A121407 Number of double eliminations of twin prime candidates within primorial intervals of p(n)#. That is, it is the number of twin prime candidates for which each half of the pair is composite, where one of those composites has p(n) for its lowest prime factor and the other composite has a prime less than p(n) for its lowest prime factor.
0, 0, 0, 2, 18, 210, 2790, 47610, 901530, 20591010, 592452630, 18202996350, 667760974650, 27146297697750, 1157142993063750, 53925515761020750, 2835489033177050250, 166057836818071448250, 10054640164031031318750
Offset: 1
Examples
The prime factors p(1) = 2 and p(2) = 3 cannot eliminate any twin prime candidates, therefore a(1) = a(2) = a(3) = 0.
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. Therefore a(4) = 2 because there are 2 double eliminations by 7 and by a prime less than 7 within each interval of p(4)# = 210.
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). Therefore a(5) = 12 + 6 = 18.
Links
- Dennis Martin, Table of n, a(n) for n = 1..100