%I
%S 2,3,7,5,19,7,29,17,19,19,43,13,103,29,31,41,103,19,191,41,67,43,137,
%T 73,149,103,109,83,317,31,311,97,181,103,191,71,439,191,233,89,379,67,
%U 463,113,181,137,967,97,613,149,197,181,607,109,331,233
%N Last position reached by winner of nth Littlewood Frog Race.
%C Related to Linnik's theorem; main sequence is A085420. [From _Charles R Greathouse IV_, Apr 16 2010]
%C a(n) is the smallest prime such that some subset of primes <= a(n) is a reduced residue system modulo n.  _Vladimir Shevelev_, Feb 19 2013
%H Charles R Greathouse IV, <a href="/A038026/b038026.txt">Table of n, a(n) for n = 1..10000</a>
%F 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.  _Charles R Greathouse IV_, Sep 08 2012
%e a(6) = 7 since the primes less than or equal to 7, {2, 3, 5, 7}, reduced modulo 6 are {2, 3, 5, 1}. This contains the reduced residue system modulo 6, which is {1, 5}, and 7 is clearly the smallest such prime.  _Vladimir Shevelev_, Feb 19 2013
%o (PARI) a(n)={
%o my(todo=(1<<n)1,r=2,q=2);
%o if(n==1, return(2));
%o for(a=0,n1,
%o if(gcd(a,n)>1,todo=bitnegimply(todo,1<<a))
%o );
%o todo=bitnegimply(todo,1<<2);
%o forprime(p=3,default(primelimit),
%o r+=pq;
%o r=r%n;
%o todo=bitnegimply(todo,1<<r);
%o if(!todo, return(p));
%o q=p;
%o );
%o error("Not enough precomputed primes")
%o }; \\ _Charles R Greathouse IV_, Feb 14 2011
%o (PARI) p(n,b)=while(!isprime(b), b+= n); b
%o a(n)=my(t=p(n,1));for(b=2,n1,if(gcd(n,b)==1,t=max(t,p(n,b))));t \\ _Charles R Greathouse IV_, Sep 08 2012
%Y This sequence is a lower bound for the related sequence A085420.
%Y Cf. A038025.
%K nonn
%O 1,1
%A _Christian G. Bower_
