A036263 -id:A036263 - cp's OEIS frontend

A001223 Prime gaps: differences between consecutive primes.

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

N. J. A. Sloane

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

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).

Vojtech Strnad, First 100000 terms [First 10000 terms from N. J. A. Sloane]
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].
Anonymous ["TheHereticAnthem20"], Prime gaps mapped to sounds, Youtube video (2018).
B. Apostol, L. Panaitopol, L Petrescu, and L. Toth, Some Properties of a Sequence Defined with the Aid of Prime Numbers, J. Int. Seq. 18 (2015) # 15.5.5.
S. Ares and M. Castro, Hidden structure in the randomness of the prime number sequence?, arXiv:cond-mat/0310148 [cond-mat.stat-mech], 2003-2005.
József Beck, Inevitable randomness in discrete mathematics, University Lecture Series, 49. American Mathematical Society, Providence, RI, 2009. xii+250 pp. ISBN: 978-0-8218-4756-5; MR2543141 (2010m:60026). See page 7.
Chris K. Caldwell, Prime k-tuple conjecture, Prime Pages' Glossary entry.
Joel E. Cohen and Dexter Senft, Gaps of size 2, 4, and (conditionally) 6 between successive odd composite numbers occur infinitely often, Notes on Number Theory and Discrete Mathematics, Volume 31, Number 3, 494-503 (2025). See p. 495.
Péter L. Erdős, Gergely Harcos, Shubha R. Kharel, Péter Maga, Tamás Róbert Mezei and Zoltán Toroczkai, The sequence of prime gaps is graphic, Mathematische Annalen 388 (2024), 2195-2215.
D. A. Goldston, S. W. Graham, J. Pintz and C. Y. Yildirim, Small gaps between primes and almost primes, arXiv:math/0506067 [math.NT], 2005.
D. A. Goldston and A. H. Ledoan, On the differences between consecutive prime numbers, I", arXiv:1111.3380v1 [math.NT], Nov 14, 2011.
D. A. Goldston, J. Pintz, and C. Y. Yildirim, Positive Proportion of Small Gaps Between Consecutive Primes, arXiv:1103.3986 [math.NT], Mar 21, 2011.
D. R. Heath-Brown and H. Iwaniec, On the difference between consecutive primes, Bull. Amer. Math. Soc. 1 (1979), 758-760.
Alexei Kourbatov, Tables of record gaps between prime constellations, arXiv preprint arXiv:1309.4053 [math.NT], 2013.
Alexei Kourbatov, The distribution of maximal prime gaps in Cramer's probabilistic model of primes, arXiv preprint arXiv:1401.6959 [math.NT], 2014.
The Polymath project, Bounded gaps between primes
Carlos Rivera, Conjecture 82. Average of log Dn / log(logPn) equal R = 0,877 08..., The Prime Puzzles & Problems Connection.
Hisanobu Shinya, On the density of prime differences less than a given magnitude which satisfy a certain inequality, arXiv:0809.3458 [math.GM], 2008-2011.
K. Soundararajan, Small gaps between prime numbers: the work of Goldston-Pintz-Yildirim, Bull. Amer. Math. Soc., 44 (2007), 1-18.
Eric Weisstein's World of Mathematics, Andrica's Conjecture
Eric Weisstein's World of Mathematics, Prime Difference Function
Yasuo Yamasaki and Aiichi Yamasaki, On the Gap Distribution of Prime Numbers, Kyoto University Research Information Repository, October 1994. MR1370273 (97a:11141).
Yitang Zhang, Bounded gaps between primes, Annals of Mathematics 179 (2014), 1121-1174.
Index entries for primes, gaps between

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.
Cf. A038664, A031131, A031165, A031166, A031167, A031168, A031169, A031170, A031171, A031172.
Cf. A174349, A029707, A029709, A320701, ..., A320720.
Sequences related to the differences between successive primes: A001223 (Delta(p)), A028334, A080378, A104120, A330556-A330561.

a001223 n = a001223_list !! (n-1)
a001223_list = zipWith (-) (tail a000040_list) a000040_list
-- Reinhard Zumkeller, Oct 29 2011

[(NthPrime(n+1) - NthPrime(n)): n in [1..100]]; // Vincenzo Librandi, Apr 02 2011

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

Differences[Prime[Range[100]]] (* Harvey P. Dale, May 15 2011 *)

diff(v)=vector(#v-1,i,v[i+1]-v[i]);
diff(primes(100)) \\ Charles R Greathouse IV, Feb 11 2011

forprime(p=1, 1e3, print1(nextprime(p+1)-p, ", ")) \\ Felix Fröhlich, Sep 06 2014

from sympy import prime
def A001223(n): return prime(n+1)-prime(n) # Chai Wah Wu, Jul 07 2022

differences(prime_range(1000)) # Joerg Arndt, May 15 2011

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

More terms from James Sellers, Feb 19 2001

A006562 Balanced primes (of order one): primes which are the average of the previous prime and the following prime.

Original entry on oeis.org

5, 53, 157, 173, 211, 257, 263, 373, 563, 593, 607, 653, 733, 947, 977, 1103, 1123, 1187, 1223, 1367, 1511, 1747, 1753, 1907, 2287, 2417, 2677, 2903, 2963, 3307, 3313, 3637, 3733, 4013, 4409, 4457, 4597, 4657, 4691, 4993, 5107, 5113, 5303, 5387, 5393

Offset: 1

N. J. A. Sloane and Robert G. Wilson v

Subsequence of A075540. - Franklin T. Adams-Watters, Jan 11 2006
This subsequence of A125830 and of A162174 gives primes of level (1,1): More generally, the i-th prime p(i) is of level (1,k) if and only if it has level 1 in A117563 and 2 p(i) - p(i+1) = p(i-k). - Rémi Eismann, Feb 15 2007
Note the similarity between plots of A006562 and A013916. - Bill McEachen, Sep 07 2009
Balanced primes U strong primes = good primes. Or, A006562 U A051634 = A046869. - Juri-Stepan Gerasimov, Mar 01 2010
Primes prime(n) such that A001223(n-1) = A001223(n). - Irina Gerasimova, Jul 11 2013
Numbers m such that A346399(m) is odd and >= 3. - Ya-Ping Lu, Dec 26 2021 and May 07 2024
"Balanced" means that the next and preceding gap are of the same size, i.e., the second difference A036263 vanishes; so these are the primes whose indices are 1 more than indices of zeros in A036263, listed in A064113. - M. F. Hasler, Oct 15 2024
Primes which are the average of three consecutive primes. - Peter Schorn, Apr 30 2025

			5 belongs to the sequence because 5 = (3 + 7)/2. Likewise 53 = (47 + 59)/2.
5 belongs to the sequence because it is a term, but not first or last, of the AP of consecutive primes (3, 5, 7).
53 belongs to the sequence because it is a term, but not first or last, of the AP of consecutive primes (47, 53, 59).
257 and 263 belong to the sequence because they are terms, but not first or last, of the AP of consecutive primes (251, 257, 263, 269).

A. Murthy, Smarandache Notions Journal, Vol. 11 N. 1-2-3 Spring 2000.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 134.

T. D. Noe, Table of n, a(n) for n = 1..10000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, p. 870.
Ernest G. Hibbs, Component Interactions of the Prime Numbers, Ph. D. Thesis, Capitol Technology Univ. (2022), see p. 33.
Shubhankar Paul, Ten Problems of Number Theory, International Journal of Engineering and Technical Research (IJETR), ISSN: 2321-0869, Volume-1, Issue-9, November 2013.
Shubhankar Paul, Legendre, Grimm, Balanced Prime, Prime triple, Polignac's conjecture, a problem and 17 tips with proof to solve problems on number theory, International Journal of Engineering and Technical Research (IJETR), ISSN: 2321-0869, Volume-1, Issue-10, December 2013.
Wikipedia, Balanced prime.

Primes A000040 whose indices are 1 more than A064113, indices of zeros in A036263 (second differences of the primes).
Cf. A082077, A082078, A082079, A096697, A096698, A096699, A096700, A096701, A096702, A096703, A096704, A096693, A051634, A051635, A054342, A117078, A117563, A125830, A117876, A125576, A046869, A173891, A173892, A173893, A006560, A075540.
Cf. A225494 (multiplicative closure); complement of A178943 with respect to A000040.
Cf. A055380, A051795, A081415, A096710 for other balanced prime sequences.

a006562 n = a006562_list !! (n-1)
a006562_list = filter ((== 1) . a010051) a075540_list
-- Reinhard Zumkeller, Jan 20 2012

a006562 n = a006562_list !! (n-1)
a006562_list = h a000040_list where
   h (p:qs@(q:r:ps)) = if 2 * q == (p + r) then q : h qs else h qs
-- Reinhard Zumkeller, May 09 2013

[a: n in [1..1000] | IsPrime(a) where a is NthPrime(n)-NthPrime(n+1)+NthPrime(n+2)]; // Vincenzo Librandi, Jun 23 2016

Transpose[ Select[ Partition[ Prime[ Range[1000]], 3, 1], #[[2]] ==(#[[1]] + #[[3]])/2 &]][[2]]
p=Prime[Range[1000]]; p[[Flatten[1+Position[Differences[p, 2], 0]]]]
Prime[#]&/@SequencePosition[Differences[Prime[Range[800]]],{x_,x_}][[All,2]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 31 2019 *)

betwixtpr(n) = { local(c1,c2,x,y); for(x=2,n, c1=c2=0; for(y=prime(x-1)+1,prime(x)-1, if(!isprime(y),c1++); ); for(y=prime(x)+1,prime(x+1)-1, if(!isprime(y),c2++); ); if(c1==c2,print1(prime(x)",")) ) } \\ Cino Hilliard, Jan 25 2005

forprime(p=1,999, p-precprime(n-1)==nextprime(p+1)-p && print1(p",")) \\ M. F. Hasler, Jun 01 2013

is(n)=n-precprime(n-1)==nextprime(n+1)-n && isprime(n) \\ Charles R Greathouse IV, Apr 07 2016

from sympy import nextprime; p, q, r = 2, 3, 5
while q < 6000:
    if 2*q == p + r: print(q, end = ", ")
    p, q, r = q, r, nextprime(r) # Ya-Ping Lu, Dec 23 2021

2*p_n = p_(n-1) + p_(n+1).
Equals { p = prime(k) | A118534(k) = prime(k-1) }. - Rémi Eismann, Nov 30 2009
a(n) = A000040(A064113(n) + 1) = (A122535(n) + A181424(n)) / 2. - Reinhard Zumkeller, Jan 20 2012
a(n) = A122535(n) + A117217(n). - Zak Seidov, Feb 14 2013
Equals A145025 intersect A000040 = A145025 \ A024675. - M. F. Hasler, Jun 01 2013
Conjecture: Limit_{n->oo} n*(log(a(n)))^2 / a(n) = 1/2. - Alain Rocchelli, Mar 21 2024
Conjecture: The asymptotic limit of the average of a(n+1)-a(n) is equivalent to 2*(log(a(n)))^2. Otherwise formulated: 2 * Sum_{n=1..N} (log(a(n)))^2 ~ a(N). - Alain Rocchelli, Mar 23 2024

Reworded comment and added formula from R. Eismann. - M. F. Hasler, Nov 30 2009

Edited by Daniel Forgues, Jan 15 2011

A046933 Number of composites between successive primes.

Original entry on oeis.org

0, 1, 1, 3, 1, 3, 1, 3, 5, 1, 5, 3, 1, 3, 5, 5, 1, 5, 3, 1, 5, 3, 5, 7, 3, 1, 3, 1, 3, 13, 3, 5, 1, 9, 1, 5, 5, 3, 5, 5, 1, 9, 1, 3, 1, 11, 11, 3, 1, 3, 5, 1, 9, 5, 5, 5, 1, 5, 3, 1, 9, 13, 3, 1, 3, 13, 5, 9, 1, 3, 5, 7, 5, 5, 3, 5, 7, 3, 7, 9, 1, 9, 1, 5, 3, 5, 7, 3, 1, 3, 11, 7, 3, 7, 3, 5, 11, 1, 17

Offset: 1

Marc LeBrun, Dec 11 1999

a(n) is odd for n>1 since all primes except 2 are odd. - Joel Brennan, Jan 02 2023

			a(1) = 0 since 2 is adjacent to 3;
a(2) = 1 since 4 is between 3 and 5;
a(4) = 3 = 11 - 7 - 1, etc.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A000040, A028334, A036263.

Cf. A008996 (record values > 0).

a046933 n = a046933_list !! (n-1)
a046933_list = map (subtract 1) a001223_list
-- Reinhard Zumkeller, Dec 12 2012

A046933:=n->ithprime(n+1)-ithprime(n)-1; seq(A046933(n), n=1..100); # Wesley Ivan Hurt, Apr 15 2014

Differences[Prime[Range[100]]] - 1 (* Arkadiusz Wesolowski, Nov 18 2011 *)
Table[Prime[n + 1] - Prime[n] - 1, {n, 100}] (* Wesley Ivan Hurt, Apr 15 2014 *)
Prepend[Drop[Length/@SequenceSplit[Range@Prime@100,{?PrimeQ}],1],0] (* _Federico Provvedi, Jul 19 2021 *)

a(n)=prime(n+1)-prime(n)-1 \\ Charles R Greathouse IV, Nov 20 2012

from sympy import prime
def A046933(n): return prime(n+1)-prime(n)-1 # Chai Wah Wu, Jan 02 2024

a(n) = prime(n+1) - prime(n) - 1 = A000040(n+1) - A000040(n) - 1.
a(n) = A001223(n) - 1.
a(n) = 2*A028334(n) - 1 for n>1. - Giovanni Teofilatto, Apr 19 2010
a(n) = Sum_{i=1..n-1} A036263(i). - Daniel Forgues, Apr 07 2014

A053289 First differences of consecutive perfect powers (A001597).

Original entry on oeis.org

3, 4, 1, 7, 9, 2, 5, 4, 13, 15, 17, 19, 21, 4, 3, 16, 25, 27, 20, 9, 18, 13, 33, 35, 19, 18, 39, 41, 43, 28, 17, 47, 49, 51, 53, 55, 57, 59, 61, 39, 24, 65, 67, 69, 71, 35, 38, 75, 77, 79, 81, 47, 36, 85, 87, 89, 23, 68, 71, 10, 12, 95, 97, 99, 101, 103, 40, 65, 107, 109, 100

Offset: 1

Labos Elemer, Mar 03 2000

nonn

Michel Waldschmidt writes: Conjecture 1.3 (Pillai). Let k be a positive integer. The equation x^p - y^q = k where the unknowns x, y, p and q take integer values, all >= 2, has only finitely many solutions (x,y,p,q). This means that in the increasing sequence of perfect powers [A001597] the difference between two consecutive terms [the present sequence] tends to infinity. It is not even known whether for, say, k=2, Pillai's equation has only finitely many solutions. A related open question is whether the number 6 occurs as a difference between two perfect powers. See Sierpiński [1970], problem 238a, p. 116. - Jonathan Vos Post, Feb 18 2008

Are there are any adjacent equal terms? - Gus Wiseman, Oct 08 2024

			Consecutive perfect powers are A001597(14) = 121, A001597(13) = 100, so a(13) = 121 - 100 = 21.

Wacław Sierpiński, 250 problems in elementary number theory, Modern Analytic and Computational Methods in Science and Mathematics, No. 26, American Elsevier, Warsaw, 1970, pp. 21, 115-116.
S. S. Pillai, On the equation 2^x - 3^y = 2^X - 3^Y, Bull, Calcutta Math. Soc. 37 (1945) 15-20.

Daniel Forgues and T. D. Noe, Table of n, a(n) for n = 1..10000
Rafael Jakimczuk, Gaps between consecutive perfect powers, International Mathematical Forum, Vol. 11, No. 9 (2016), pp. 429-437.
Holly Krieger and Brady Haran, Catalan's Conjecture, Numberphile video (2018).
Michel Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004.

For non-perfect-powers (A007916) we have A375706.
The union is A023055.
For prime-powers (A000961 or A246655) we have A057820.
Sorted positions of first appearances are A376268, complement A376519.
For second differences we have A376559.
Ascending and descending points are A376560 and A376561.
A001597 lists perfect-powers.
A112344 counts integer partitions into perfect-powers, factorizations A294068.
A333254 gives run-lengths of differences between consecutive primes.
Cf. A025475, A069623, A219551.
Cf. A007921, A036263, A045542, A052410, A053707, A174965, A336416, A375735, A375736, A375740, A376562.

Differences@ Select[Range@ 3200, # == 1 || GCD @@ FactorInteger[#][[All, 2]] > 1 &] (* Michael De Vlieger, Jun 30 2016, after Ant King at A001597 *)

from sympy import mobius, integer_nthroot
def A053289(n):
    if n==1: return 3
    def f(x): return int(n-2+x+sum(mobius(k)*(integer_nthroot(x,k)[0]-1) for k in range(2,x.bit_length())))
    kmin, kmax = 1,2
    while f(kmax)+1 >= kmax:
        kmax <<= 1
    rmin, rmax = 1, kmax
    while True:
        kmid = kmax+kmin>>1
        if f(kmid)+1 < kmid:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    while True:
        rmid = rmax+rmin>>1
        if f(rmid) < rmid:
            rmax = rmid
        else:
            rmin = rmid
        if rmax-rmin <= 1:
            break
    return kmax-rmax # Chai Wah Wu, Aug 13 2024

a(n) = A001597(n+1) - A001597(n). - Jonathan Vos Post, Feb 18 2008
From Amiram Eldar, Jun 30 2023: (Start)
Formulas from Jakimczuk (2016):
Lim sup_{n->oo} a(n)/(2*n) = 1.
Lim inf_{n->oo} a(n)/(2*n)^(2/3 + eps) = 0. (End)
Can be obtained by inserting 0 between 3 and 6 in A375702 and then adding 1 to all terms. In particular, for n > 2, a(n+1) - 1 = A375702(n). - Gus Wiseman, Sep 14 2024

cp's OEIS Frontend

A295973 Primes introducing new second differences in A036263.

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A293154 Record high values in A036263.

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A293155 Indices of record high values in A036263.

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A376656 Sorted positions of first appearances in the second differences (A036263) of consecutive primes (A000040).

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A094900 a(n) = floor( prime(n-1)*A036263(n-2)/ A001223(n-1)).

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A216094 a(1) = 2; a(n+1) = a(n) + |A036263(n)| + 2.

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A001223 Prime gaps: differences between consecutive primes.

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A006562 Balanced primes (of order one): primes which are the average of the previous prime and the following prime.

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