

A001223


Prime gaps: differences between consecutive primes.
(Formerly M0296 N0108)


687



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
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OFFSET

1,2


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
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
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
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
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 ktuple 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.
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).


LINKS

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].


FORMULA

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
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


MAPLE

with(numtheory): for n from 1 to 500 do printf(`%d, `, ithprime(n+1)  ithprime(n)) od:


MATHEMATICA



PROG

(Sage) differences(prime_range(1000)) # Joerg Arndt, May 15 2011
(PARI) diff(v)=vector(#v1, i, v[i+1]v[i]);
(PARI) forprime(p=1, 1e3, print1(nextprime(p+1)p, ", ")) \\ Felix Fröhlich, Sep 06 2014
(Magma) [(NthPrime(n+1)  NthPrime(n)): n in [1..100]]; // Vincenzo Librandi, Apr 02 2011
(Haskell)
a001223 n = a001223_list !! (n1)
a001223_list = zipWith () (tail a000040_list) a000040_list
(Python)
from sympy import prime


CROSSREFS



KEYWORD

nonn,nice,easy,hear,changed


AUTHOR



EXTENSIONS



STATUS

approved



