
COMMENTS

The sequence of k values is 2, 4, 8, 10, 14, 20, 28, 32, 34, 40, 50, etc. a(n) is necessarily congruent to 10 modulo 30 starting with n=4 (and is coincidentally so for n=2 and 3). Each successive unknown term's existence is only conjectural, but is supported by standard heuristics. The values in the lead sentence come through considerations modulo the smallest primes.
Concatenation of 22 numbers is ruled out by consideration modulo 11 in concert with the prior need for the stem to be 10 mod 30: A prime concatenation of 22 values necessarily involves passage from one length number to another, with a power of 10 not leading. Beginning with a(11), however, with each time a multiple of 22 for the number of concatenated values is passed there is a measure of uncertainty. Theoretically, it seems a(11) could include for largest prime a concatenation of 44 (rather than 50) values, but with these primes very sparse it is a near certainty heuristically that this is not so. Mathematically, a conjectured a(11) would have a higher bar for strict proof than a(n) for n < 11, and the same holds repeatedly for even less accessible terms.
The alternative sequence where only the rapidity of arrival of the nth prime determines a(n) (k minimal for the largest prime, with no constraint on k for the smaller prime concatenations) necessarily shares its first 5 terms in common with this one. It also shares its 6th by virtue of the fact that this sequence's a(6) is the only value less than 10^12 producing its 6th prime with the attachment of the 20th value, whether alternative length possibilities for primes are allowed or not (i.e., the first cases giving 5 other smaller primes  in addition to one of 20 concatenated values  where there is a prime concatenation of 16 values, in place of one of either 2 or 8 values, are both at least this large). However, it does necessarily differ at a(7) and a(8) (but then not necessarily at a(9)), as the resolution of the theoretical problem for the twin sequence is given for a(7) by the possibility of 5, 7, 11, 13, 17, 23 and 25 numbers being concatenated to give primes, and for a(8) by the replacement of 2 concatenated values with concatenations of both 16 and 26 of them (with result that a(8) for this alternative sequence appears already with concatenation of 28 values, while here that corresponds to a(7)).
This necessity to distinguish between whether or not only the last of the primes comes as quickly as possible does not arise as an issue in the situation where the number itself is required to be prime (A172257). [Comments reedited from Feb 2014]  James G. Merickel, Aug 07 2015


EXAMPLE

43 is prime while 32 and 21 are not, so a(1)=4; 109 and 10987 are both prime, and like concatenations for values 4 through 9 do not produce 2 primes, so a(2)=10; 1000999, 1000999998997 and 1000999998997996995994993 are all prime and no smaller value produces 3 primes so quickly, so a(3)=1000.
