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 A175764 Number of iterations of the mapping k->f(k) to reach one of 2, 5, or 29, starting with k=n, and with f(k)=(k^2+4)/d, where d is the next-to-largest divisor of k^2+4, or -1 if the sequence never reaches one of the required values. 0
 1, 0, 9, 1, 0, 1, 2, 1, 1, 1, 1, 1, 8, 1, 2, 1, 4, 1, 1, 1, 1, 1, 9, 1, 5, 1, 3, 1, 0, 1, 1, 1, 3, 1, 2, 1, 6, 1, 1, 1, 1, 1, 5, 1, 2, 1, 10, 1, 1, 1, 1, 1, 1, 1, 9, 1, 10, 1, 1, 1, 1, 1, 1, 1, 2, 1, 6, 1, 1, 1, 1, 1, 10, 1, 9, 1, 5, 1, 1, 1, 1, 1, 2, 1, 2, 1, 6, 1, 1, 1, 1, 1, 5, 1, 2, 1, 3, 1, 1, 1, 1, 1, 11 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS It appears that the sequence always reaches 2, 5, or 29 for any initial value n. Is this easy to prove? It appears that a(n) is 1 whenever n>29 and n mod 10 is one of {0,1,2,4,6,8,9}. This has been verified to n=5000. Also, it appears that a(n) is 9 whenever n mod 130 is one of {3,23,55,75,107,127}. This has also been verified to n=5000. Are these conjectures easy to prove? LINKS EXAMPLE For n=3, we have 3 -> (3^2+4)/d = 13/1 -> (13^2+4)/d = 173/1 -> (173^2+4)/d = 29933/809 = 37, since the divisors of 29933 are {1,37,809,29933}. Continuing, we get the orbit {3,13,173,37,1373,1217,97,9413,89,5,29,5,29,...}, showing that 5 is reached after 9 steps, after which the orbit is periodic {...,5,29,5,29,...}. Thus a(3)=9. CROSSREFS Cf. A076423, A087717. Sequence in context: A155783 A257097 A256667 * A269948 A121935 A070060 Adjacent sequences:  A175761 A175762 A175763 * A175765 A175766 A175767 KEYWORD nonn AUTHOR John W. Layman, Aug 30 2010 STATUS approved

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Last modified May 26 09:27 EDT 2019. Contains 323579 sequences. (Running on oeis4.)