A343118 Length of the longest sequence of equidistant primes among the first n primes.
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, 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, 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
Offset: 2
Keywords
Examples
For the first 2 primes {2,3}, the sequence is itself a list of two equidistant primes, so a(2) = 2.
For the first 3 primes {2,3,5}, there is at most two equidistant primes, so a(3) = 2.
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.
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.
Links
- Ben Green and Terence Tao, The primes contain arbitrarily long arithmetic progressions, Annals of Mathematics, Vol 167 (2008), pp. 481-547.
- Wikipedia, Primes in arithmetic progression
- Wikipedia, Green-Tao theorem
Programs
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Mathematica
nmax = 128; (* Last n *) maxlen = 11 ; (* Maximum exploratory length of sequences of equidistant primes. "maxlen" must be larger than the maximum term obtained with "nmax" *) (* 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 > 2 *) Do[ Do[Do[ If[a[n, period, seqlen] != {}, AppendTo[atab, seqlen]] , {period, 2, Ceiling[Prime[n]/(seqlen - 1)], 2}] , {seqlen, 2, maxlen}]; (* Saving the pairs {n, corresponding maximum lengths} *) AppendTo[aterms, {n, Max[atab]}] , {n, 3, nmax}]; (* Prepending the first term corresponding to the trivial case of first two primes {2,3} *) Join[{2}, (Transpose[aterms][[2]])]
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