to the exact formula for S(n) = n/12[3n^3+2n^2-3n-2]. This leads to a one line Clojure definition with much less room for error to creep in. One should always check the math first as opposed to diving straight into algorithmics.
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