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).
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, 33, 20, 31, 36, 23, 34, 39, 26, 37, 42, 29, 40, 45, 32, 43, 48, 35, 46, 51, 38, 49, 54, 41, 52, 57, 44, 55, 60, 47, 58, 63, 50, 61, 66, 53, 64, 69, 56, 67, 72, 59, 70, 75, 62, 73, 78
Offset: 1
Keywords
Examples
The first 5 terms of the sequence can be plotted on the number line as: 1,2,*,*,5,*,7,*,9,*,*,*. 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).
Links
- Diana Mecum, Table of n, a(n) for n = 1..1011 [From _Diana L. Mecum_, Aug 15 2008]
Extensions
Terms a(14) through a(1011) from Diana L. Mecum, Aug 15 2008
Comments