User:Emlightened/xkcd Notations/emlightened's Brackets

For those that have been watching the other thread, I'm only going to use this one for a bit.

My notation works like this:

If you've got a number, and it's got a {0} to the left of it (but nothing to the right), you delete the {0} and add one to the number (so {0}3 = 4). If you've got a number, and it's got a {1} to the left of it (but nothing to the right), you delete the {1} and replace it with that number of {0}'s (so {1}3 = {0}{0}{0}3). If you've got a number, and it's got a {2} to the left of it (but nothing to the right), you delete the {2} and replace it with that number of {1}'s (so {2}3 = {1}{1}{1}3). Generally, if you've got a number with a {n+1} to the left of it (but nothing to the right), you delete the {n+1} and replace it with that number of {n}'s (so {n+1}3 = {n}{n}{n}3). To make it more concise, instead of writing out {1}{1}{1}, you can write out 3{1}, for instance. i.e. a number to the left of {n} represents that number of copies of {n} (so 6{1}6 = {1}{1}{1}{1}{1}{1}6).

For {0}, {1} and {2}, we can write their values explicitly, easily. {0}n = n+1 {1}n = {0}{0}...{0}n (with n {0}'s) = 2n {2}n = {1}{1}...{1}n (with n {1}'s) = n*2^n

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