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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. 7
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Linnik proved that there are n0 and L such that a(n) < n^L for all n > n0. It has been conjectured that a(n) < n^2. The sequence A034694 has the primes p(n,1).
a(n) is also the maximum term in row n of A060940, which defines a(1). A007918(n+1) is the minimum term in row n of A060940. - Seiichi Manyama, Apr 02 2018
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
EXAMPLE
a(5) = 19 because p(5,1) = 11, p(5,2) = 7, p(5,3) = 13 and p(5,4) = 19.
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}]
PROG
(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
CROSSREFS
A038026 is a lower bound.
Cf. A034694.
Sequence in context: A179894 A060215 A309666 * A027672 A322138 A104138
KEYWORD
nonn
AUTHOR
T. D. Noe, Jun 29 2003
EXTENSIONS
a(1) defined via A060940 by Seiichi Manyama, Apr 02 2018
STATUS
approved

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Last modified April 24 22:17 EDT 2024. Contains 371964 sequences. (Running on oeis4.)