A072752 -id:A072752 - cp's OEIS frontend

A048670 Jacobsthal function A048669 applied to the product of the first n primes (A002110).

2, 4, 6, 10, 14, 22, 26, 34, 40, 46, 58, 66, 74, 90, 100, 106, 118, 132, 152, 174, 190, 200, 216, 234, 258, 264, 282, 300, 312, 330, 354, 378, 388, 414, 432, 450, 476, 492, 510, 538, 550, 574, 600, 616, 642, 660, 686, 718, 742, 762, 798, 810, 834, 858, 876, 908, 926, 954

Offset: 1

Jan Kristian Haugland

Pintz shows that j(x#) >= (2*e^gamma + o(1)) x log x log log log x / (log log x)^2 and hence a(n) >= (2*e^gamma + o(1)) n log^2 n log log log n / (log log n)^2 by the Prime Number Theorem. - Charles R Greathouse IV, Sep 08 2012
Jacobsthal conjectures that a(n) >= j(k) := A048669(k) for any k with n prime factors, which would make this the RECORDS transform of A048669. Hajdu & Saradha disprove the conjecture, showing that this fails for n = 24 where j(k) = 236 > 234 = a(24) for any k divisible by 76964283982898776138308824190 and with 24 prime factors in total. - Charles R Greathouse IV, Sep 08 2012 / Edited by Jan Kristian Haugland, Feb 02 2019
Ford, Green, Konyagin, Maynard, & Tao show that j(x#) >> x log x log log log x / log log x and hence a(n) >> n log^2 n log log log n / log log n. - Charles R Greathouse IV, Mar 29 2018
Computation of a(62)-a(64) was supported by Google Cloud. - Andrzej Bozek, Mar 14 2021

L. E. Dickson, History of the Theory of Numbers, Vol. 1, p. 439, Chelsea, 1952.

Andrzej Bozek, Table of n, a(n) for n = 1..64
Andrzej Bozek, Gerbicz's table with added a(58)-a(64)
Fintan Costello and Paul Watts, A computational upper bound on Jacobsthal's function, arXiv:1208.5342 [math.NT], 2012.
Kevin Ford, Ben Green, Sergei Konyagin, James Maynard, and Terence Tao, Long gaps between primes, arXiv:1412.5029 [math.NT], 2014-2016; Journal of the American Mathematical Society 31:1 (2018), pp. 65-105.
Robert Gerbicz, Table of n, a(n), u(n) for n=1..57, where every integer from [u(n),u(n)+a(n)-2] is divisible by at least one of the first n primes. Note that u(n) is not unique, the smallest is given by A049300(n).
Thomas R. Hagedorn, Computation of Jacobsthal's function h(n) for n < 50, Math. Comp. 78 (2009) 1073-1087.
L. Hajdu and N. Saradha, Disproof of a conjecture of Jacobsthal, Mathematics of Computation 81 (2012), pp. 2461-2471.
H. Iwaniec, On the error term in the linear sieve, Acta Arithmetica 19 (1971), pp. 1-30.
Helmut Maier and Carl Pomerance, Unusually large gaps between consecutive primes, Transactions of the American Mathematical Society 322:1 (1990), pp. 201-237.
János Pintz, Very large gaps between consecutive primes, Journal of Number Theory 63 (1997), pp. 286-301.
Stéphane Vinatier and Reg Alcorn, Mathématiques et Art, Gaz. Soc. Math. de France (in French, 2024) No. 181, 44-58. See p. 57.
Stéphane Vinatier and William Reginald Alcorn, Mathematics as seen by an artist: Inspiring mathematical objects, Eur. Math. Soc. Mag. (2025) Vol. 135, 20-31. See p. 30.
Mario Ziller, New computational results on a conjecture of Jacobsthal, arXiv:1903.11973 [math.NT], 2019.
Mario Ziller, On differences between consecutive numbers coprime to primorials, arXiv:2007.01808 [math.NT], 2020.
Mario Ziller and John F. Morack, Algorithmic concepts for the computation of Jacobsthal's function, arXiv:1611.03310 [math.NT], 2016.

Cf. A048669, A002110, A005867, A049300, A058989, A072752, A331118.

(* This program is not suitable to compute more than a few terms *) primorial[n_] := Product[Prime[k], {k, 1, n}]; j[n_] := Module[{L = 1, m = 1}, For[k = 2, k <= n + 1, k++, If[GCD[k, n] == 1, If[L + m < k, m = k - L]; L = k]]; m]; a[n_] := a[n] = j[primorial[n]]; Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 1, 10}] (* Jean-François Alcover, Sep 27 2013, after M. F. Hasler *)

a(n) = A058989(n) + 1.
a(n) << n^2*(log n)^2, see Iwaniec. - Charles R Greathouse IV, Sep 08 2012
a(n) >= (2*e^gamma + o(1)) n log^2 n log log log n / (log log n)^2, see Pintz.
a(n) = 2 * A072752(n) + 2. - Mario Ziller, Dec 08 2016
Maier & Pomerance conjecture that Max_{n <= x} A048669(n) = log(x)*(log log x)^(2+o(1)) which suggests a(n) = n*(log n)^(3+o(1)). - Charles R Greathouse IV, Mar 29 2018
a(n) = largest (or last) term in row n of A331118. - Michael De Vlieger, Dec 11 2020

a(21)-a(24) from Max Alekseyev, Apr 09 2006
a(25)-a(49) from Thomas Hagedorn, Feb 21 2007
a(46) corrected (published value in Hagedorn's 2009 Mathematics of Computation article was correct) and a(50)-a(54) added by Mario Ziller, Dec 08 2016
a(55)-a(57) from Robert Gerbicz, Apr 10 2017
a(58)-a(64) from Andrzej Bozek, Mar 14 2021

A058989 Largest number of consecutive integers such that each is divisible by a prime <= the n-th prime.

Original entry on oeis.org

1, 3, 5, 9, 13, 21, 25, 33, 39, 45, 57, 65, 73, 89, 99, 105, 117, 131, 151, 173, 189, 199, 215, 233, 257, 263, 281, 299, 311, 329, 353, 377, 387, 413, 431, 449, 475, 491, 509, 537, 549, 573, 599, 615, 641, 659, 685, 717, 741, 761, 797, 809, 833, 857, 875, 907, 925, 953, 977, 1001, 1029, 1057, 1097, 1109

Offset: 1

Jud McCranie, Jan 16 2001

Marty Weissman conjectured that a(n)=2q-1, where q is the largest prime smaller than the n-th prime. The conjecture holds for the first few terms, but then a(n) is larger than 2q-1. Phil Carmody proved a(n)>=2q-1. Terms were calculated by Weissman, Carmody and McCranie.
A049300(n) is the smallest value of the mentioned consecutive integers. - Reinhard Zumkeller, Jun 14 2003
By Lemma 5 of Maynard, there is a constant C > 0 such that, for any n, there is a prime p <= C*exp(prime(n)) such that q - p >= a(n) where q is the smallest prime larger than p. (Probably C = 7/e^3 = 0.35... is admissible.) - Charles R Greathouse IV, Aug 01 2024

			The 4th prime is 7. Nine is the maximum number of consecutive integers such that each is divisible by 2, 3, 5 or 7. (Example: 2 through 10) So a(4)=9.

Dickson, L. E., History of the Theory of Numbers, Vol. 1, p. 439, Chelsea, 1952.

Thomas R. Hagedorn, Computation of Jacobsthal's function h(n) for n < 50, Math. Comp. 78 (2009) 1073-1087.
H. Iwaniec, On the error term in the linear sieve, Acta. Arith. 19 (1971), pp. 1-30.
J. D. Laison and M. Schick, Seeing dots: visibility of lattice points, Mathematics Magazine, Vol. 80 (2007), pp. 274-282. See page 281 reference 13.
James Maynard, Gaps between primes, arXiv preprint arXiv:1910.13450 [math.NT], 2019.
Michael Penn, a nice divisibility problem, YouTube video, 2024.
János Pintz, Very large gaps between consecutive primes, Journal of Number Theory 63 (1997), pp. 286-301.
Mario Ziller and John F. Morack, Algorithmic concepts for the computation of Jacobsthal's function, arXiv:1611.03310 [math.NT], 2016.

Cf. A048669, A048670, A072752.

(* This program is not suitable to compute more than a few terms *)
primorial[n_] := Product[Prime[k], {k, 1, n}];
j[n_] := Module[{L = 1, m = 1}, For[k = 2, k <= n+1, k++, If[GCD[k, n] == 1, If[L+m < k, m = k-L]; L = k]]; m];
a[1] = 1;
a[n_] := a[n] = j[primorial[n]] - 1;
Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 1, 10}] (* Jean-François Alcover, Sep 05 2017 *)

do(n,P,R)=for(i=1,#R, if(n==R[i], return(do(n+1,P,R)))); if(#P==0, return(n-1)); my(b=0); for(i=1,#P,my(t=do(n+1,setminus(P,[P[i]]), concat(R,Mod(n,P[i])))); if(t>b,b=t)); b
a(n)=do(1,primes(n),[]) \\ Charles R Greathouse IV, Aug 08 2024

a(n) = A048670(n) - 1. See that entry for additional information.
Iwaniec proved that a(n) << n^2*(log n)^2. - Charles R Greathouse IV, Sep 08 2012
a(n) >= (2e^gamma + o(1)) n log^2 n log log log n / (log log n)^2, see A048670. - Charles R Greathouse IV, Sep 08 2012
a(n) = 2 * A072752(n) + 1. - Mario Ziller, Dec 08 2016
See A048669 for many other bounds and references. - N. J. A. Sloane, Apr 19 2017

Laison and Schick reference from Parthasarathy Nambi, Oct 19 2007
More terms from A048670 added by Max Alekseyev, Feb 07 2008
a(46) corrected and a(50)-a(54) added by Mario Ziller, Dec 08 2016
a(55)-a(64) from A048670 added by Constantino Calancha, Aug 05 2023

A072753 Maximum gap in two-stage prime-sieves.

Original entry on oeis.org

2, 4, 10, 24, 31, 42, 60, 74, 94, 117, 148, 173, 213, 236, 275, 316, 364, 409, 436

Offset: 3

Mario Ziller, Jul 10 2002

From John F. Morack, Dec 20 2012: (Start)
There is a relationship between this sequence and Goldbach's Conjecture.
If you consider that you are "sieving" from the even number down and from 1 up, then if the even number is a power of 2 this is exactly a two-stage sieve, but you must include the 3, so each term is multiplied by 3. For other even numbers, all two-stage sieves including 3 will over-exclude numbers, so if anything is left after the sieve, it must be prime, excluding the number 1.
Consider the following: Let p be the largest prime < sqrt(2N), further let p be the m-th prime. Let's deal with the m-th term in the sequence a(m). If a(m)*3 < (N - 2) for all N > K for some constant K, there is a set of two primes that sum to 2N. (End)
From Giovanni Resta, Aug 06 2015: (Start)
a(n) is the maximal value m such that there exist n-2 pairs 0 <= a_i, b_i < prime(i) for each 3 <= i <= n, such that each number between 1 and m is either a_i or b_i mod prime(i). The condition a(k) >= m can be verified by solving an integer linear programming problem with binary variables x_i_j, where i ranges between 3 and k and j ranges between 0 and prime(k)-1, and x_i_j = 1 if a_i = j or b_i = j.
For a(11) = 148 the pairs are 5 (1,3) 7 (4,5) 11 (2,4) 13 (9,10) 17 (1,12) 19 (1,15) 23 (4,7) 29 (7,26) 31 (11,14) 37 (17,21) 41 (3,23).
For a(12) = 173 they are 5 (1,3) 7 (1,2) 11 (5,7) 13 (2,4) 17 (1,8) 19 (7,14) 23 (1,20) 29 (10,19) 31 (1,12) 37 (13,18) 41 (29,34) 43 (1,36).
For a(13) = 213 they are 5 (1,3), 7 (1,2), 11 (7,9) 13 (1,11) 17 (2,9) 19 (4,14) 23 (3,9) 29 (16,25) 31 (4,5) 37 (10,34) 41 (17,28) 43 (36,39) 47 (12,14).
For a(14) = 236 they are 5 (1,3) 7 (3,5) 11 (4,5) 13 (4,9) 17 (5,14) 19 (7,10) 23 (11,17) 29 (3,26) 31 (3,19) 37 (5,25) 41 (2,31) 43 (1,34) 47 (3,41) 53 (20,32). (End)
a(17) and a(18) were calculated using Giovanni Resta's ILP approach with models generated from all combinations of primes to 17. We are looking for a GPU approach. GLPK was used to process the auto-generated models. - John F. Morack is solely responsible for the computation work. - John F. Morack, Jan 03 2016
a(19) verification computation was completed Jan 08 2016. An interesting point about the term 355 is that it was found through trial and error using sole prime placement counts, and only after finding it as a candidate was it verified using exhaustive computation. The computation again used all combinations of the primes to 17 and the rejection of all first-occurring sole prime placements of the prime 5 using Giovanni Resta's ILP approach with all binary variables. - John F. Morack, Jan 08 2016

			a(5) = 10 because c(3)=2, d(3)=4, c(4)=1, d(4)=3, c(5)=4, d(5)=6 satisfy the requirements: 1 == 1 (mod 7), 2 == 2 (mod 5), 3 == 3 (mod 7), 4 == 4 (mod 5), 5 == 5 (mod 11), 6 == 6 (mod 11), 7 == 2 (mod 5), 8 == 1 (mod 7), 9 == 4 (mod 5), 10 == 3 (mod 7).

John F. Morack, Sequences and Mods for a(17) and a(18)
John F. Morack, Mod Sets and Sequences for Terms a(17)-a(19)
Mario Ziller, John F. Morack, Divisibility in paired progressions, Goldbach’s conjecture, and the infinitude of prime pairs, arXiv:1706.00317 [math.NT], 2017.
Mario Ziller, John F. Morack, On the computation of the generalised Jacobsthal function for paired progressions, arXiv:1706.03668 [math.NT], 2017.

Cf. A072752, A288815.

Let p(n) be the sequence of primes, i.e., p(1)=2. For n>=3 we define a(n) = max { m IN N | EXIST c(k), d(k) IN N, k=3, .., n : FOR ALL i IN {1, .., m} EXISTS j IN {3, .., n} : i == c(j) (mod p(j)) OR i == d(j) (mod p(j)) }

a(n) = (A288815(n) - 6)/6. - Mario Ziller, Jun 19 2017

a(11) from Mario Ziller, May 30 2005
a(12) from Mario Ziller, Jun 20 2013
a(13) from Mario Ziller, Sep 26 2014
a(14)-a(15) from Mario Ziller, Aug 06 2015
a(16) from Giovanni Resta, Aug 06 2015
a(17)-a(18) from John F. Morack Jan 03 2016
a(19) from John F. Morack Jan 08 2016
a(19) corrected and a(20)-a(21) added by Mario Ziller, Jun 17 2017

A234299 a(n) = |A| is the smallest order of a set A of consecutive integers which has Euler-phi(35711...Pn) members coprime to 357..*Pn, where Pn is the n-th odd prime.

Original entry on oeis.org

2, 13, 101, 1149, 15005, 255243, 4849829, 111546416, 3234846593, 100280245037, 3710369067373, 152125131763569, 6541380665834971, 307444891294245656, 16294579238595022313, 961380175077106319477, 58644190679703485491570, 3929160775540133527939470

Offset: 1

John F. Morack, Dec 22 2013

The sequence is strongly associated with A072752.
In A072752, you are looking for a maximum sized set of consecutive numbers where none are counted by Euler's phi(3*5*7*...*Pn); this sequence looks for a minimum sized set of consecutive numbers where all the numbers counted by Euler are included.
One candidate for A (not necessarily of minimum size) is the set {1, 2, 3,..., 3*5*..*Pn}, which has the requested number of coprime elements. This yields the simple upper bound a(n) <= A070826(n+1). - R. J. Mathar, May 03 2017

			a(1)=2,  phi(3) = 2, A={1,2 }, B={1,2}, |B|=2  gcd(1,3) = 1; gcd(2,3) = 1; minimum(|A|) = 2.
a(2)=13, phi(3*5) = 8, A={7,8,9,10,...,19}, B={7, 8, 11, 13, 14, 16, 17, 19}, |B|=8, A was chosen so |A| is a minimum.

Cf. A005867, A070826, A072752.

Let A = { set of any k consecutive integers}, |A| its size.
Let B = {x IN A | gcd(x, 3*5*7...Prime(n))=1}.
Condition: |B| = phi(3*5*7...Prime(n))= A005867(n+1).
a(n) = minimum(|A|) which meets the above condition.
a(n) = A070826(n+1) - A072752(n+1).

Edited by R. J. Mathar, May 03 2017

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A048670 Jacobsthal function A048669 applied to the product of the first n primes (A002110).

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A058989 Largest number of consecutive integers such that each is divisible by a prime <= the n-th prime.

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A072753 Maximum gap in two-stage prime-sieves.

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A234299 a(n) = |A| is the smallest order of a set A of consecutive integers which has Euler-phi(3*5*7*11*...*Pn) members coprime to 3*5*7*..*Pn, where Pn is the n-th odd prime.

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A234299 a(n) = |A| is the smallest order of a set A of consecutive integers which has Euler-phi(35711...Pn) members coprime to 357..*Pn, where Pn is the n-th odd prime.