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%I A242453 #15 Dec 01 2014 00:22:21
%S A242453 1,2,3,3,4,5,5,5,6,7,7,7,7,8,8,9,9,9,10,10,11,11,11,11,11,11,12,13,13,
%T A242453 13,13,13,13,14,14,15,15,16,16,16,16,17,17,17,17,17,17,17,17,18,19,19,
%U A242453 19,19,19,19,19,19,19,20,20,21,21,21,21,22,22,22,23,23
%N A242453 (Conjectured) infinite nondecreasing sequence with a(1) = 1 such that the divisors of n appear a total of a(n) times in the sequence.
%C A242453 Can it be proved or disproved that this sequence is valid? It is invalid if, for some value of n, it is impossible for a(n) to be as great as the total number of appearances that the aliquot divisors of n have already made in the sequence.
%C A242453 A near-example: The combination of the sequence definition and 2) the values of terms a(1)-a(29) eliminates all possible values of a(30) except for the number 13.  If the aliquot divisors of 30 (1, 2, 3, 5, 6, 10 and 15) appeared in the sequence a total 14 or more times, 13 would also be eliminated as a possible value, and this sequence would be invalid. Since 30's aliquot divisors appear in the sequence a total of only 12 times, no contradiction occurs.
%C A242453 The first 2500 terms appear to be free of any contradiction that would "annihilate" the sequence.
%e A242453 a(2) cannot be 1, because then it would be the second term in the sequence to divide 1. (Since a(1) = 1, it is necessary that exactly 1 term in the sequence is a divisor of 1.)
%e A242453 Nor can a(2) be 3 or greater, because then a(2) would be greater than the number of terms in the sequence that divided 2 (there would be only 1 such term). Since, by definition, a(2) must equal the number of terms in the sequence that divide 2, this is a contradiction.
%e A242453 Since 2 is the only possible value for a(2) that does not create a contradiction, a(2) = 2.
%Y A242453 Cf. A001462.
%K A242453 nonn
%O A242453 1,2
%A A242453 _Matthew Vandermast_, Jul 24 2014

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