Smallest number m such that for 0 < k < n+1, np(m+k-1) = np(m)-k+1, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).

A246792

Smallest number m such that for 0 < k < n+1, np(m+k-1) = np(m)-k+1, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).