%I A111698
%S A111698 1,5,9,2,7,12,3,10,15,4,13,18,6,16,21,8,19,24,11,22,27,14,25,30,17,28,
%T A111698 33,20,31,36,23,34,39,26,37,42,29,40,45,32,43,48,35,46,51,38,49,54,41,
%U A111698 52,57,44,55,60,47,58,63,50,61,66,53,64,69,56,67,72,59,70,75,62,73,78
%N A111698 a(1)=1. Skipping over integers occurring earlier in the sequence, count 
               down a composite from a(n) to get a(n+1) so that a(n+1) is the smallest 
               possible positive integer arrived at this way. If there are no positive 
               integers at a distance of a composite number of yet unused integers, 
               instead count up from a(n) 4 (the lowest composite positive integer) 
               positions (skipping already occurring integers) to get a(n+1).
%C A111698 I have found two patterns for this sequence. The first is that there 
               is a pattern 0,3,6,0,3,6,0,3,6,... which states the lengths of the 
               "LessThanList" for each term. In other words, a(6) = 12. There are 
               six integers less than 12 which are not already listed in the sequence 
               at this point, {3,4,6,8,10,11}. a(7) = 3. There are no integers not 
               already on the list which are less than 3 at this point. a(8) = 10. 
               There are three integers less than 10 which are not already on the 
               list at this point, {4,6,8}. Also, after the 14th term, the sequence 
               becomes regular in the following way. The difference between successive 
               terms is as follows: 5,-13,11,5,-13,11,... [From Diana Mecum (diana.mecum(AT)gmail.com), 
               Aug 15 2008]
%H A111698 Diana Mecum, <a href="b111698.txt">Table of n, a(n) for n = 1..1011</
               a> [From Diana Mecum (diana.mecum(AT)gmail.com), Aug 15 2008]
%H A111698 Leroy Quet, <a href="http://www.prism-of-spirals.net/">Home Page</a> 
               (listed in lieu of email address)
%e A111698 The first 5 terms of the sequence can be plotted on the number line as:
%e A111698 1,2,*,*,5,*,7,*,9,*,*,*.
%e A111698 Now a(5) is 7. Counting down from 7 gets a noncomposite (1,2, or 3) number 
               of steps to arrive at each yet unused positive integer. So we instead 
               count up 4 positions - skipping the 9 as we count - to arrive at 
               12 (which is at the right-most * of the number-line above).
%Y A111698 Cf. A111453, A111118.
%Y A111698 Sequence in context: A021632 A011494 A030125 this_sequence A021948 A154265 
               A111453
%Y A111698 Adjacent sequences: A111695 A111696 A111697 this_sequence A111699 A111700 
               A111701
%K A111698 nonn
%O A111698 1,2
%A A111698 Leroy Quet Nov 17 2005
%E A111698 Terms a(14) through a(1011) from Diana Mecum (diana.mecum(AT)gmail.com), 
               Aug 15 2008