OFFSET
1,1
COMMENTS
All but the first term is odd because A104272 has only one even term, 2. Because of all primes > 2 are odd, 1 can be subtracted from each term.
If this sequence has an infinite number of terms in which a(n) = 3, then the twin prime conjecture can be proved.
By comparing the fractions we can see that (p_(i+1)-p_i)/(2*sqrt(p_i)) and a(n)/(2*sqrt(p_k)) are < 1 for all n > 0, in fact a(n)/(1.8*sqrt(p_k)) < 1 for all n > 0. When taking into account numbers in A182873(n) and A190874(n) to sqrt(R_n) we see that A182873(n)/(A190874(n)*sqrt(R_n)) < 1 for all n > 1.
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
J. Sondow, Ramanujan primes and Bertrand's postulate, arXiv:0907.5232 [math.NT], 2009-201; Amer. Math. Monthly 116 (2009) 630-635.
J. Sondow, J. W. Nicholson, and T. D. Noe, Ramanujan Primes: Bounds, Runs, Twins, and Gaps, arXiv:1105.2249 [math.NT], 2011; J. Integer Seq. 14 (2011) Article 11.6.2.
Wikipedia, Ramanujan Prime
Marek Wolf, A Note on the Andrica Conjecture, arXiv:1010.3945 [math.NT], 2010.
EXAMPLE
A168421(19) = 127, A104272(19) = 227; so a(19) = 2*A168421(19) - A104272(19) = 254 - 227 = 27. Note: for n = 20, 21, 22, 23 A168421(n) = 127. Because A168421 remains the same for these n and A104272 increases, the size of the range for a(n) for these n decreases. Note: a(18) = 2*97 - 181 = 194 - 181 = 13. This is nearly half a(19). The actual gap betweens A104272(19) and the next prime, 229, is 2.
MATHEMATICA
CROSSREFS
KEYWORD
nonn
AUTHOR
John W. Nicholson, Sep 12 2011
STATUS
approved