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A336531
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A sieve: start with the positive integers. Let a(1)=1. Mark out the following numbers: a(1)+1, a(1)+1+2, a(1)+1+2+3, a(1)+1+2+3+4, ... . The next integer in the list not marked out is 3, so a(2)=3. Mark out the following numbers: a(2)+1, a(2)+1+2, a(2)+1+2+3, a(2)+1+2+3+4, ... . Repeat the procedure for a(3), a(4), a(5), ... .
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1
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1, 3, 5, 10, 12, 14, 19, 21, 23, 28, 30, 32, 52, 54, 61, 63, 70, 72, 86, 95, 102, 104, 111, 113, 142, 144, 151, 153, 160, 162, 169, 171, 212, 221, 230, 246, 268, 270, 293, 300, 302, 309, 311, 318, 320, 327, 349, 358, 360
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OFFSET
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1,2
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COMMENTS
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Are there infinitely many pairs of the form (a(n), a(n)+2)? Let b(m) be the number of pairs less than m that differ by 2, and let s be the sum of reciprocals of consecutive terms of these pairs:
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m | b(m)| s
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10^2 | 11 | 2.627931
10^3 | 34 | 2.788503
10^4 | 64 | 2.807758
10^5 | 95 | 2.809793
10^6 | 151 | 2.810210
10^7 | 241 | 2.810273
10^8 | 386 | 2.810284
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Does the sum of these reciprocals ((1/1 + 1/3) +(1/3 + 1/5) + (1/10 + 1/12) + (1/12 + 1/14) + (1/19 + 1/21) + ...) converge to a finite number?
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LINKS
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FORMULA
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EXAMPLE
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The first few sieving stages are as follows:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 ...
1 X 3 X 5 6 X 8 9 10 X 12 13 14 15 X 17 18 19 20 21 X 23 ...
1 X 3 XX 5 X X 8 X 10 X 12 X 14 15 X 17 X 19 20 21 X 23 ...
1 X 3 XX 5 XX X X X 10 XX 12 X 14 X X 17 X 19 X 21 X 23 ...
1 X 3 XX 5 XX X X X 10 XXX 12 XX 14 X XX 17 X 19 XX 21 X 23 ...
1 X 3 XX 5 XX X X X 10 XXX 12 XXX 14 XX XX 17 XX 19 XX 21 XX 23 ...
1 X 3 XX 5 XX X X X 10 XXX 12 XXX 14 XXX XX X XX 19 XXX 21 XX 23 ...
1 X 3 XX 5 XX X X X 10 XXX 12 XXX 14 XXX XX X XX 19 XXXX 21 XXX 23 ...
1 X 3 XX 5 XX X X X 10 XXX 12 XXX 14 XXX XX X XX 19 XXXX 21 XXXX 23 ...
... Continue forever and the numbers not crossed off give the sequence.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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