All terms are even, and the sequence is strictly increasing, and therefore also yields the maximal gap between n-digit primes (unless a gap containing 10^k would be larger than all gaps up to 10^(k+1), which does not happen). Therefore also a subsequence of
A005250, which is a subsequence of
A001223. -
M. F. Hasler, Dec 29 2014
For 3 < n < 19, a(n) <= 6 (n - 1)(n - 2). Conjecture: for any n > 3, a(n) <= 6 (n - 1)(n - 2). Let q = 6 (n - 1)(n - 2) and d = (10^n) - (10^(n/2) - 1)^2. Since for any even n, d is the smallest difference between two consecutive squares of the form a^2 - b^2, where a^2 = 10^n, b = a - 1, for any even n > 2, d > 5q (where 3q is, according to the conjecture, not less than the sum of the three largest gaps between 4 consecutive primes p1...p4, or 3 * a(n), and 2q is, respectively, not less than the sum of the two largest gaps (p1 - p0) + (p5 - p4), or 2 * a(n)). In the same way, we can state that for any odd n > 3, if a^2 is the smallest square such that a^2 has (n+1) digits, b = a - 1, and d = a^2 - b^2, then d > 5q. The correctness of the above conjecture would establish the well-known Brocard's and Legendre's conjectures (see the link below for both definitions), since they are proved for the first 10000 primes. -
Sergey Pavlov, Jan 30 2017
