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%I #21 Sep 10 2019 21:22:58
%S 3,5,8,12,14,18,20,23,27,30,32,35,38,42,44,48,50,53,59,62,68,72,74,78,
%T 80,83,87,90,92,95,98,102,104,108,110,114,117,120,122,128,132,134,138,
%U 140,143,147,150,152,158,164,168,170,173,179,182,188,192
%N Write down the numbers from 3 to infinity. Take next number, M say, that has not been crossed off and cross off all the numbers i*M + 1 for i >= 1. Repeat. The numbers that are left form the sequence.
%C Equivalently, each term is the lowest natural number n such that n mod k is not 1 for any number k in the sequence so far. Also, the average difference between terms grows as O(2 log log n). - _Trevor Cappallo_, Sep 10 2019
%H Popular Computing (Calabasas, CA), <a href="/A003309/a003309.pdf">Sieves: Problem 43</a>, Vol. 2 (No. 13, Apr 1974), pp. 6-7. This is Sieve #3. [Annotated and scanned copy]
%H <a href="/index/Si#sieve">Index entries for sequences generated by sieves</a>
%e The first few sieving stages are as follows:
%e 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
%e 3 X 5 6 X 8 9 XX 11 12 XX 14 15 XX 17 18 XX 20 ...
%e 3 X 5 X X 8 9 XX XX 12 XX 14 15 XX 17 18 XX 20 ...
%e 3 X 5 X X 8 X XX XX 12 XX 14 15 XX XX 18 XX 20 ...
%e 3 X 5 X X 8 X XX XX 12 XX 14 XX XX XX 18 XX 20 ...
%Y Cf. A003309, A003310.
%K nonn
%O 1,1
%A _N. J. A. Sloane_, Nov 23 2004
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