cp's OEIS Frontend

The old name was: "Schinzel's rhobar(n), number of distinct lengths of a block of consecutive integers on which a maximum of n primes occurs infinitely often (under the k-tuple conjecture)." [Note that "rhobar" is A023193.]

Soundararajan states that, on average, there is one prime in the interval [k,k+log(k)] for any number k. Is there an upper limit to the number of primes in such an interval? Not if the prime k-tuple conjecture is true, in which case a(n) exists for all n. Note that a(n) > e^A008407(n). See A120935 for the largest prime in the interval.

a(n) begins a sequence of n primes whose prime pattern is one of the patterns in the n-th row of A186634. For example, the sequence of four consecutive primes beginning with 3251 is (3251, 3253, 3257, 3259), which has pattern (0, 2, 6, 8), which is in the 4th row of A186634.

Soundararajan states that, on average, there is one prime in the interval [n,n+log(n)] for any number n. See A120934 for the prime n that yield new records.

The row lengths are given by A023193 (the rhobar function of Schinzel and Sierpiński called rho* by Hensley and Richards).

This irregular triangle has already been considered by Engelsma, see Table 2, for n=1..56, p. 27.

A k-tuple of integers B_k = [b_1, ..., b_k] with 0 = b_1 < b_2 < ... < b_k <= n-1 is called admissible if for each prime p there exists at least one congruence class modulo p which contains none of the B_k elements. (This corresponds to the alternative definition of Hensley and Richards, p. 378 (*) or Richards, p. 423, 1.5 Definition and (*).) Note that the definition of "admissibility" is translation invariant (see the Note by Richards, p. 424, which is obvious from the translation equivalence of complete residue systems modulo p). Therefore the interval I_n = [0, n-1] of length n has been chosen. The b_1 = 0 choice is conventional. Without this choice other admissible k-tuples are obtained by translation as long as b_k + a < n-1. E.g., for n = 8 and k = 3 the tuple [1, 5, 7] is admissible and a translation of the considered tuple [0, 4, 6].

Only primes p <= k have to be tested to decide on the admissibility of a B_k tuple because for larger k there is always some residue class which contains none of the k members of B_k.

Because p = 2 already forbids even and odd numbers to appear in B_k for k >= 2, one can for the admissibility test eliminate all odd numbers in the chosen I_n. Therefore, only Ieven_n:= [0, 2, ..., 2*floor((n-1)/2)] =: 2*[0, 1, ..., floor((n-1)/2)] need be considered. B_1 = [0] is admissible for all n >= 1.

Because only the interval Ieven_n is of relevance, there will occur repetitions for admissible tuples for n if n = 2*k+1 and n = 2*k+2.

With the set B_k(p) = B_k (mod p) := {0, b_1 (mod p), ..., b_k (mod p)} the criterion for admissibility can be written as p - #(B_k(p)) > 0, for all primes 3 <= p <= k (because there are p congruence classes defined by smallest nonnegative complete residue system [0, 1, ..., p-1]).

Admissible tuples (starting with 0) with least b_k - b_1 = b_k value give rise to prime k-constellations of diameter b_k. E.g., for k = 2 the admissible tuple [0, 4] does not lead to a prime 2-constellation for n >= 5; [0, 6] is out for n >= 7, ... . But there are two prime 3-constellations given by [0, 2, 6] and [0, 4, 6] for n >= 7.

Row sums are in A292225, that is, total number of admissible tuples starting with 0 from the interval I_n = [0, n-1].

The ">n+1" rules out n-tuples like (2,2), which only occurs for the primes 3, 5, 7. All terms from a(19) on equal 0.

An n-tuple (a_1,a_2,...,a_n) is counted iff the partial sums 0, a_1, a_1+a_2, ..., a_1+...+a_n do not contain a complete residue system (mod p) for any prime p.

Similar to A023192. (Here we ignore the empty pattern and start at 0.) These are called "admissible constellations" of primes. - Don Reble, Jun 12 2004

For n>1, it appears that each prime is a cluster prime. See A038134.

All terms are conjectural. A prime n-tuplet is defined as the densest permissible prime constellation containing n primes. The term a(2) corresponds to twin primes, a(3) to prime triplets, a(4) to prime quadruplets, etc. Extensive computational evidence suggests that these terms are valid. However, there is no proof that the greatest integer M exists - not even for a subset of values of n. If one could find a constructive existence proof, then Twin Prime Conjecture as well as Legendre's Conjecture would require just a trivial additional step. - Edited by Hugo Pfoertner, Sep 15 2021

Note that, for some n, a prime (n+1)-tuple must include a prime n-tuple; e.g., prime quadruplets include prime triples. Thus, if any term is -1, subsequent terms may be -1, too. - Franklin T. Adams-Watters and Alexei Kourbatov, Jul 14 2011

However, for other n, a prime (n+1)-tuple does NOT include a prime n-tuple; e.g. 7-tuples {p, p + 2, p + 6, p + 8, p + 12, p + 18, p + 20} do not contain 6-tuples {p-4, p, p + 2, p + 6, p + 8, p + 12}; see List of all possible patterns of prime k-tuplets by Tony Forbes.

Assuming the Hardy-Littlewood k-tuple conjecture, the average distance between k-tuples grows slower than the distance between consecutive squares. This is an indication (but not a proof) that the maximum integer M in this sequence does exist for all n.

a(6) > 3005845357, because there is a gap of size 7191214380 between consecutive sextuplets, enclosing in its interior the interval between the two squares 3005845357^2 and 3005845358^2, found with a fast sieving program provided by Martin Raab, i.e., 9035106309825245467 < 3005845357^2 = 9035106310198457449 < 3005845358^2 = 9035106316210148164 < 9035106317016459847. Statistical considerations based on the observed scatter of maximum gap sizes (see A200503 and data provided at N. Luhn link) suggest a range of 3*10^9 < a(6) < 6*10^9. In a private communication Martin Raab provided the following estimates for the order of magnitude of the next terms: a(7) ~= 1.1*10^11, a(8) ~= 1.6*10^12, a(9) ~= 6*10^13, a(10) ~= 3*10^16. - Hugo Pfoertner, Jul 31 2023, corrected Oct 23 2023

Each constellation is encoded by means of dividing each of the increments to p in the k-tuple by two, raising two to the power of each and then summing the result. E.g.:

(p,p+2,p+6) -> p+(0,2,6) => (0,1,3) -> 2^0 + 2^1 + 2^3 = 11.

Each encoding is unique and so can be reversed e.g.:

89 = 2^0 + 2^3 + 2^4 + 2^6 -> (0,3,4,6) => (p,p+6,p+8,p+12).

Those constellations that represent all moduli for all their matching primes p are not counted; for example, encoding #7, which implies (p,p+2,p+4) only matches the prime triple (3,5,7) which is (0,2,1) mod 3, and so is not a valid constellation, and thus 7 is not in the list. Encoding #155 is the first that fails modulo 5, and is also not in the list.

a(25) = 0, and a(24) cannot exist. The same is true with a(k) and k>25. From A020497, we see that a range of 101 numbers is required to find 24 primes. It is an open question if a(23) exists.

From Robert Israel, Jan 18 2017: (Start)

Dickson's conjecture implies that a(23) does exist.

Let Q = 27926129625869590, and R = 614889782588491410 the product of all primes < 50.

Then for any k, the 23 numbers Q+i+k*R for i = 1, 3, 7, 9, 13, 19, 21, 27, 31, 33, 37, 43, 49, 51, 57, 63, 69, 73, 79, 87, 91, 97, 99 have no prime divisors < 50.

Dickson's conjecture would indicate that there are infinitely many k for which these numbers are all prime, and thus there are 23 primes between Q+k*R and Q+k*R+100. (End)

Heuristics suggest a(23) exists (see above) and has between 20 and 30 digits. There are 192 residue classes mod 23# = 223092870 in which a(23) might fall, all of which are 11 mod 30 and either 3 or 4 mod 7. - Charles R Greathouse IV, Jul 12 2017