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%I #10 Nov 06 2017 12:06:17
%S 1,1,2,0,2,1,1,4,3,5,1,0,2,2,2,1,1,4,3,1,1,1,2,2,2,1,1,2,2,1,1,1,1,2,
%T 2,1,1,3,3,3,1,1,1,2,3,1,1,2,4,3,1,1,2,2,2,1,1,4,1,4,1,1,2,2,2,1,1,5,
%U 1,3,1,1,2,3,2,1,1,4,2,2,1,1,2,2,1,1
%N Number of steps required to reach either 5 or 13, starting with n, when iterating the map A125256: x -> smallest odd prime divisor of n^2+1; or a(n) = -1 in case 5 is never reached.
%C The orbit or trajectory under A125256 appears to end in the cycle 5 -> 13 -> 5 -> etc. for any initial value n.
%C 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).
%H Ray Chandler, <a href="/A294658/b294658.txt">Table of n, a(n) for n = 2..20001</a>
%F a(n) = A294656(n) - 2.
%e For n = 1 the map A125256 is not defined.
%e 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.
%e 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.
%e 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.
%e a(5) = a(13) = 0 because for these initial values 5 and 13, no iteration is needed until either 5 or 13 is reached.
%o (PARI) A294658(n)=for(k=0,oo,bittest(8224,n)&&return(k);n=A125256(n)) \\ 8224 = 2^5 + 2^13. One could add 2^0 + 2^1 = 3 to avoid an error message for initial values 0 and 1, for which A125256 is not defined.
%Y Cf. A125256, A294656, A294657.
%K nonn
%O 2,3
%A _M. F. Hasler_, Nov 06 2017
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