A1 = 22/47
A2 = (22/47 - 1)/(22/47
A3 = (-25/22 - 1)/(-25/22)
A4 = (47/25 - 1)/(47/25)
A5 = -25/22
A6 = 47/25
A7 = 22/47
looks like if the subscript is divisible by 4, the answer is 22/47
since 2012 divides evenly by 4, we get
22/47 as a/b, so a = 22, b = 47
f(A) = 1 - 1/A
f(m/n) = (m/n - 1)/(m/n)
f((m-n)/m) = (m-n-m)/(m-n) = -n/(m-n)
f(-n/(m-n)) = (-n+n-m)/(-n) = m/n
so, f(f(f(A))) = A
every 3rd iteration, we are back to A.
2012/3 = 670 with remainder 2.
Thus, f2012(A) = f2(A)
so, take a look:
f(22/47) = 1 - 47/22 = -25/22
f(-25/22) = 1 + 22/25 = 47/25
Steve is right, it is a cycle of 3 terms, not 4 terms like I hastily concluded.
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