OFFSET
2,3
COMMENTS
The orbit or trajectory under A125256 appears to end in the cycle 5 -> 13 -> 5 -> etc. for any initial value n.
Sequence A294656 gives the size of the complete orbit of n under the map A125256, including the two elements 5 and 13 of the terminating cycle. Thus a(n) is 2 less than A294656(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) = A294656(n) - 2.
EXAMPLE
For n = 1 the map A125256 is not defined.
a(2) = 1 because under A125256, 2 -> 2^2+1 = 5 (= its smallest odd prime factor), so 5 is reached after just a(2) = 1 iteration of this map.
a(3) = 1 because A125256(3) = 5, least odd prime factor of 3^2+1 = 10 = 2*5, so here again 5 is reached after just a(2) = 1 iteration of A125256.
a(4) = 2 because A125256(4) = 4^2 + 1 = 17, and A125256(17) = 5 = least odd prime factor of 17^2 + 1 = 289 + 1 = 2*5*29, so 5 is reached after a(4) = 2 iterations of A125256.
a(5) = a(13) = 0 because for these initial values 5 and 13, no iteration is needed until either 5 or 13 is reached.
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Nov 06 2017
STATUS
approved