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 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 2,3 COMMENTS It seems that most terms are primorials (see comments in A338869 and A338238). LINKS 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: A112968 A263407 A221838 * A338869 A104588 A157279 Adjacent sequences: A343119 A343120 A343121 * A343123 A343124 A343125 KEYWORD nonn AUTHOR Andres Cicuttin, Apr 05 2021 STATUS approved

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Last modified December 7 01:51 EST 2022. Contains 358649 sequences. (Running on oeis4.)