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A194943
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Greatest d such that d*n+b is the least prime in the arithmetic progression k*n+b for some 0 < b < n with gcd(b, n) = 1.
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3
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1, 2, 1, 3, 1, 4, 2, 4, 1, 6, 1, 7, 3, 2, 2, 6, 2, 10, 2, 4, 3, 10, 3, 10, 3, 6, 2, 10, 2, 18, 4, 6, 5, 6, 4, 11, 5, 5, 3, 14, 2, 10, 5, 8, 6, 20, 3, 12, 5, 8, 11, 12, 3, 6, 4, 7, 5, 12, 2, 24, 9, 6, 5, 6, 3, 15, 5, 8, 3, 18, 4, 24, 8, 8, 6, 10
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OFFSET
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2,2
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COMMENTS
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a(n) exists due to Linnik's theorem; thus a(n) < c * n^4.2 for some constant c.
Heath-Brown's conjecture on Linnik's theorem implies that a(n) < n.
On the GRH, a(n) << phi(n) * log(n)^2 * phi(n)/n.
Pomerance shows that a(n) > (e^gamma + o(1)) log(n) * phi(n)/n, and Granville & Pomerance conjecture that a(n) >> log(n)^2 * phi(n)/n.
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LINKS
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FORMULA
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MATHEMATICA
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p[b_, d_] := Module[{k = b+d}, While[ !PrimeQ[k], k += d]; (k-b)/d]; a[n_] := Module[{r = p[1, n]}, For[b = 2, b <= n-1, b++, If[GCD[b, n] > 1, Null, r = Max[r, p[b, n]]]]; r]; Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Oct 02 2013, translated from Pari *)
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PROG
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(PARI) p(b, d)=my(k=d+b); while(!isprime(k), k+=d); (k-b)/d
a(n)=my(r=p(1, n)); for(b=2, n-1, if(gcd(b, n)>1, next); r=max(r, p(b, n))); r
(Python)
from math import gcd
from gmpy2 import is_prime
def p(b, d):
k = d + b
while not is_prime(k):
k += d
return (k-b)//d
return max(p(b, n) for b in range(1, n) if gcd(b, n) == 1)
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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STATUS
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approved
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