A085420 For each n, let p(n,b) be the smallest prime in the arithmetic progression k*n+b, with k > 0. Then a(n) = max(p(n,b)) with 0 < b < n and gcd(b,n) = 1.
2, 3, 7, 7, 19, 11, 29, 23, 43, 19, 71, 23, 103, 53, 43, 43, 103, 53, 191, 59, 97, 79, 233, 73, 269, 103, 173, 83, 317, 79, 577, 151, 227, 193, 239, 157, 439, 191, 233, 157, 587, 107, 467, 257, 389, 307, 967, 191, 613, 269, 421, 601, 659, 199, 353, 233, 433, 317, 709
Offset: 1
Keywords
Examples
a(5) = 19 because p(5,1) = 11, p(5,2) = 7, p(5,3) = 13 and p(5,4) = 19.
References
- P. Ribenboim, The New Book of Prime Number Records, Springer, 1996, pp. 277-284.
Links
- T. D. Noe, Table of n, a(n) for n = 1..10000 (corrected by Michel Marcus, Jan 19 2019)
- A. Granville, Least primes in arithmetic progressions, Théorie des nombres / Number Theory (ed. J. M. De Koninck & C. Levesque), (de Gruyter, New York, 1989), 306-321.
- Eric Weisstein's World of Mathematics, Linnik's Theorem
Programs
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Mathematica
minP[n_, a_] := Module[{k, p}, If[GCD[n, a]>1, p=0, k=1; While[ !PrimeQ[k*n+a], k++ ]; p=k*n+a]; p]; Table[Max[Table[minP[n, i], {i, n-1}]], {n, 2, 100}]
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PARI
p(n,b)=while(!isprime(b+=n),); ba(n)=my(t=p(n,1));for(b=2,n-1,if(gcd(n,b)==1,t=max(t,p(n,b))));t \\ Charles R Greathouse IV, Sep 08 2012
Extensions
a(1) defined via A060940 by Seiichi Manyama, Apr 02 2018
Comments