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

%I A343118 #35 May 21 2021 19:19:12
%S A343118 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,
%T A343118 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,
%U A343118 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
%N A343118 Length of the longest sequence of equidistant primes among the first n primes.
%C A343118 This sequence is unbounded as stated by the Green-Tao theorem.
%H A343118 Ben Green and Terence Tao, <a href="https://doi.org/10.4007/annals.2008.167.481">The primes contain arbitrarily long arithmetic progressions</a>, Annals of Mathematics, Vol 167 (2008), pp. 481-547.
%H A343118 Wikipedia, <a href="https://en.wikipedia.org/wiki/Primes_in_arithmetic_progression">Primes in arithmetic progression</a>
%H A343118 Wikipedia, <a href="https://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem">Green-Tao theorem</a>
%F A343118 a(A000720(A005115(n))) = n. - _Rémy Sigrist_, Apr 15 2021
%e A343118 For the first 2 primes {2,3}, the sequence is itself a list of two equidistant primes, so a(2) = 2.
%e A343118 For the first 3 primes {2,3,5}, there is at most two equidistant primes, so a(3) = 2.
%e A343118 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.
%e A343118 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.
%t A343118 nmax = 128; (* Last n *)
%t A343118 maxlen = 11 ; (* Maximum exploratory length of sequences of equidistant primes. "maxlen" must be larger than the maximum term obtained with "nmax" *)
%t A343118 (* 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 {} *)
%t A343118 a[n_, period_, seqlen_] := Module[{tab, test},
%t A343118 (* Building sequences of equidistant numbers ending with prime(n) *)
%t A343118   tab = Table[Prime[n] - k*period, {k, 0, seqlen - 1}];
%t A343118 (* Checking if all elements are primes and greater than 2 *)
%t A343118   test = (And @@ PrimeQ@tab) && (And @@ Map[(# > 2 &), tab]);
%t A343118   Return[If[test, tab, {}]]];
%t A343118 atab = {}; aterms = {};
%t A343118 (* For every n, exploring all sequences of equidistant primes among the first n primes with n > 2 *)
%t A343118 Do[
%t A343118   Do[Do[
%t A343118     If[a[n, period, seqlen] != {}, AppendTo[atab, seqlen]]
%t A343118     , {period, 2, Ceiling[Prime[n]/(seqlen - 1)], 2}]
%t A343118    , {seqlen, 2, maxlen}];
%t A343118 (* Saving the pairs {n, corresponding maximum lengths} *)
%t A343118   AppendTo[aterms, {n, Max[atab]}]
%t A343118   , {n, 3, nmax}];
%t A343118 (* Prepending the first term corresponding to the trivial case of first two primes {2,3} *)
%t A343118 Join[{2}, (Transpose[aterms][[2]])]
%Y A343118 Cf. A000720, A005115, A338869, A343122.
%K A343118 nonn
%O A343118 2,1
%A A343118 _Andres Cicuttin_, Apr 05 2021