cp's OEIS Frontend

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).

%I A111698 #12 May 11 2014 22:39:47
%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,... . - _Diana L. Mecum_, Aug 15 2008
%H A111698 Diana Mecum, <a href="/A111698/b111698.txt">Table of n, a(n) for n = 1..1011</a> [From _Diana L. Mecum_, Aug 15 2008]
%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 rightmost * of the number line above).
%Y A111698 Cf. A111453, A111118.
%K A111698 nonn
%O A111698 1,2
%A A111698 _Leroy Quet_, Nov 17 2005
%E A111698 Terms a(14) through a(1011) from _Diana L. Mecum_, Aug 15 2008