(idea) by jt Tue Apr 27 2004 at 4:44:01
Here are the solutions to the questions posed under the node Friedman Numbers:
25 = 5^2
121 = 11^2
125 = 5^(1+2)
126 = 6 * 21
127 = 2^7 - 1
128 = 2^(8-1)
153 = 3 * 51
123456789 = ((86 + 2 * 7)^5 - 91) / 3^4
987654321 = (8 * (97 + 6/2)^5 + 1) / 3^4
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