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 A190874 First differences of A179196, pi(R_(n+1)) - pi(R_n) where R_n is A104272(n). 10
 4, 2, 3, 3, 2, 2, 2, 1, 5, 1, 2, 3, 4, 1, 3, 2, 1, 7, 1, 1, 1, 1, 3, 1, 3, 3, 1, 5, 1, 3, 1, 5, 1, 1, 2, 1, 1, 4, 4, 1, 2, 8, 1, 2, 1, 1, 1, 1, 5, 1, 1, 1, 1, 3, 5, 1, 2, 2, 3, 4, 2, 1, 1, 3, 1, 4, 7, 1, 1, 2, 3, 3, 2, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 5, 2, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The count of primes of the interval(R_n,R_(n+1)] where R_n is A104272(n). The sequence A182873 is the first difference of Ramanujan primes R_(n+1)- R_n. While each non-Ramanujan prime is bound by Ramanujan primes, the maximal non-Ramanujan prime gap is less than the maximal Ramanujan prime gap, A182873, and the ratio of a(n)/A182873(n) is the average gap size at R_n. Record terms of n, a(n) are in A202186, A202187. Each record term value of a(n) - 1 is the index m of A168425(m). A202188 is the index of A168425 when A174641(n) = A168425(m), it has repeated values of A202187. Starting at index n = A191228(A174602(m)) in this sequence, the first instance of a count of m - 1 consecutive 1's is seen. Limit inferior of a(n) is positive, because there are infinitely many Ramanujan primes and each term of the sequence is >= 1. Limit superior of a(n)/log(pi(R_n)) is positive infinity. Equivalently, there are infinitely many n > 0 such that pi(R_(n+1)) > pi(R_n) + t log(pi(R_n)), for every t > 0. For all n > 3, a(n) < n. a(n) = rho(n+1) - rho(n) using rho(x) as defined in Sondow, Nicholson, Noe. LINKS T. D. Noe, Table of n, a(n) for n = 1..10000 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. FORMULA a(n) = pi(R_(n+1)) - pi(R_n) or a(n) = A000720(A104272(n+1)) - A000720(A104272(n)). a(n) = A179196(n+1) - A179196(n). EXAMPLE R(4) = 29, the fourth Ramanujan prime, the next Ramanujan prime is a(4) = 3 primes away or R(5) = 41. MATHEMATICA nn = 100; R = Table[0, {nn}]; s = 0; Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s--]; If[s

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