A111118 a(1) = 1; skipping over integers occurring earlier in the sequence, count down c(n) (c(n) = n-th composite) from a(n) to get a(n+1). If this is <= 0, instead count up from a(n) c(n) positions (skipping already occurring integers) to get a(n+1).

1, 5, 11, 2, 13, 23, 9, 26, 7, 28, 3, 31, 52, 29, 55, 25, 57, 22, 59, 19, 62, 17, 64, 15, 66, 10, 68, 6, 71, 115, 69, 117, 63, 119, 60, 121, 56, 124, 53, 126, 50, 128, 47, 131, 45, 133, 43, 135, 40, 137, 38, 140, 35, 142, 33, 144, 30, 147, 24, 149, 18, 151, 14, 153, 8, 156

Offset: 1

Author

Leroy Quet, Oct 15 2005

Keywords

• nonn

Comments

If we did not skip earlier occurring integers when counting, we would instead have sequence A100298.

Examples

			The first 4 terms of the sequence can be plotted on the number line as:
1,2,*,*,5,*,*,*,*,*,11,*,*.
Now a(4) is 2. Counting c(4) = 9 down from 2 gets a negative integer. So we instead count up 9 positions, skipping the 5 and 11 as we count, to arrive at 13 (which is at the rightmost * of the number line above).
		

Crossrefs

Cf. A100298, A110080, A002808.

Extensions

More terms from Klaus Brockhaus, Oct 17 2005