A336531 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), ... .

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

Offset: 1

Author

Lechoslaw Ratajczak, Oct 04 2020

Keywords

• nonn

Comments

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?

Examples

			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.
		

Crossrefs

Cf. A000217, A030194.

Formula

a(n) = A030194(n-1) + 1. - Hugo Pfoertner, Oct 05 2020