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A085420
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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.
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7
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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
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OFFSET
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1,1
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COMMENTS
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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).
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REFERENCES
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P. Ribenboim, The New Book of Prime Number Records, Springer, 1996, pp. 277-284.
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LINKS
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EXAMPLE
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a(5) = 19 because p(5,1) = 11, p(5,2) = 7, p(5,3) = 13 and p(5,4) = 19.
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MATHEMATICA
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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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PROG
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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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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