4 ms·

9876543210 If I'm also allowed an operator, 3↑⁹⁸⁷⁶⁵⁴¹⁰2

• rookie 16y ago
  ha, yes allowed operators, as many as youd like.
  • sesqu 16y ago
    Then the best I can come up with on no notice is 2 ↑^(4 ↑^(6 ↑^(8↑^(9↑10)) 7) 5) 3. That would be trivial to increase substantially by using some other notation for the hyperoperation, but I'm going with the presumably most familiar one. I'm sure there are bigger numbers - specifically, Wikipedia mentions Conway chains, where a→b→c = a ↑^c b, and so 2→3→4→5→6→7→8→9→10 should be quite large indeed.
    • baddox 16y ago
      Excellent! I hadn't thought about it, but at a glance I'm pretty sure your ordering and grouping does generate the maximal result. However, the parent comment said as much punctuation as you like, which allows for unbounded expressions with unary operators like the factorial!
• deleted 16y ago
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