A343122 - OEIS

%I #42 Jun 11 2022 09:33:25

%S 1,1,2,2,2,2,2,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,

%T 6,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,

%U 30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30

%N Consider the longest arithmetic progressions of primes from among the first n primes; a(n) is the smallest constant difference of these arithmetic progressions.

%C It seems that most terms are primorials (see comments in A338869 and A338238).

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Primes_in_arithmetic_progression">Primes in arithmetic progression</a>

%e For n=2, the first two primes are 2 and 3, the only subsequence of equidistant primes. The constant difference is 1, so a(2) = 1.

%e For n=3, there are three sequences of equidistant primes: {2,3} with constant difference 1, {3,5} with difference 2, and {2,5} with difference 3, so a(3) = 1 because 1 is the smallest constant difference among the three longest sequences.

%t nmax=100; (* Last n *)

%t maxlen=11 ; (* Maximum exploratory length of sequences of equidistant primes *)

%t (* 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 a[n_,period_,seqlen_]:=Module[{tab,test},

%t (* Building sequences of equidistant numbers ending with prime(n) *)

%t tab=Table[Prime[n]-k*period,{k,0,seqlen-1}];

%t (* Checking if all elements are primes and greater than 2 *)

%t test=(And@@PrimeQ@tab)&&(And@@Map[(#>2&),tab]);

%t Return[If[test,tab,{}]]];

%t atab={}; aterms={}; (* For every n, exploring all sequences of equidistant primes among the first n primes with n > 3 *)

%t Do[

%t Do[Do[

%t If[a[n,period,seqlen]!={},AppendTo[atab,{seqlen,period}]]

%t ,{period,2,Ceiling[Prime[n]/(seqlen-1)],2}]

%t ,{seqlen,2,maxlen}];

%t (* "longmax" is the length of the longest sequences *)

%t longmax=Sort[atab,#1[[1]]>#2[[1]]&][[1]][[1]];

%t (* Selecting the elements corresponding to the longest sequences *)

%t atab=Select[atab,#[[1]]==longmax&];

%t (* Saving the pairs {n, corresponding minimum periods} *)

%t AppendTo[aterms,{n,Min[Transpose[atab][[2]]]}]

%t ,{n,4,nmax}];

%t (* Prepending the first two terms corresponding to the simple cases of first primes {2,3} and {2,3,5} *)

%t Join[{1,1},(Transpose[aterms][[2]])]

%Y Cf. A338869, A338238, A002110 (Primorials), A343118, A033188.

%K nonn

%O 2,3

%A _Andres Cicuttin_, Apr 05 2021