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%I #15 Jan 01 2021 14:37:08
%S 1,3,5,7,13,17,31,35,41,43,47,63,101,105,107,131,175,177,185,211,235,
%T 237,267,301,305,315,323,397,407,451,571,631,633,683,757,841,877,947,
%U 987,1043,1221,1251,1431,1501,1655,1781,1961,1981,2023,2067,2157,2197,2253,2367,2457,2505,2615
%N Numbers surviving a repeated sieving process for pseudo-lucky numbers (A249876).
%C Start with the positive integers as the 1st starting sequence. The 1st full sieving process for the pseudo-lucky numbers begins with the 2nd term in the 1st starting sequence and generates A249876 (the 2nd starting sequence). The n-th full sieving process begins with the (n+1)-th term in the n-th starting sequence and generates the (n+1)-th starting sequence. The numbers that are left form the final sequence.
%C Let b(m) be the number of elements of this sequence <= m. Let c(m) = round(square(s*m/log(s*m))), where s = 11.
%C --------------------------------
%C m | b(m) | c(m) | b(m)-c(m)
%C --------------------------------
%C 10^2 | 12 | 13 | -1
%C 10^3 | 39 | 34 | +5
%C 10^4 | 103 | 97 | +6
%C 5*10^4 | 210 | 204 | +6
%C 6*10^4 | 228 | 222 | +6
%C 7*10^4 | 236 | 238 | -2
%C 8*10^4 | 256 | 254 | +2
%C 9*10^4 | 270 | 268 | +3
%C 10^5 | 282 | 281 | +1
%C --------------------------------
%C Is c(m) an approximation to b(m)?
%e The 1st full sieving process:
%e 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 ...
%e 1 X 3 X 5 X 7 X 9 X 11 X 13 X 15 X 17 X 19 X 21 X 23 X 25 ...
%e 1 3 5 7 X 11 13 15 17 X 21 23 25 ...
%e 1 3 5 7 11 13 X 17 21 23 25 ...
%e Continuing the above procedure generates the 2nd starting sequence (the pseudo-lucky numbers) to begin the 2nd full sieving process:
%e 1 3 5 7 11 13 17 21 23 25 31 35 41 43 45 47 55 57 63 65 73 75 83 87 95 ...
%e 1 3 5 7 X 13 17 21 23 X 31 35 41 43 X 47 55 57 63 X 73 75 83 87 X ...
%e 1 3 5 7 13 17 X 23 31 35 41 43 47 X 57 63 73 75 83 87 ...
%e 1 3 5 7 13 17 23 31 35 41 43 47 X 63 73 75 83 87 ...
%e 1 3 5 7 13 17 23 31 35 41 43 47 63 73 75 83 X ...
%e Continuing the above procedure generates the 3rd starting sequence to begin the 3rd full sieving process:
%e 1 3 5 7 13 17 23 31 35 41 43 47 63 73 75 83 101 105 107 123 127 131 151 ...
%e 1 3 5 7 13 17 X 31 35 41 43 47 63 X 75 83 101 105 107 123 X 131 151 ...
%e 1 3 5 7 13 17 31 35 41 43 47 63 X 83 101 105 107 123 131 151 ...
%e 1 3 5 7 13 17 31 35 41 43 47 63 83 101 105 107 X 131 151 ...
%e Continuing the above procedure generates the 4th starting sequence to begin the 4th full sieving process:
%e 1 3 5 7 13 17 31 35 41 43 47 63 83 101 105 107 131 151 153 175 177 185 ...
%e 1 3 5 7 13 17 31 35 41 43 47 63 X 101 105 107 131 151 153 175 177 185 ...
%e 1 3 5 7 13 17 31 35 41 43 47 63 101 105 107 131 X 153 175 177 185 ...
%e Continuing the above procedure generates the 5th starting sequence to begin the 5th full sieving process:
%e 1 3 5 7 13 17 31 35 41 43 47 63 101 105 107 131 153 175 177 185 211 235 ...
%e 1 3 5 7 13 17 31 35 41 43 47 63 101 105 107 131 X 175 177 185 211 235 ...
%e ...
%e Continue forever and the numbers not crossed off give the sequence.
%Y Cf. A000959, A249876.
%K nonn
%O 1,2
%A _Lechoslaw Ratajczak_, Dec 07 2020
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