A326823 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.

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, 22, 20, 5, 10, 16, 8, 19, 16, 4, 4, 10, 16, 14, 14, 2, 14, 12, 12, 10, 31, 17, 38, 23, 29, 17, 27, 8, 34, 21, 34, 23, 15, 21, 15, 6, 19, 11, 19, 17, 9, 9, 7

Offset: 1

Author

Ali Sada, Oct 20 2019

Keywords

• nonn

Comments

It is conjectured that this algorithm will always terminate at 1 or 11.

Matthijs Coster verified the conjecture for n <= 100000.

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).

Jim Nastos verified the conjecture for n <= 43*10^6 (Oct 21 2019).

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.

Examples

			The trajectory of 3 is 3 -> 3+4 = 7 -> 7+9 = 16 -> 8 -> 4 -> 2 -> 1, taking a(3) = 6 steps to reach 1. - _M. F. Hasler_, May 08 2025
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
The trajectory of 22 reaches 11 in a single iteration, so a(22) = 1. - _Jon E. Schoenfield_, Oct 20 2019
		

Links

• M. F. Hasler, Table of n, a(n) for n = 1..1000, May 08 2025

Crossrefs

Cf. A006577, A326825.

Programs

PARI

M326823=Map([1,0;11,0]); A326823(n)=if(mapisdefined(M326823, n, &n), n, mapput(M326823, n, 1+n=A326823(if(n%2, n+(sqrtint(n)+1)^2, n\2))); 1+n) \\ M. F. Hasler, May 08 2025

Extensions

Edited and data corrected by Jon E. Schoenfield and N. J. A. Sloane, Oct 20 2019