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EXAMPLE
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a(3) = 3. Every distinct row is periodic with a period dividing 6. b=1 generates 0, 2, 2, 1, 1, 0 repeating, b=2 generates 0, 0, 2, 2, 1, 1 repeating, and b=3 generates 0, 1, 2 repeating. All other values of b give one of these.
Using the dynamic Mathematica program provided with the paper by Dearden et al. (2013) (but taking the transpose of the output table), we see that for all b >= 1 the first row is always 0, 0, 0, 0, ..., so a(1) = 1.
By looking at the second rows, we see that for all b >= 1 the 2nd row is always 0, 1, 0, 1, ..., so a(2) = 1.
By looking at the 3rd rows, we see that for all b with 1 = b mod 3, we get 0, 2, 2, 1, 1, 0 repeating (with period 6); for all b with 2 = b mod 3, we get 0, 0, 2, 2, 1, 1 repeating (with period 6); and with 0 = b mod 3, we get 0, 1, 2 repeating (with period 3). (See also the example above.) Thus, a(3) = 3.
By looking at the 4th rows, we see that for all b with 1 = b mod 12, we get 0, 3, 0, 0, 1, 1, 2, 1, 2, 2, 3, 3 repeating (with period 12); for 2 = b mod 12, we get 0, 1, 2, 3, 2, 3, 2, 3, 0, 1, 0, 1 repeating (with period 12); for 3 = b mod 12, we get 0, 0, 0, 3, 3, 3, 2, 2, 2, 1, 1, 1 repeating (with period 12); for (b mod 12) = 0, 4, or 8, we get 0, 1, 2, 3 repeating (with period 4); for 5 = b mod 12, we get 0, 1, 0, 1, 1, 2, 2, 3, 2, 3, 3, 0 repeating (with period 12); for 6 = b mod 12, we get 0, 3, 2, 3, 2, 1, 2, 1, 0, 1, 0, 3 repeating (with period 12); for 7 = b mod 12, we get 0, 3, 0, 2, 3, 1, 2, 1, 2, 0, 1, 3 repeating (with period 12); for 9 = b mod 12, we get 0, 2, 0, 3, 1, 3, 2, 0, 2, 1, 3, 1 repeating (with period 12); with 10 = b mod 12, we get 0, 1, 2, 1, 2, 1, 2, 3, 0, 3, 0, 3 repeating (with period 12); and for 11 = b mod 12, we get 0, 1, 0, 3, 0, 2, 3, 2, 3, 1, 2 repeating (with period 12). Thus, a(4) = 10. (End)
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