A056957 In repeated iterations of function m->m/2 if m even, m->3m-1 if m odd, a(n) is minimum value achieved if starting from n.

1, 1, 1, 1, 5, 1, 5, 1, 5, 5, 1, 1, 5, 5, 1, 1, 17, 5, 5, 5, 17, 1, 17, 1, 17, 5, 5, 5, 1, 1, 17, 1, 17, 17, 5, 5, 17, 5, 1, 5, 17, 17, 1, 1, 17, 17, 5, 1, 17, 17, 5, 5, 1, 5, 17, 5, 1, 1, 1, 1, 17, 17, 5, 1, 1, 17, 17, 17, 1, 5, 1, 5, 17, 17, 5, 5, 1, 1, 1, 5, 5, 17, 17, 17, 1, 1, 1, 1, 5, 17

Offset: 1

Author

Henry Bottomley, Jul 18 2000

Keywords

• nonn

Comments

At least for n<10000, the only possible cycles reached include 1,2,1,..., 5,14,7,20,10,5,... and 17,50,25,74,37,110,55,164,82,41,122,61,182,91,272,136,68,34,17,... For n<5 only the first occurs, while for n<17 only the first two occur.

Examples

			a(9)=5 since iteration starts: 9, 26, 13, 38, 19, 56, 28, 14, 7, 20, 10, 5, 14, 7, 20, 10, 5, ... and 5 is the smallest value
		

Crossrefs

Cf. A001281. If n is in A039500 then a(n)=1, if n is in A039501 then a(n)=5, if n is in A039502 then a(n)=17. If n is negative then this becomes the 3x+1 problem and the minimum values become those which are most negative (i.e. maximum absolute values) as in A056959.

Formula

a(2n) = a(n)

Extensions

Edited by Bryce Herdt (mathidentity(AT)yahoo.com), Apr 18 2010