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%I #12 Mar 30 2012 18:35:59
%S 5,5,5,7,17,5,16,22,5,16,20,8,8,16,18,20,22,6,16,8,16,18,20,23,34,16,
%T 6,27,11,16,18,22,21,32,16,9,23,25,9,9,28,16,20,39,19,30,16,21,21,21,
%U 23,23,35,7,26,37,16,18,37,9,28,28,16,43,14,19,19,34,21,21,33,24
%N Number of iterations of the map n->n/3 if n == 0 (mod 3), n->4*n+a if 4*n+a == 0 (mod 3) where a = 1 or 2, before reaching the end of the cycle.
%C The problem is as follows: start with any number n. If n is divisible by 3, divide it by 3, otherwise multiply it by 4 and add 1 or 2 in order to find a new integer divisible by 3. Do we always reach the end of a cycle? It is conjectured that the answer is yes.
%C On the set of positive integers, the orbit of any number seems to end in the orbit of 1, or of another integer.
%C This problem has a resemblance with the Collatz problem.
%H Alois P. Heinz, <a href="/A194428/b194428.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1) = 5 because 1 -> 6 -> 2 -> 9 -> 3 -> 1 with 5 iterations ;
%e a(2) = 5 because 2 -> 9 -> 3 -> 1-> 6 -> 2 with 5 iterations ;
%e a(3) = 5 because 3 -> 1 -> 6 -> 2 -> 9 -> 3 with 5 iterations ;
%e a(4) = 7 because 4 -> 18 -> 6 -> 2 -> 9 -> 3 -> 1 -> 6 with 7 iterations.
%p T:=array(1..2000):for n from 1 to 100 do: T[1]:=n:n0:=n:k:=2:for it from 1 to 50 do: z:=irem(n0,3):if z=0 then n0:=n0/3:T[k]:=n0:k:=k+1:else n0:=4*n0 + 1:if irem(n0,3)=0 then T[k]:=n0:k:=k+1:else n0:=n0+1:T[k]:=n0:k:=k+1:fi:fi:od:U:=convert(T,set):n1:=nops(U): printf(`%d, `, n1):od:
%Y Cf. A001281.
%K nonn
%O 1,1
%A _Michel Lagneau_, Aug 23 2011
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