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%I #49 Oct 21 2019 14:13:14
%S 0,1,6,2,7,7,5,3,17,8,0,8,21,6,21,4,15,18,3,9,13,1,11,9,16,22,16,7,20,
%T 22,20,5,10,16,8,19,16,4,4,10,16,14,14,2,14,12,12,10,31,17,38,23,29,
%U 17,27,8,34,21,34,23,15,21,15,6,19,11,19,17,9,9,7
%N a(n) is the number of iterations needed to reach 1 or 11 starting at n and using the map k -> (k/2 if k is even, otherwise k + (smallest square > k)). Set a(n) = -1 if the trajectory never reaches 1 or 11.
%C It is conjectured that this algorithm will always terminate at 1 or 11.
%C _Matthijs Coster_ verified the conjecture for n <= 100000.
%C _Neil Fernandez_ checked for n <= 2*10^6 and always reached either the cycle (1,5,14,7,16,8,4,2,1) or the cycle (11,27,63,127,271,560,280,140,70,35,71,152,76,38,19,44,22,11).
%C _Jim Nastos_ verified the conjecture for n <= 43*10^6 (Oct 21 2019).
%C Generalization: The algorithm also appears to terminate when replacing the least perfect square greater than n with the greatest perfect square less than n. It also seems to terminate when square is replaced by any power.
%e The trajectory of 22 reaches 11 in a single iteration, so a(22) = 1. - _Jon E. Schoenfield_, Oct 20 2019
%e The trajectory of 9 is [9, 25, 61, 125, 269, 558, 279, 568, 284, 142, 71, 152, 76, 38, 19, 44, 22, 11], taking 17 steps to reach 11. So a(9) = 17. - _N. J. A. Sloane_, Oct 20 2019
%Y Cf. A006577, A326825.
%K nonn
%O 1,3
%A _Ali Sada_, Oct 20 2019
%E Edited and data corrected by _Jon E. Schoenfield_ and _N. J. A. Sloane_, Oct 20 2019
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