A343118 Length of the longest sequence of equidistant primes among the first n primes.

2, 2, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 2

Author

Andres Cicuttin, Apr 05 2021

Keywords

• nonn

Comments

This sequence is unbounded as stated by the Green-Tao theorem.

Examples

			For the first 2 primes {2,3}, the sequence is itself a list of two equidistant primes, so a(2) = 2.
For the first 3 primes {2,3,5}, there is at most two equidistant primes, so a(3) = 2.
For the first 4 primes {2,3,5,7}, the subsequence {3,5,7} is the longest subsequence with 3 equidistant primes, so a(4) = 3.
For the first 10 primes {2,3,5,7,11,13,17,19,23,29}, the subsequence {5,11,17,23,29} is the longest subsequence with 5 equidistant primes, so a(10) = 5.
		

Links

• Ben Green and Terence Tao, The primes contain arbitrarily long arithmetic progressions, Annals of Mathematics, Vol 167 (2008), pp. 481-547.
• Wikipedia, Primes in arithmetic progression
• Wikipedia, Green-Tao theorem

Crossrefs

Cf. A000720, A005115, A338869, A343122.

Programs

Mathematica

nmax = 128; (* Last n *)
maxlen = 11 ; (* Maximum exploratory length of sequences of equidistant primes. "maxlen" must be larger than the maximum term obtained with "nmax" *)
(* a[n,p,s] returns the sequence of "s" equidistant primes with period "p" and last prime prime(n) if it exists, otherwise it returns {} *)
a[n_, period_, seqlen_] := Module[{tab, test},
(* Building sequences of equidistant numbers ending with prime(n) *)
  tab = Table[Prime[n] - k*period, {k, 0, seqlen - 1}];
(* Checking if all elements are primes and greater than 2 *)
  test = (And @@ PrimeQ@tab) && (And @@ Map[(# > 2 &), tab]);
  Return[If[test, tab, {}]]];
atab = {}; aterms = {};
(* For every n, exploring all sequences of equidistant primes among the first n primes with n > 2 *)
Do[
  Do[Do[
    If[a[n, period, seqlen] != {}, AppendTo[atab, seqlen]]
    , {period, 2, Ceiling[Prime[n]/(seqlen - 1)], 2}]
   , {seqlen, 2, maxlen}];
(* Saving the pairs {n, corresponding maximum lengths} *)
  AppendTo[aterms, {n, Max[atab]}]
  , {n, 3, nmax}];
(* Prepending the first term corresponding to the trivial case of first two primes {2,3} *)
Join[{2}, (Transpose[aterms][[2]])]

Formula

a(A000720(A005115(n))) = n. - Rémy Sigrist, Apr 15 2021