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The set of the three digits of such a number can be arranged to form an increasing arithmetic sequence. There are 8 possible sequences with a common difference of 1, since the first term can be any of the digits 0 through 7. There are 6 possible sequences with a common difference of 2, 4 with a common difference of 3, and 2 with a common difference of 4. Hence there are 20 possible arithmetic sequences. Each of the 4 sets that contain 0 can be arranged to form $2\cdot 2!=4$ different numbers, and the 16 sets that do not contain 0 can be arranged to form $3!=6$ different numbers. Thus there are a total of $4\cdot 4+16\cdot 6=\boxed{112}$ numbers with the required properties.