A135311 A greedy sequence of prime offsets.
0, 2, 6, 8, 12, 18, 20, 26, 30, 32, 36, 42, 48, 50, 56, 62, 68, 72, 78, 86, 90, 96, 98, 102, 110, 116, 120, 128, 132, 138, 140, 146, 152, 156, 158, 162, 168, 176, 182, 186, 188, 198, 200, 210, 212, 216, 230, 240, 242, 246, 252, 260, 266, 270, 272, 278, 282
Offset: 1
Keywords
Examples
Given a(1) through a(5), a(6) can't be 14 since the set {0,2,6,8,12,14} contains elements from every residue class (mod 5). a(6) can't be 16 because {0,2,6,8,12,16} contains elements from every residue class (mod 3). a(6)=18 is possible, since the residues (mod 2) are all 0, the residues (mod 3) are all 0 or 2 and the residues (mod 5) are all 0, 1, 2, or 3.
Links
- Alessandro Languasco, Table of n, a(n) for n = 1..2089
- Thomas J. Engelsma, K-Tuple Permissible Patterns.
- Anthony D. Forbes, Prime k-tuplets.
- Kevin Ford, Florian Luca and Pieter Moree, Values of the Euler phi-function not divisible by a given odd prime, and the distribution of Euler-Kronecker constants for cyclotomic fields, arXiv preprint arXiv:1108.3805 [math.NT], 2011.
- Alessandro Languasco, Efficient computation of the Euler-Kronecker constants of prime cyclotomic fields, Research in Number Theory, 7, 2021, paper n. 2 (preliminary version, arXiv:1903.05487 [math.NT], 2019-2020).
- Alessandro Languasco, Pieter Moree, Sumaia Saad Eddin, and Alisa Sedunova, Computation of the Kummer ratio of the class number for prime cyclotomic fields, arXiv:1908.01152 [math.NT], 2019.
- Pieter Moree, Irregular behaviour of class numbers and Euler-Kronecker constants of cyclotomic fields: the log log log devil at play, arXiv:1711.07996 [math.NT], 2017. Mentions this sequence.
- Eric Weisstein's World of Mathematics, Prime Constellation
- Wikipedia, Schinzel's hypothesis H.
Programs
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Mathematica
a[1]=0;a[n_]:=a[n]=Module[{v,set,ok,p},For[v=a[n-1]+2,True,v+=2,set=Append[a/@Range[n-1],v]; For[p=3;ok=True,p<=n,p+=2,If[PrimeQ[p]&&Length[Union[Mod[set,p]]]==p,ok=False;Break[]]];If[ok,Return[v]]]]
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PARI
{greedy()=local(A, L, B, n, v , ok , R, setR, p, k); A=vector(2089); \\ 2089 is the length to get Sum_{i>=2}(1/A[i])>2; see Ford, Luca, Moree paper, p. 1454 L=length(A); B = 10^(5); \\ upper bound for the number of primes used; enough for the first 2089 terms A[1]=0; \\ first trivial term; for (n=2, L, R=vector(n); forstep (v=A[n-1]+2, B, 2 , ok=1; forprime(p = 2, v, for(k=1,n-1, R[k]=A[k]%p); R[n]=v%p; setR=Set(R); if (length(setR) > p-1, ok=0; break); \\ v is not good ); if (ok==1, A[n]=v; break); ); ); return(A) } \\ Alessandro Languasco, Aug 11 2019
Extensions
Edited by Dean Hickerson, Dec 07 2007
Comments