Logo

Greetings from The On-Line Encyclopedia of Integer Sequences!

Hints

Search: id:A121407
Displaying 1-1 of 1 results found. page 1
     Format: long | short | internal | text      Sort: relevance | references | number      Highlight: on | off
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
2
0, 0, 0, 2, 18, 210, 2790, 47610, 901530, 20591010, 592452630, 18202996350, 667760974650, 27146297697750, 1157142993063750, 53925515761020750, 2835489033177050250, 166057836818071448250, 10054640164031031318750 (list; graph; listen)
OFFSET

1,4

LINKS

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

FORMULA

a(1) = a(2) = 0; for n >= 3, a(n) = t(n) - e'(n) = (p(n-1)-1)*t(n-1) - (p(n-1)-2)*e'(n-1), where p(n) is n-th prime, t(n) is given by sequence A005867 and e'(n) is given by sequence A121406.

EXAMPLE

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.

CROSSREFS

Cf. A002110, A005867, A121406.

Sequence in context: A116072 A092882 A123855 this_sequence A153647 A052726 A155666

Adjacent sequences: A121404 A121405 A121406 this_sequence A121408 A121409 A121410

KEYWORD

easy,frac,nonn

AUTHOR

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

page 1

Search completed in 0.002 seconds

Lookup | Welcome | Find friends | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Puzzles | Hot | Classics
More pages | Superseeker | Maintained by N. J. A. Sloane (njas@research.att.com)

Last modified December 15 00:47 EST 2009. Contains 170825 sequences.


AT&T Labs Research