cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A190874 First differences of A179196, pi(R_(n+1)) - pi(R_n) where R_n is A104272(n).

Original entry on oeis.org

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
Offset: 1

Views

Author

John W. Nicholson, May 22 2011

Keywords

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.

Examples

			R(4) = 29, the fourth Ramanujan prime, the next Ramanujan prime is a(4) = 3 primes away or R(5) = 41.
		

Crossrefs

Programs

  • Mathematica
    nn = 100;
    R = Table[0, {nn}]; s = 0; Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s--]; If[sJean-François Alcover, Nov 11 2018, after T. D. Noe in A104272 *)

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).

A001223 Prime gaps: differences between consecutive primes.

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4, 6, 2, 10, 6, 6, 6, 2, 6, 4, 2, 10, 14, 4, 2, 4, 14, 6, 10, 2, 4, 6, 8, 6, 6, 4, 6, 8, 4, 8, 10, 2, 10, 2, 6, 4, 6, 8, 4, 2, 4, 12, 8, 4, 8, 4, 6, 12
Offset: 1

Views

Author

Keywords

Comments

There is a unique decomposition of the primes: provided the weight A117078(n) is > 0, we have prime(n) = weight * level + gap, or A000040(n) = A117078(n) * A117563(n) + a(n). - Rémi Eismann, Feb 14 2008
Let rho(m) = A179196(m), for any n, let m be an integer such that p_(rho(m)) <= p_n and p_(n+1) <= p_(rho(m+1)), then rho(m) <= n < n + 1 <= rho(m + 1), therefore a(n) = p_(n+1) - p_n <= p_rho(m+1) - p_rho(m) = A182873(m). For all rho(m) = A179196(m), a(rho(m)) < A165959(m). - John W. Nicholson, Dec 14 2011
A solution (modular square root) of x^2 == A001248(n) (mod A000040(n+1)). - L. Edson Jeffery, Oct 01 2014
There exists a constant C such that for n -> infinity, Cramer conjecture a(n) < C log^2 prime(n) is equivalent to (log prime(n+1)/log prime(n))^n < e^C. - Thomas Ordowski, Oct 11 2014
a(n) = A008347(n+1) - A008347(n-1). - Reinhard Zumkeller, Feb 09 2015
Yitang Zhang proved lim inf_{n -> infinity} a(n) is finite. - Robert Israel, Feb 12 2015
lim sup_{n -> infinity} a(n)/log^2 prime(n) = C <==> lim sup_{n -> infinity}(log prime(n+1)/log prime(n))^n = e^C. - Thomas Ordowski, Mar 09 2015
a(A038664(n)) = 2*n and a(m) != 2*n for m < A038664(n). - Reinhard Zumkeller, Aug 23 2015
If j and k are positive integers then there are no two consecutive primes gaps of the form 2+6j and 2+6k (A016933) or 4+6j and 4+6k (A016957). - Andres Cicuttin, Jul 14 2016
Conjecture: For any positive numbers x and y, there is an index k such that x/y = a(k)/a(k+1). - Andres Cicuttin, Sep 23 2018
Conjecture: For any three positive numbers x, y and j, there is an index k such that x/y = a(k)/a(k+j). - Andres Cicuttin, Sep 29 2018
Conjecture: For any three positive numbers x, y and j, there are infinitely many indices k such that x/y = a(k)/a(k+j). - Andres Cicuttin, Sep 29 2018
Row m of A174349 lists all indices n for which a(n) = 2m. - M. F. Hasler, Oct 26 2018
Since (6a, 6b) is an admissible pattern of gaps for any integers a, b > 0 (and also if other multiples of 6 are inserted in between), the above conjecture follows from the prime k-tuple conjecture which states that any admissible pattern occurs infinitely often (see, e.g., the Caldwell link). This also means that any subsequence a(n .. n+m) with n > 2 (as to exclude the untypical primes 2 and 3) should occur infinitely many times at other starting points n'. - M. F. Hasler, Oct 26 2018
Conjecture: Defining b(n,j,k) as the number of pairs of prime gaps {a(i),a(i+j)} such that i < n, j > 0, and a(i)/a(i+j) = k with k > 0, then
lim_{n -> oo} b(n,j,k)/b(n,j,1/k) = 1, for any j > 0 and k > 0, and
lim_{n -> oo} b(n,j,k1)/b(n,j,k2) = C with C = C(j,k1,k2) > 0. - Andres Cicuttin, Sep 01 2019

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 870.
  • GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See page 92.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 186-192.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000040 (primes), A001248 (primes squared), A000720, A037201, A007921, A030173, A036263-A036274, A167770, A008347.
Second difference is A036263, first occurrence is A000230.
For records see A005250, A005669.
Sequences related to the differences between successive primes: A001223 (Delta(p)), A028334, A080378, A104120, A330556-A330561.

Programs

  • Haskell
    a001223 n = a001223_list !! (n-1)
    a001223_list = zipWith (-) (tail a000040_list) a000040_list
    -- Reinhard Zumkeller, Oct 29 2011
    
  • Magma
    [(NthPrime(n+1) - NthPrime(n)): n in [1..100]]; // Vincenzo Librandi, Apr 02 2011
    
  • Maple
    with(numtheory): for n from 1 to 500 do printf(`%d,`,ithprime(n+1) - ithprime(n)) od:
  • Mathematica
    Differences[Prime[Range[100]]] (* Harvey P. Dale, May 15 2011 *)
  • PARI
    diff(v)=vector(#v-1,i,v[i+1]-v[i]);
    diff(primes(100)) \\ Charles R Greathouse IV, Feb 11 2011
    
  • PARI
    forprime(p=1, 1e3, print1(nextprime(p+1)-p, ", ")) \\ Felix Fröhlich, Sep 06 2014
    
  • Python
    from sympy import prime
    def A001223(n): return prime(n+1)-prime(n) # Chai Wah Wu, Jul 07 2022
  • Sage
    differences(prime_range(1000)) # Joerg Arndt, May 15 2011
    

Formula

G.f.: b(x)*(1-x), where b(x) is the g.f. for the primes. - Franklin T. Adams-Watters, Jun 15 2006
a(n) = prime(n+1) - prime(n). - Franklin T. Adams-Watters, Mar 31 2010
Conjectures: (i) a(n) = ceiling(prime(n)*log(prime(n+1)/prime(n))). (ii) a(n) = floor(prime(n+1)*log(prime(n+1)/prime(n))). (iii) a(n) = floor((prime(n)+prime(n+1))*log(prime(n+1)/prime(n))/2). - Thomas Ordowski, Mar 21 2013
A167770(n) == a(n)^2 (mod A000040(n+1)). - L. Edson Jeffery, Oct 01 2014
a(n) = Sum_{k=1..2^(n+1)-1} (floor(cos^2(Pi*(n+1)^(1/(n+1))/(1+primepi(k))^(1/(n+1))))). - Anthony Browne, May 11 2016
G.f.: (Sum_{k>=1} x^pi(k)) - 1, where pi(k) is the prime counting function. - Benedict W. J. Irwin, Jun 13 2016
Conjecture: Limit_{N->oo} (Sum_{n=2..N} log(a(n))) / (Sum_{n=2..N} log(log(prime(n)))) = 1. - Alain Rocchelli, Dec 16 2022
Conjecture: The asymptotic limit of the average of log(a(n)) ~ log(log(prime(n))) - gamma (where gamma is Euler's constant). Also, for n tending to infinity, the geometric mean of a(n) is equivalent to log(prime(n)) / e^gamma. - Alain Rocchelli, Jan 23 2023
It has been conjectured that primes are distributed around their average spacing in a Poisson distribution (cf. D. A. Goldston in above links). This is the basis of the last two conjectures above. - Alain Rocchelli, Feb 10 2023

Extensions

More terms from James Sellers, Feb 19 2001

A104272 Ramanujan primes R_n: a(n) is the smallest number such that if x >= a(n), then pi(x) - pi(x/2) >= n, where pi(x) is the number of primes <= x.

Original entry on oeis.org

2, 11, 17, 29, 41, 47, 59, 67, 71, 97, 101, 107, 127, 149, 151, 167, 179, 181, 227, 229, 233, 239, 241, 263, 269, 281, 307, 311, 347, 349, 367, 373, 401, 409, 419, 431, 433, 439, 461, 487, 491, 503, 569, 571, 587, 593, 599, 601, 607, 641, 643, 647, 653, 659
Offset: 1

Views

Author

Jonathan Sondow, Feb 27 2005

Keywords

Comments

Referring to his proof of Bertrand's postulate, Ramanujan (1919) states a generalization: "From this we easily deduce that pi(x) - pi(x/2) >= 1, 2, 3, 4, 5, ..., if x >= 2, 11, 17, 29, 41, ..., respectively." Since the a(n) are prime (by their minimality), I call them "Ramanujan primes."
See the additional references and links mentioned in A143227.
2n log 2n < a(n) < 4n log 4n for n >= 1, and prime(2n) < a(n) < prime(4n) if n > 1. Also, a(n) ~ prime(2n) as n -> infinity.
Shanta Laishram has proved that a(n) < prime(3n) for all n >= 1.
a(n) - 3n log 3n is sometimes positive, but negative with increasing frequency as n grows since a(n) ~ 2n log 2n. There should be a constant m such that for n >= m we have a(n) < 3n log 3n.
A good approximation to a(n) = R_n for n in [1..1000] is A162996(n) = round(k*n * (log(k*n)+1)), with k = 2.216 determined empirically from the first 1000 Ramanujan primes, which approximates the {k*n}-th prime number which in turn approximates the n-th Ramanujan prime and where abs(A162996(n) - R_n) < 2 * sqrt(A162996(n)) for n in [1..1000]. Since R_n ~ prime(2n) ~ 2n * (log(2n)+1) ~ 2n * log(2n), while A162996(n) ~ prime(k*n) ~ k*n * (log(k*n)+1) ~ k*n * log(k*n), A162996(n) / R_n ~ k/2 = 2.216/2 = 1.108 which implies an asymptotic overestimate of about 10% (a better approximation would need k to depend on n and be asymptotic to 2). - Daniel Forgues, Jul 29 2009
Let p_n be the n-th prime. If p_n >= 3 is in the sequence, then all integers (p_n+1)/2, (p_n+3)/2, ..., (p_(n+1)-1)/2 are composite numbers. - Vladimir Shevelev, Aug 12 2009
Denote by q(n) the prime which is the nearest from the right to a(n)/2. Then there exists a prime between a(n) and 2q(n). Converse, generally speaking, is not true, i.e., there exist primes outside the sequence, but possess such property (e.g., 109). - Vladimir Shevelev, Aug 14 2009
The Mathematica program FasterRamanujanPrimeList uses Laishram's result that a(n) < prime(3n).
See sequence A164952 for a generalization we call a Ramanujan k-prime. - Vladimir Shevelev, Sep 01 2009
From Jonathan Sondow, May 22 2010: (Start)
About 46% of primes < 19000 are Ramanujan primes. About 78% of the lesser of twin primes < 19000 are Ramanujan primes.
About 15% of primes < 19000 are the lesser of twin primes. About 26% of Ramanujan primes < 19000 are the lesser of twin primes.
A reason for the jumps is in Section 7 of "Ramanujan primes and Bertrand's postulate" and in Section 4 of "Ramanujan Primes: Bounds, Runs, Twins, and Gaps". (See the arXiv link for a corrected version of Table 1.)
See Shapiro 2008 for an exposition of Ramanujan's proof of his generalization of Bertrand's postulate. (End)
The (10^n)-th R prime: 2, 97, 1439, 19403, 242057, 2916539, 34072993, 389433437, .... - Robert G. Wilson v, May 07 2011, updated Aug 02 2012
The number of R primes < 10^n: 1, 10, 72, 559, 4459, 36960, 316066, 2760321, .... - Robert G. Wilson v, Aug 02 2012
a(n) = R_n = R_{0.5,n} in "Generalized Ramanujan Primes."
All Ramanujan primes are in A164368. - Vladimir Shevelev, Aug 30 2011
If n tends to infinity, then limsup(a(n) - A080359(n-1)) = infinity; conjecture: also limsup(a(n) - A080359(n)) = infinity (cf. A182366). - Vladimir Shevelev, Apr 27 2012
Or the largest prime x such that the number of primes in (x/2,x] equals n. This equivalent definition underlines an important analogy between Ramanujan and Labos primes (cf. A080359). - Vladimir Shevelev, Apr 29 2012
Research questions on R_n - prime(2n) are at A233739, and on n-Ramanujan primes at A225907. - Jonathan Sondow, Dec 16 2013
The questions on R_n - prime(2n) in A233739 have been answered by Christian Axler in "On generalized Ramanujan primes". - Jonathan Sondow, Feb 13 2014
Srinivasan's Lemma (2014): prime(k-n) < prime(k)/2 if R_n = prime(k) and n > 1. Proof: By the minimality of R_n, the interval (prime(k)/2,prime(k)] contains exactly n primes and so prime(k-n) < prime(k)/2. - Jonathan Sondow, May 10 2014
For some n and k, we see that A168421(k) = a(n) so as to form a chain of primes similar to a Cunningham chain. For example (and the first example), A168421(2) = 7, links a(2) = 11 = A168421(3), links a(3) = 17 = A168421(4), links a(4) = 29 = A168421(6), links a(6) = 47. Note that the links do not have to be of a form like q = 2*p+1 or q = 2*p-1. - John W. Nicholson, Feb 22 2015
Extending Sondow's 2010 comments: About 48% of primes < 10^9 are Ramanujan primes. About 76% of the lesser of twin primes < 10^9 are Ramanujan primes. - Dana Jacobsen, Sep 06 2015
Sondow, Nicholson, and Noe's 2011 conjecture that pi(R_{m*n}) <= m*pi(R_n) for m >= 1 and n >= N_m (see A190413, A190414) was proved for n > 10^300 by Shichun Yang and Alain Togbé in 2015. - Jonathan Sondow, Dec 01 2015
Berliner, Dean, Hook, Marr, Mbirika, and McBee (2016) prove in Theorem 18 that the graph K_{m,n} is prime for n >= R_{m-1}-m; see A291465. - Jonathan Sondow, May 21 2017
Okhotin (2012) uses Ramanujan primes to prove Lemma 8 in "Unambiguous finite automata over a unary alphabet." - Jonathan Sondow, May 30 2017
Sepulcre and Vidal (2016) apply Ramanujan primes in Remark 9 of "On the non-isolation of the real projections of the zeros of exponential polynomials." - Jonathan Sondow, May 30 2017
Axler and Leßmann (2017) compute the first k-Ramanujan prime for k >= 1 + epsilon; see A277718, A277719, A290394. - Jonathan Sondow, Jul 30 2017

Examples

			a(1) = 2 is Bertrand's postulate: pi(x) - pi(x/2) >= 1 for all x >= 2.
a(2) = 11 because a(2) < 8 log 8 < 17 and pi(n) - pi(n/2) > 1 for n = 16, 15, ..., 11 but pi(10) - pi(5) = 1.
Consider a(9)=71. Then the nearest prime > 71/2 is 37, and between a(9) and 2*37, that is, between 71 and 74, there exists a prime (73). - _Vladimir Shevelev_, Aug 14 2009 [corrected by _Jonathan Sondow_, Jun 17 2013]
		

References

  • Srinivasa Ramanujan, Collected Papers of Srinivasa Ramanujan (Ed. G. H. Hardy, S. Aiyar, P. Venkatesvara and B. M. Wilson), Amer. Math. Soc., Providence, 2000, pp. 208-209.
  • Harold N. Shapiro, Ramanujan's idea, Section 9.3B in Introduction to the Theory of Numbers, Dover, 2008.

Crossrefs

Cf. A006992 (Bertrand primes), A056171 (pi(n) - pi(n/2)).
Cf. A162996 (Round(kn * (log(kn)+1)), with k = 2.216 as an approximation of R_n = n-th Ramanujan Prime).
Cf. A163160 (Round(kn * (log(kn)+1)) - R_n, where k = 2.216 and R_n = n-th Ramanujan prime).
Cf. A178127 (Lesser of twin Ramanujan primes), A178128 (Lesser of twin primes if it is a Ramanujan prime).
Cf. A181671 (number of Ramanujan primes less than 10^n).
Cf. A174635 (non-Ramanujan primes), A174602, A174641 (runs of Ramanujan and non-Ramanujan primes).
Cf. A189993, A189994 (lengths of longest runs).
Cf. A190124 (constant of summation: 1/a(n)^2).
Cf. A192820 (2- or derived Ramanujan primes R'_n), A192821, A192822, A192823, A192824, A225907.
Cf. A193761 (0.25-Ramanujan primes), A193880 (0.75-Ramanujan primes).
Cf. A185004 - A185007 ("modular" Ramanujan primes).
Not to be confused with the Ramanujan numbers or Ramanujan tau function, A000594.

Programs

  • Maple
    A104272 := proc(n::integer)
        local R;
        if n = 1 then
            return 2;
        end if;
        R := ithprime(3*n-1) ; # upper limit Laishram's thrm Thrm 3 arXiv:1105.2249
        while true do
            if A056171(R) = n then # Defn. 1. of Shevelev JIS 14 (2012) 12.1.1
                return R ;
            end if;
            R := prevprime(R) ;
        end do:
    end proc:
    seq(A104272(n),n=1..200) ; # slow downstream search <= p(3n-1) R. J. Mathar, Sep 21 2017
  • Mathematica
    (RamanujanPrimeList[n_] := With[{T=Table[{k,PrimePi[k]-PrimePi[k/2]}, {k,Ceiling[N[4*n*Log[4*n]]]}]}, Table[1+First[Last[Select[T,Last[ # ]==i-1&]]],{i,1,n}]]; RamanujanPrimeList[54]) (* Jonathan Sondow, Aug 15 2009 *)
    (FasterRamanujanPrimeList[n_] := With[{T=Table[{k,PrimePi[k]-PrimePi[k/2]}, {k,Prime[3*n]}]}, Table[1+First[Last[Select[T,Last[ # ]==i-1&]]],{i,1,n}]]; FasterRamanujanPrimeList[54])
    nn=1000; R=Table[0,{nn}]; s=0; Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s--]; If[sT. D. Noe, Nov 15 2010 *)
  • PARI
    ramanujan_prime_list(n) = {my(L=vector(n), s=0, k=1); for(k=1, prime(3*n)-1, if(isprime(k), s++); if(k%2==0 && isprime(k/2), s--); if(sSatish Bysany, Mar 02 2017
  • Perl
    use ntheory ":all"; my $r = ramanujan_primes(1000); say "[@$r]"; # Dana Jacobsen, Sep 06 2015
    

Formula

a(n) = 1 + max{k: pi(k) - pi(k/2) = n - 1}.
a(n) = A080360(n-1) + 1 for n > 1.
a(n) >= A080359(n). - Vladimir Shevelev, Aug 20 2009
A193761(n) <= a(n) <= A193880(n).
a(n) = 2*A084140(n) - 1, for n > 1. - Jonathan Sondow, Dec 21 2012
a(n) = prime(2n) + A233739(n) = (A233822(n) + a(n+1))/2. - Jonathan Sondow, Dec 16 2013
a(n) = max{prime p: pi(p) - pi(p/2) = n} (see Shevelev 2012). - Jonathan Sondow, Mar 23 2016
a(n) = A000040(A179196(n)). - R. J. Mathar, Sep 21 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = A190303. - Amiram Eldar, Nov 20 2020

A168421 Small Associated Ramanujan Prime, p_(i-n).

Original entry on oeis.org

2, 7, 11, 17, 23, 29, 31, 37, 37, 53, 53, 59, 67, 79, 79, 89, 97, 97, 127, 127, 127, 127, 127, 137, 137, 149, 157, 157, 179, 179, 191, 191, 211, 211, 211, 223, 223, 223, 233, 251, 251, 257, 293, 293, 307, 307, 307, 307, 307, 331, 331, 331
Offset: 1

Views

Author

John W. Nicholson, Nov 25 2009

Keywords

Comments

a(n) is the smallest prime p_(k+1-n) on the left side of the Ramanujan Prime Corollary, 2*p_(i-n) > p_i for i > k, where the n-th Ramanujan Prime R_n is the k-th prime p_k. [Comment clarified and shortened by Jonathan Sondow, Dec 20 2013]
Smallest prime number, a(n), such that if x >= a(n), then there are at least n primes between x and 2x exclusively.
This is very useful in showing the number of primes in the range [p_k, 2*p_(i-n)] is greater than or equal to 1. By taking into account the size of the gaps between primes in [p_(i-n),p_k], one can see that the average prime gap is about log(p_k) using the following R_n / (2*n) ~ log(R_n).
Proof of Corollary: See Wikipedia link
The number of primes until the next Ramanujan prime, R_(n+1), can be found in A190874.
Not the same as A124136.
A084140(n) is the smallest integer where ceiling ((A104272(n)+1)/2), a(n) is the next prime after A084140(n). - John W. Nicholson, Oct 09 2013
If a(n) is in A005382(k) then A005383(k) is a twin prime with the Ramanujan prime, A104272(n) = A005383(k) - 2, and A005383(k) = A168425(n). If this sequence has an infinite number of terms in A005382, then the twin prime conjecture can be proved. - John W. Nicholson, Dec 05 2013
Except for A000101(1)=3 and A000101(2)=5, A000101(k) = a(n). Because of the large size of a gap, there are many repeats of the prime number in this sequence. - John W. Nicholson, Dec 10 2013
For some n and k, we see that a(n) = A104272(k) as to form a chain of primes similar to a Cunningham chain. For example (and the first example), a(2) = 7, links A104272(2) = 11 = a(3), links A104272(3) = 17 = a(4), links A104272(4) = 29 = a(6), links A104272(6) = 47. Note that the links do not have to be of a form like q = 2*p+1 or q = 2*p-1. - John W. Nicholson, Dec 14 2013
Srinivasan's Lemma (2014): p_(k-n) < (p_k)/2 if R_n = p_k and n > 1. Proof: By the minimality of R_n, the interval ((p_k)/2,p_k] contains exactly n primes, so p_(k-n) < (p_k)/2. - Jonathan Sondow, May 10 2014
In spite of the name Small Associated Ramanujan Prime, a(n) is not a Ramanujan prime for many values of n. - Jonathan Sondow, May 10 2014
Prime index of a(n), pi(a(n)) = i-n, is equal to A179196(n) - n + 1. - John W. Nicholson, Sep 15 2015
All maximal prime pairs in A002386 and A000101 are bounded by, for a particular n and i, the prime A104272(n) and twice a prime in A000040() following a(n). This means the gap between maximal prime pair cannot be more than twice the prior maximal prime gap. - John W. Nicholson, Feb 07 2019

Examples

			For n=10, the n-th Ramanujan prime is A104272(n)= 97, the value of k = 25, so i is >= 26, i-n >= 16, the i-n prime is 53, and 2*53 = 106. This leaves the range [97, 106] for the 26th prime which is 101. In this example, 53 is the small associated Ramanujan prime.
		

Crossrefs

Cf. A165959 (range size), A230147 (records).

Programs

Formula

a(n) = prime(primepi(A104272(n)) + 1 - n).
a(n) = nextprime(A084139(n+1)), where nextprime(x) is the next prime > x. Note: some A084139(n) may be prime, therefore nextprime(x) not equal to x. - John W. Nicholson, Oct 11 2013
a(n) = nextprime(A084140(n)). - John W. Nicholson, Oct 11 2013

Extensions

Extended by T. D. Noe, Nov 22 2010

A168425 Large Associated Ramanujan Prime, p_i.

Original entry on oeis.org

3, 13, 19, 31, 43, 53, 61, 71, 73, 101, 103, 109, 131, 151, 157, 173, 181, 191, 229, 233, 239, 241, 251, 269, 271, 283, 311, 313, 349, 353, 373, 379, 409, 419, 421, 433, 439, 443, 463, 491, 499, 509, 571, 577, 593, 599, 601, 607, 613, 643, 647, 653, 659, 661
Offset: 1

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Author

John W. Nicholson, Nov 25 2009

Keywords

Comments

a(n) is the smallest prime on the right side of the Ramanujan Prime Corollary, 2*p_(i-n) > p_i, for i > k where k = pi(p_k) = pi(R_n) That is, p_k is the n-th Ramanujan Prime, R_n and the k-th prime.
a(n) = nextprime(R_n) = nextprime(p_k), where nextprime(x) is the next prime larger than x.
This is very useful in showing the number of primes in the range [p_k, 2*p_(i-n)] is greater than or equal to 1. By taking into account the size of the gaps between primes in [p_(i-n),p_k], one can see that the average prime gap is about log(p_k) using the following R_n / (2*n) ~ log(R_n).
Proof of Corollary: See Wikipedia link.
The number of primes until the next Ramanujan prime, R_(n+1), can be found in A190874.
Srinivasan's Lemma (2014): p_(k-n) < (p_k)/2 if R_n = p_k and n > 1. Proof: By the minimality of R_n, the interval ((p_k)/2,p_k] contains exactly n primes, so p_(k-n) < (p_k)/2. - Jonathan Sondow, May 10 2014
In spite of the name Large Associated Ramanujan Prime, a(n) is not a Ramanujan prime for many values of n. - Jonathan Sondow, May 10 2014

Examples

			For n=10, the n-th Ramanujan prime is A104272(n)= 97, the value of k = 25, so i is >= 26, i-n >= 16, the i-n prime is 53, and 2*53 = 106. This leaves the range [97, 106] for the 26th prime which is 101. In this example, 101 is the large associated Ramanujan prime.
		

Crossrefs

Programs

  • Mathematica
    nn = 100; R = Table[0, {nn}]; s = 0;
    Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s--];
    If[s < nn, R[[s+1]] = k], {k, Prime[3 nn]}
    ];
    RamanujanPrimes = R + 1;
    Prime[PrimePi[#]+1]& /@ RamanujanPrimes (* Jean-François Alcover, Nov 03 2018, after T. D. Noe in A104272 *)
  • PARI
    genit(n=100)={my(L=vector(n),s=0,k=1,z);for(k=1,prime(3*n)-1,if(ispseudoprime(k),s++);if(k%2==0&&ispseudoprime(k/2),s--);if(snextprime(x+1),L);v} \\ Bill McEachen, Jun 24 2023 (incorporates code from A104272)
  • Perl
    use ntheory ":all"; say next_prime(nth_ramanujan_prime($)) for 1..100; # _Dana Jacobsen, Dec 25 2015
    

Formula

a(n) = prime(primepi(A104272(n)) + 1).
a(n) = A151800(A104272(n)). - Michel Marcus, Jun 27 2023

A214756 a(n) = largest Ramanujan prime R_k in A104272 that is <= A002386(n).

Original entry on oeis.org

2, 2, 2, 17, 71, 107, 503, 881, 1103, 1301, 9521, 15671, 19543, 31387, 155849, 360289, 370061, 492067, 1349147, 1356869, 2010553, 4652239, 17051297, 20831119, 47326519, 122164649, 189695483, 191912659
Offset: 1

Views

Author

John W. Nicholson, Jul 27 2012

Keywords

Comments

While many values in A214757(n) are equal to A000101(n), here it seems the only value such that A002386(n) is equal to a(n) is a(1) = R_k = A002386(1) = 2.
See "Let rho(m) = A179196(m)" comment at A001223.

Examples

			A104272(94) = 1301 < 1327 = A002386(10), so a(10) = 1301.
		

Crossrefs

Programs

  • Perl
    use ntheory ":all";  sub a_from_2386 { my $n = shift; $n = prev_prime($n) while !is_ramanujan_prime($n); $n } # Dana Jacobsen, Jul 13 2016
    
  • Perl
    perl -Mntheory=:all -nE 'my $n=$1 if /(\d+)$/; $r=ramanujan_primes($n>1e6 ? $n-1e6 : 2, $n); say ++$x," ",$r->[-1];' b002386.txt # Dana Jacobsen, Jul 13 2016

Extensions

Edited by N. J. A. Sloane, Aug 06 2012
a(16)-a(28) from Donovan Johnson, Nov 04 2012

A214757 a(n) = smallest Ramanujan prime R_k in A104272 that is >= A000101(n).

Original entry on oeis.org

11, 11, 11, 29, 97, 127, 569, 937, 1151, 1367, 9613, 15727, 19681, 31481, 156007, 360769, 370387, 492251, 1349669, 1357333, 2010881, 4652507, 17051981, 20831639, 47326913, 122165059, 189695893, 191913047
Offset: 1

Views

Author

John W. Nicholson, Jul 27 2012

Keywords

Comments

While many values in a(n) are equal to A000101(n), for A214756 it seems the only value such that A002386(n) is equal to A214756(n) is A214756(1) = R_k = A002386(1) = 2.
See "Let rho(m) = A179196(m)" comment at A001223.

Examples

			A104272(95) = R_k = 1367 > 1361 = A000101(10), so a(10) = 1367.
		

Crossrefs

Extensions

Edited by N. J. A. Sloane, Aug 06 2012
a(16)-a(28) from Donovan Johnson, Nov 04 2012

A214926 Difference A214925(n) - A214924(n), prime count between Ramanujan primes bounding maximal gap primes.

Original entry on oeis.org

4, 4, 4, 3, 5, 3, 8, 7, 5, 7, 7, 3, 10, 6, 8, 24, 19, 6, 24, 25, 16, 8, 30, 17, 12, 13, 12, 11
Offset: 1

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Author

John W. Nicholson, Aug 06 2012

Keywords

Comments

Conjecture: For every n > 0, a(n) > 1.
Let rho(m) = A179196(m), for any n, let m be an integer such that p_(rho(m)) <= p_n and p_(n+1) <= p_(rho(m+1)), then rho(m) <= n < n + 1 <= rho(m + 1), therefore A001223(n) = p_(n+1) - p_n <= p_rho(m+1) - p_rho(m) = A182873(m). For all rho(m) = A179196(m), A001223(rho(m)) < A165959(m). (Comment copied from A001223). John W. Nicholson, Nov 17 2013

Examples

			a(4) = pi(A214757(4)) - pi(A214756(4)) = 10 - 7 = 3
		

Crossrefs

Formula

a(n) = pi(A214757(n)) - pi(A214756(n)).
a(n) = rho(A214757(n)) - rho(A214756(n)).

Extensions

Extension to a(28) added by John W. Nicholson, Nov 11 2013

A190413 primepi(R_{n*m}) <= n*primepi(R_m) for m >= a(n), where R_k is the k-th Ramanujan prime (A104272).

Original entry on oeis.org

1, 1245, 189, 189, 85, 85, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
Offset: 1

Views

Author

T. D. Noe, May 11 2011

Keywords

Comments

This is Conjecture 1 in the paper by Sondow, Nicholson, and Noe. The conjecture has been verified for n <= 20 and Ramanujan primes less than 10^9.
A restatement is rho(n*m) <= n*rho(m) for m >= a(n), where rho = A179196.
The conjecture has been proven for n > 10^300 by Shichun Yang and Alain Togbé. - Jonathan Sondow, Jan 21 2016
The conjecture has been proven for n > 38 and m > 9 by Christian Axler. Complete exception list can be found in remark of paper. - John W. Nicholson, Aug 04 2019

Crossrefs

Formula

For all n >= 20, a(n) = 2.

A190414 primepi(R_m) <= i*primepi(R_j) for any factorization m=i*j if j >= a(i), where R_k is the k-th Ramanujan prime (A104272).

Original entry on oeis.org

1, 2490, 567, 756, 425, 510, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100
Offset: 1

Views

Author

John W. Nicholson, May 10 2011

Keywords

Comments

This is another interpretation of Conjecture 1 in the paper by Sondow, Nicholson, and Noe. The conjecture has been verified for i <= 20 and Ramanujan primes less than 10^9.
The conjecture has been proven for i > 38 and j > 9 by Christian Axler. Complete exception list can be found in remark of paper. - John W. Nicholson, Aug 04 2019

Crossrefs

Formula

For all n >= 20, a(n) = 2*n.
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