1,3

Sum_{ n >= 1 } a(n)/A002110(n) converges to 1/6. That is, (2 / 30) + (6 / 210) + (30 / 2310) + (270 / 30030) + (2970 / 510510) + ... = (1 / 6).

Dennis Martin, Table of n, a(n) for n = 1..100

Dennis R. Martin, On the Infinite Series Characterizing the Elimination of Twin Prime Candidates.

a(1) = a(2) = 0; a(3) = 2; for n >= 4, a(n) = (p(n-1)-2)*a(n-1), where p(n) is the n-th prime.

The prime factors p(1) = 2 and p(2) = 3 cannot eliminate any twin prime candidates, therefore a(1) = a(2) = 0. The prime factor p(3) = 5 will eliminate a(3) = 2 twin prime candidates in every interval of p(3)# = 30 starting after p(3). For example, the composites 25 and 35 eliminate the twin prime candidate pairs centered at 24 and 36, respectively, while the composites 55 and 65 eliminate the twin prime candidates centered at 54 and 66.

For the prime factor p(4) = 7, there will be 8 composites having p(4) for their lowest prime factor within every interval of p(4)# = 210 starting after 7. For instance, the composites {49, 77, 91, 119, 133, 161, 203, 217} are adjacent to and eliminate the twin prime candidates centered at {48, 78, 90, 120, 132, 162, 204, 216}. However, 2 of those 8 are already eliminated by p(3), those being the candidates centered at 204 and 216, since 205 and 215 obviously are composites having 5 for their lowest prime factor. And in the next interval of p(4)# = 210 the pattern repeats. The composites {259, 287, 301, 329, 343, 371, 413, 427} all have 7 for their lowest prime factor and they eliminate the twin prime candidate pairs centered at {258, 288, 300, 330, 342, 372, 414, 426}. But the ones centered at 414 and 426 are also adjacent to 415 and 425, which have 5 for their lowest prime factor and thus can be considered to have already been eliminated. a(4) = 8 - 2 = 6.

For p(5) = 11, there are 48 composites that have 11 for their lowest prime factor over any interval of p(5)# = 2310 starting after 11. Those 48 composites are all adjacent to a twin prime candidate center post, but 12 of those candidates are eliminated by p(3) (the ones corresponding to the centers 144, 186, 474, 516, 804, 1134, 1176, 1506, 1794, 1836, 2124 and 2166) and 6 are eliminated by p(4) (those corresponding to the candidate centered at 120, 342, 582, 1728, 1968 and 2190). That is a total of 18 out of those 48 in every interval of 2310 that are eliminated by a prime factor less than p(5), therefore a(5) = 48 - 18 = 30.

But then 30 = 6(7-2) and 6 = 2(5-2). By continuing to count the twin prime eliminations in this manner, it can be deduced that each subsequent term is found by multiplying the previous term by the previous prime minus 2.

Cf. A002110, A005867, A121407.

Sequence in context: A003266 A097385 A066068 this_sequence A028361 A106339 A120295

Adjacent sequences: A121403 A121404 A121405 this_sequence A121407 A121408 A121409

easy,frac,nonn

Dennis R. Martin (dennis.martin(AT)dptechnology.com), Jul 28 2006, Dec 05 2006