cp's OEIS Frontend

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).

%I A111118 #10 May 12 2014 10:22:35
%S A111118 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,
%T A111118 10,68,6,71,115,69,117,63,119,60,121,56,124,53,126,50,128,47,131,45,
%U A111118 133,43,135,40,137,38,140,35,142,33,144,30,147,24,149,18,151,14,153,8,156
%N 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).
%C A111118 If we did not skip earlier occurring integers when counting, we would instead have sequence A100298.
%e A111118 The first 4 terms of the sequence can be plotted on the number line as:
%e A111118 1,2,*,*,5,*,*,*,*,*,11,*,*.
%e A111118 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).
%Y A111118 Cf. A100298, A110080, A002808.
%K A111118 nonn
%O A111118 1,2
%A A111118 _Leroy Quet_, Oct 15 2005
%E A111118 More terms from _Klaus Brockhaus_, Oct 17 2005