I posted this in another math forum, but have received no response: Let sopfr(n) be the sum of prime factors of n (see http://mathworld.wolfram.com/SumofPrimeFactors.html ). There have been observations that certain iterations involving the sopfr function seem to lead invariably to cycles, or closed loops (see for instance http://www.mathpages.com/home/kmath006/part6/part6.htm ). I looked at iterations of the form F(n)=sopfr(F(n-1)+F(n-2)), where F(1) and F(2) are any positive integers. These sequences also appear to end in repeating cycles sooner or later, in fact I have found the following four cycles (of length 1, 4, 5, and 13): 8 (eg F(1) = 1, F(2) = 9, F(n) = sopfr(F(n-1)+F(n-2)) 10,14,9,23 (eg F(1) = 1, F(2) = 12, F(n) = sopfr(F(n-1)+F(n-2)) 10,10,9,19,11 (eg F(1) = 1, F(2) = 1, F(n) = sopfr(F(n-1)+F(n-2)) 39,43,43,45,17,33,12,11,23,19,12,31,43. Interestingly, there seem to be similar results when taking the sum of prime factors of more than two preceding terms. For instance, with F(n)=sopfr(F(n-1)+F(n-2)+F(n-3)+F(n-4)+F(n-5)), a possible cycle is one that has a length of 31196 terms. You can start that cycle with, for instance, 49,93,435,98,92. With F(n)=sopfr(F(n-1)+...+F(n-7)), a possible cycle has a length of 28274564 (if my computation was right). Again similar results can be obtained by checking sequences of the type F(n) = sopfr(F(n-1)*F(n-2)) = sopfr(F(n-1)) + sopfr(F(n-2)), and so on. Is there a proof that these sequences should always end in a cycle?