

A162180


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.


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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, 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, 19, 14, 22, 20, 28, 14, 83, 15, 18, 9, 11, 2, 4, 2, 653, 3, 6, 4, 17, 13, 14, 12, 13, 1
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OFFSET

1,2


COMMENTS

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.
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, ...}.


LINKS



EXAMPLE

a(1) = 1 ;
a(2) = 2 because the divisors of 2 are {1, 2} and 2 is the first successor of 1;
a(3) = 1 because the divisors of 3 are {1, 3} and 1 is the first predecessor of 2;
a(4) = 5 because the divisors of 5 are {1, 5} and 5 is the first successor of 1;
a(5) = 2 because the divisors of 10 are {1, 2, 5, 10} and 2 is the first predecessor of 5;
a(6) = 3 because the divisors of 12 are {1, 2, 3, 4, 6, 12} and 3 is the first successor of 2.


MAPLE

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):


CROSSREFS

Cf.


KEYWORD

nonn


AUTHOR



STATUS

approved



