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

2,3

Comments

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

Links

Table of n, a(n) for n=2..100.

Wikipedia, Primes in arithmetic progression

Example

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.

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]])]

Crossrefs

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

Sequence in context: A263407 A377059 A221838 * A338869 A104588 A157279

Adjacent sequences: A343119 A343120 A343121 * A343123 A343124 A343125

Keyword

nonn

Author

Andres Cicuttin, Apr 05 2021

Status

approved