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Given n>=0, create an array with the pattern {1, 1, 2, 1, 2, 3, ... 1, 2, 3 .. n} (spaces added to show the grouping). Note that the length of the array will be 1 + 2 + 3 ... + n, which is known to sum to exactly n*(n + 1)/2. seriesUp(3) → [1, 1, 2, 1, 2, 3] seriesUp(4) → [1, 1, 2, 1, 2, 3, 1, 2, 3, 4] seriesUp(2) → [1, 1, 2] ...Save, Compile, Run (ctrl-enter) |
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Difficulty: 321.0
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