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Approximation of A020497(n) by function of the form prime(n+k) - c, where k,c are constants. The choice k=4 is stipulated by removing 2 and 5 from the considered prime constellations and 11 is chosen as the first prime (for n=1), since the second constellation {11,13} should be a twin pair. The choice c=10 corresponds to a(0) = 0.

a(n) = 0 for n=1,2,3,4,5,7,8,9,10,11,12,13,14,15,16,22,36,37,43,...

Version 2 of the "next prime" function is "smallest prime > n". This produces A151800.

Maple uses version 2.

According to the "k-tuple" conjecture, a(n) is the initial term of the lexicographically earliest increasing arithmetic progression of n primes; the corresponding common differences are given by A061558. - David W. Wilson, Sep 22 2007

It is easy to show that the initial term of an increasing arithmetic progression of n primes cannot be smaller than a(n). - N. J. A. Sloane, Oct 18 2007

Also, smallest prime bounded by n and 2n inclusively (in accordance with Bertrand's theorem). Smallest prime >n is a(n+1) and is equivalent to smallest prime between n and 2n exclusively. - Lekraj Beedassy, Jan 01 2007

Run lengths of successive equal terms are given by A125266. - Felix Fröhlich, May 29 2022

Conjecture: if n > 1, then a(n) < n^(n^(1/n)). - Thomas Ordowski, Feb 23 2023

Tony Forbes defines a prime k-tuplet (distinguished from a prime k-tuple) to be a maximally possible dense cluster of primes (a prime constellation) which will necessarily involve consecutive primes whereas a prime k-tuple is a prime cluster which may not necessarily be of maximum possible density (in which case the primes are not necessarily consecutive.)

a(n) >> n log log n; in particular, for any eps > 0, there is an N such that a(n) > (e^gamma - eps) n log log n for all n > N. Probably N can be chosen as 1; the actual rate of growth is larger. Can a larger growth rate be established? Perhaps a(n) ~ n log n. - Charles R Greathouse IV, Apr 19 2012

Conjecture: (i) The sequence a(n)^(1/n) (n=3,4,...) is strictly decreasing (to the limit 1). (ii) We have 0 < a(n)/n - H_n < (gamma + 2)/(log n) for all n > 4, where H_n denotes the harmonic number 1+1/2+1/3+...+1/n, and gamma refers to the Euler constant 0.5772... [The second inequality has been verified for n = 5, 6, ..., 5000.] - Zhi-Wei Sun, Jun 28 2013.

Conjecture: For any integer n > 2, there is 1 < k < n such that 2*n - a(k)- 1 and 2*n - a(k) + 1 are twin primes. Also, every n = 3, 4, ... can be written as p + a(k)/2 with p a prime and k an integer greater than one. - Zhi-Wei Sun, Jun 29-30 2013.

The number of configurations that realize this minimal diameter, is A083409(n). - Jeppe Stig Nielsen, Jul 26 2018

Engelsma points out that the values he lists past a(342)=2328 are only the "best known" values, and are not confirmed as minimal. - Brian Kehrig, Mar 21 2025

It is known that a(25)=0. Terms for n = 22 and 23 are unknown. Glaisher tabulates the number of centuries having 0, 1, 2, ... primes for numbers up to 9000000. Glaisher's 1883 book is still in print!

a(24) does not exist because the only century having 24 primes is 0 to 99 -- the same century having 25 primes. From A020497, we see that a range of 101 numbers is required to find 24 primes. Dickson's conjecture implies that a(n) exists for n=18..23. - Charles R Greathouse IV, Feb 24 2011

To see that Dickson's conjecture is applicable to the preceding statement, the appropriate general sequence to consult is A364678, which affirms that 23 primes are permissible between adjacent multiples of 100, as opposed to in an arbitrary interval of 99 integers. - Peter Munn, Sep 04 2023

a(n) for n = 18..23 is greater than 10^10. Ribenboim discusses Dickson's conjecture in two books. - T. D. Noe, Feb 24 2011

a(19) <= 1108851311300675700427. - Donovan Johnson, Feb 28 2011

a(20) <= 394338677302163715754576644. - Tim Johannes Ohrtmann, Aug 27 2015

The (wrong) old name was: Largest number of pairwise coprime numbers that can occur in an interval of length n. - Wolfdieter Lang, Oct 10 2017

Conjecturally, a(n) is the largest number of primes that occurs on an infinite number of intervals of n consecutive integers. The conjecture is apparently due to Dickson; Hardy & Littlewood's Conjecture B concerns only pairs (p, p + 2n).

According to the link at www.opertech.com, a(3159) >= 447 > 446 = pi(3159). The k-tuples conjecture then implies that, for an infinitude of n, the interval [n+1, n+3159] includes 447 primes. For these n, pi(n+3159) >= pi(n)+447 > pi(n)+446 = pi(n)+pi(3159), contradicting the conjecture that pi(x+y) <= pi(x)+pi(y). - David W. Wilson, May 23 2005

From Wolfdieter Lang, Oct 10 2017: (Start)

The following admissibility definition is adapted from the Hensley and Richards [H-R] or Richards [R] links. A k-tuple B_k = [b_1, b_2, ..., b_k] of integers with 0 <= b_1 < b_2 < ... < b_k is admissible if, for each prime p, there exists at least one congruence class modulo p which contains none of the B_k members. Because complete residue systems modulo p are equivalent under translation one can consider the length n interval [0, 1, ..., n-1] and admissible k-tuples starting with 0. The prime p = 2 allows then only even tuple numbers from I_n = [0, 2, ..., floor((n-1)/2)]. Only primes p <= k have to be tested.

a(n) is then the maximal k for which there is at least one such admissible B_k tuple from the interval I_n. This function a(n) is called rho^* in (H-R) and (R). It has been given as rhobar in the Schinzel - Sierpiński link, Théorème 1, p. 201.

Note that there are also admissible k-tuples from members of [0, 1, ..., n-1] which do not start with 0. Such tuples are translations of the ones starting with 0. E.g., [1, 3] is an admissible 2-tuple for any [0, 1,..., n-1] interval with n >= 3, but it is a translation of the considered [0, 2] tuple.

For the multiplicities of k see A047947(k), for k >= 1.

For the smallest k such that a(k) = n see A020497(n), for k >= 1.

For the number of all admissible k-tuples from the interval I_n starting with 0 see the array A292224(n, k), with k = 1..a(n), which has been given in the Engelsma link, Table 2, p. 27.

One of the Hardy-Littlewood conjectures (the prime tuple conjecture, see also conjecture (B) given by [H-R] and [R], and Ribenboim, hypothesis (D_1), p. 373, from the Dickson conjecture) is that there are infinitely many primes with gaps defined by any admissible B_k tuple, that is, all p, p + b_2, ..., p + b_k are prime for infinitely many primes p, for k >= 2. For k = 1 this is well known.

(End)

For a proper definition see the cross-references.

Given a(i) for 1 <= i < n, a(n) is the smallest number > a(n-1) such that, for every prime p, the set {a(i) mod p : 1<=i<=n} has at most p-1 elements. Assuming Schinzel's hypothesis H, an equivalent statement is that a(n) is minimal such that there are infinitely many primes p with p+a(i) prime for 1 <= i <= n.

For every n, a(n) is not congruent to 1 (mod 2), nor to 1 (mod 3), nor to 4 (mod 5), nor to 3 (mod 7), ...

Note that this sequence does not always give the minimal difference between the first and last of n consecutive large primes, A008407. E.g., a(6)=18 but the 6 consecutive primes 97, 101, 103, 107, 109, 113 give the minimal difference of 16.

Conjecturally, a(n) is the smallest number such that n primes occur infinitely often among (x+a(1), ...,x+a(n)).

From M. F. Hasler, Nov 25 2024: (Start)

For a given prime p, if r is the only residue (mod p) not among {a(1), ..., a(n)} (mod p) for some index n, then no term of the sequence can be congruent to r (mod p).

(Instead of a(1...n), one can consider any collection of terms.) - Examples:

(1) p = 2, r = 0, n = 1: No term can be congruent to 0 (mod 2), i.e., even.

(2) p = 3, r = 2, n = 2: No term may be congruent to 2 (mod 3).

(3) p = 5, r = 0, n = 4: No term may be a multiple of 5.

(4) p = 7, r = 4, n = 6: No term may be congruent to 4 (mod 7).

(5) p = 11, r = 6, n = 11: No term may be congruent to 6 (mod 11). (End)

The first pattern (0,2) is for twin primes (p,p+2). Row n contains A083409(n) patterns, each one consisting of 0 followed by n-1 terms. In each row the patterns are in lexicographic order.

These numbers (in a slightly different order) appear in Table 1 of the paper by Tony Forbes. Sequence A186702 gives the least prime starting a given pattern.