(In reply to
re: Solution--another guess... by Brian Smith)
Ok, if 1855 is correct, then I was probably heading in the right direction with the idea of taking on 1's to the front and back side of the previous binary number.
I guess I should have known better than to guess 63 since you already admitted that the 12th number in the series was in the range of 100-500. Telly made mention of the fourth in each seies being all 1s and appearing to double each time. If that's the case, then number 12 in the series should be 255.
A slight modification to the algorithm I gave earlier so that A' is not related to A, but to D is to tack on a 101 to D to get A'. To get A'' from D' (which is 1111 base2) is to tack on a 10101 in order to get 101011111 (=351). The next A (A''') would get a 1010101 (think the same number of 0s as there are apostrophes).
If this is the case, then numbers 13-16 in the sequence should be: 22015, 29183, 123903, and 65536.
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Posted by Erik O.
on 2004-06-10 16:14:00 |
re(2): Solution--another guess...
