As an example, consider b = 2^63 = 9223372036854775808(decimal) = 8000000000000000(hexadecimal) = 1000000000000000000000000000000000000000000000000000000000000000(binary) t = 5836946587753906177(dec) = 5100FFFF00000001(hex) = 101000100000000111111111111111100000000000000000000000000000001(binary) ContinuedFraction(t/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2] I have found that there are many such examples, and they are easy to find by computer search because t is a sum of few signed powers of 2. In the present example, t is a sum of 6 of them, and since binomial(64,6) * 2^5 = 2399179776 = 2.4 * 10^9 it would be feasible to consider all possibilities. Another example has t' = 101000100000000111111111111111011111111111111111111111111111111(binary) which yields a very "nearby" continued fraction to the first example: ContinuedFraction(t'/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2] -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)