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.

Showing 1-5 of 5 results.

A230147 Record values in A165959.

Original entry on oeis.org

2, 3, 5, 11, 15, 27, 33, 37, 65, 67, 75, 77, 95, 137, 147, 151, 153, 169, 191, 219, 247, 249, 251, 291, 297, 303, 307, 319, 415, 429, 441, 465, 495
Offset: 1

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Author

John W. Nicholson, Nov 20 2013

Keywords

Comments

The index value A230146(n) is also the index of A104272(n).
Because of the bounds on a(n), A182873(n), A192820, and A190661, old conjectures about prime numbers may be proved.

Crossrefs

Cf. A230146 = indices of record values of A165959.

Programs

A230146 a(n) = k, index of A165959(k) at record values.

Original entry on oeis.org

1, 2, 3, 6, 17, 19, 77, 121, 167, 1413, 1550, 1882, 1905, 2917, 12584, 14990, 20640, 25730, 27336, 28023, 93554, 106600, 127318, 134871, 295621, 669047, 1000875, 1178689, 1208481, 2620546, 6427781, 9755037, 9863644
Offset: 1

Views

Author

John W. Nicholson, Nov 19 2013

Keywords

Comments

The index value a(n) is also the index of A104272.

Crossrefs

Cf. A230147 holds record values of A165959.

Programs

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

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

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

Views

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
Showing 1-5 of 5 results.