%I #22 Sep 16 2017 00:27:42
%S 1,2,1,5,2,3,1,2,1,19,2,4,2,23,3,4,2,3,1,2,1,5,3,31,4,8,4,5,1,2,1,149,
%T 2,3,1,2,1,307,2,4,2,311,3,4,2,3,1,2,1,13,2,4,2,5,3,6,4,13,11,22,11,
%U 19,14,22,20,28,14,83,15,18,9,11,2,4,2,653,3,6,4,17,13,14,12,13,1
%N Let a(n) = d[m(n),i] be the ith divisor of the integer m(n) and the index j such that d[m(n),i] = d[m(n+1),j], m(n+1) is the first integer > m(n). The sequence a(n), starting with m(1) = 1, i = 1 is given by the conditions: a(1) = 1 and a(n+1) is the first predecessor of a(n) if n even, or the first successor of a(n) if n odd.
%C Conjecture: the sequence is the union of an infinity subsets of the form {1, 2}, {1, 5, 2, 3}, ..., {1, p, ..., q}, ... where the number 1 is the first element of each subset.
%C The corresponding values of m(n) are {1, 2, 4, 5, 10, 12, 15, 16, 18, 19, 38, 40, 44, 46, 69, 72, 76, 78, 81, 82, 84, 85, 90, 93, 124, ...}.
%e a(1) = 1 ;
%e a(2) = 2 because the divisors of 2 are {1, 2} and 2 is the first successor of 1;
%e a(3) = 1 because the divisors of 3 are {1, 3} and 1 is the first predecessor of 2;
%e a(4) = 5 because the divisors of 5 are {1, 5} and 5 is the first successor of 1;
%e a(5) = 2 because the divisors of 10 are {1, 2, 5, 10} and 2 is the first predecessor of 5;
%e a(6) = 3 because the divisors of 12 are {1, 2, 3, 4, 6, 12} and 3 is the first successor of 2.
%p with(numtheory):T:=array(1..162):T[1]:=1:d:=1:i:=2:for n from 2 to 5000 do:x:=divisors(n):n1:=nops(x):j:=0:for m from 1 to n1 while(j=0) do:if x[m]= d then z:=x[m+((1)^i)]:T[i]:=z:d:=z:i:=i+1:j:=1:else fi:od:od:print(T):
%Y Cf.
%K nonn
%O 1,2
%A _Michel Lagneau_, Sep 10 2011
