

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 nexttolargest 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
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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

Table of n, a(n) for n=1..103.


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



