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A variation on Flavius's sieves (A000960, A099207): Start with the Chen primes; at the k-th sieving step, remove every (k+1)-st term of the sequence remaining after the (k-1)-st sieving step; iterate.
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%I #11 Jun 23 2020 11:01:18

%S 2,5,17,41,83,137,233,317,467,617,761,941,1259,1427,1913,2281,2531,

%T 2957,3511,3797,4447,5153,5351,6481,6863,7669,8581,9533,10259,11497,

%U 12569,13441,14081,15737,16187,17657,19541,19991,21587,23017,24317

%N A variation on Flavius's sieves (A000960, A099207): Start with the Chen primes; at the k-th sieving step, remove every (k+1)-st term of the sequence remaining after the (k-1)-st sieving step; iterate.

%H <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>

%e Start with

%e 2 3 5 7 11 13 17 19 23 29 31 37 41 47 53 59 67 71 83 89 101 107 109 113 127 131 ... and delete every second term, giving

%e 2 5 11 17 23 31 41 53 67 83 101 109 127 ... and delete every 3rd term, giving

%e 2 5 17 23 41 53 83 101 127 ... and delete every 4th term, giving

%e .... Continue forever and what's left is the sequence.

%p ts_chen:= proc(n) local i, ans; ans:=[ ]: for i from 1 to n do if ( isprime(i) = 'true') then if ( isprime(i+2) = 'true' or numtheory[bigomega](i+2) = 2) then ans:=[ op(ans), i ] fi fi od: return ans end: S[1]:=convert(ts_chen(25000), set): for n from 2 to 2500 do S[n]:=S[n-1] minus {seq(S[n-1][n*i], i=1..nops(S[n-1])/n)} od: convert(S[2100],list);

%Y Cf. A000960, A099207, A109611.

%K nonn

%O 0,1

%A _Jani Melik_, May 05 2006