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  • \$\begingroup\$ Nice! You can save one more character by changing "s=s*..." to "s*=..." \$\endgroup\$ Commented Oct 25, 2011 at 19:41
  • \$\begingroup\$ @DerekKurth: I had thought that when I was first doing optimizations, but that would mess up the logic because it needs to be s*(i-j)/j, not s*((i-j)/j). \$\endgroup\$ Commented Oct 25, 2011 at 19:49
  • \$\begingroup\$ Hmm, I tried it as s*=... in the jsfiddle and it seemed to work. Maybe I did something wrong, though. \$\endgroup\$ Commented Oct 25, 2011 at 20:16
  • 1
    \$\begingroup\$ @DerekKurth: Technically it is the same, but the idea is that if you multiply by (i-j) before dividing by j, then there is no need for floating point arithmetic because the results should always be an integer. If you do ((i-j)/j) first, this will result in decimal values which can be a source of error, and at the very least will require extra code for rounding/truncating. You don't begin to see this until you get to about n>11, and you'll see decimal values in the output, i.e., 1 11 55 165 330 461.99999999999994 461.99999999999994... \$\endgroup\$ Commented Oct 25, 2011 at 21:47
  • \$\begingroup\$ Ah, that makes sense! \$\endgroup\$ Commented Oct 25, 2011 at 22:10