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%I #12 Jun 10 2018 04:21:09
%S 2,5,89,53,157,173,1597,15233,8803,106753,1570927,5296771
%N a(n) = the smallest prime p such that there are exactly n sets of consecutive primes, each of which has an arithmetic mean of p.
%e a(4) = 53 because there are exactly four sets of consecutive primes which have means of 53: {53}, {47,53,59}, {41,...,67} and {31,...,73},
%o (PARI) {a(n)= m=2; starting_index=1; k=starting_index; sum_of_primes=0; prime_count=0; sets=0; until( (prime(starting_index)>m) && (sets==n), if( (prime(starting_index)>m) || (sets>n), m=nextprime(m+1); sets=0; starting_index=1; k=starting_index); sum_of_primes=sum_of_primes+prime(k); prime_count++; mean=sum_of_primes/prime_count; if(mean<m, k++, sum_of_primes=0; prime_count=0; starting_index++; k=starting_index; if(mean==m, sets++))); return(m)} \\ _Rick L. Shepherd_, Jun 14 2004
%Y Cf. A050221, A050237, A060863, A082370.
%K more,nonn
%O 1,1
%A _Naohiro Nomoto_, May 11 2003
%E Edited by _Don Reble_, Jun 17 2003
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