A343122 Consider the longest arithmetic progressions of primes from among the first n primes; a(n) is the smallest constant difference of these arithmetic progressions.

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, 6, 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, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30

Offset: 2

Author

Andres Cicuttin, Apr 05 2021

Keywords

• nonn

Comments

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

Examples

			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.
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.
		

Links

• Wikipedia, Primes in arithmetic progression

Crossrefs

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

Programs

Mathematica

nmax=100; (* Last n *)
maxlen=11 ; (* Maximum exploratory length of sequences of equidistant primes *)
(* 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 > 3 *)
Do[
Do[Do[
If[a[n,period,seqlen]!={},AppendTo[atab,{seqlen,period}]]
,{period,2,Ceiling[Prime[n]/(seqlen-1)],2}]
,{seqlen,2,maxlen}];
(* "longmax" is the length of the longest sequences *)
longmax=Sort[atab,#1[[1]]>#2[[1]]&][[1]][[1]];
(* Selecting the elements corresponding to the longest sequences *)
atab=Select[atab,#[[1]]==longmax&];
(* Saving the pairs {n, corresponding minimum periods} *)
AppendTo[aterms,{n,Min[Transpose[atab][[2]]]}]
,{n,4,nmax}];
(* Prepending the first two terms corresponding to the simple cases of first primes {2,3} and {2,3,5} *)
Join[{1,1},(Transpose[aterms][[2]])]