OFFSET
2,1
COMMENTS
The orbit or trajectory under A125256 appears to end in the cycle 5 -> 13 -> 5 -> etc. for any initial value n.
Sequence A294658 gives the number of steps to reach either 5 or 13, i.e. an element of this terminating cycle. Therefore a(n) (which counts these two elements as well as the initial value) is 2 more than A294658(n) for all n. This is confirmed by careful examination of special cases - assuming, of course, that all trajectories end in the cycle (5, 13).
LINKS
Ray Chandler, Table of n, a(n) for n = 2..20001
FORMULA
a(n) = A294658(n) + 2.
EXAMPLE
For n = 1 the map A125256 is not defined.
a(2) = 3 = # { 2, 5, 13 }, because under A125256, 2 -> 2^2+1 = 5 (= its smallest odd prime factor), 5 -> least odd prime factor(5^2+1 = 26) = 13, 13 -> least odd prime factor(13^2 + 1 = 170 = 2*5*17) = 5, etc.
a(3) = 3 = # { 3, 5, 13 }, because under A125256, 3 -> smallest odd prime factor(3^2+1 = 10) = 5, 5 -> 13, 13 -> 5 etc.
a(4) = 4 = # { 4, 17, 5, 13 }, because under A125256, 4 -> 4^2+1 = 17 (= its smallest odd prime factor), 17 -> smallest odd prime factor(17^2+1 = 290 = 2*5*29) = 5, 5 -> 13, 13 -> 5 etc.
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Nov 06 2017
STATUS
approved