
COMMENTS

The same prime may arise in more than one way, and if so all occurrences are counted.
For n = 34, a(34) = 420: Divisors(420) = {1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210, 420}. We obtain 34 prime numbers 11, 13, 17, 23, 31, 37, 41, 43, 47, 53, 61, 67, 71, 73, 101, 103, 107, 127, 151, 157, 211, 281, 283, 307, 353, 421, 421, 521, 601, 607, 701, 1021, 1051, 4201, as follows: 1_1, 1_3, 1_7, 2_3, 3_1, 3_7, 4_1, 4_3, 4_7, 5_3, 6_1, 6_7, 7_1, 7_3, 10_1, 10_3, 10_7, 12_7_ 15_1, 15_7, 21_1, 28_1, 28_3, 30_7, 35_3, 4_21, 42_1, 5_21, 60_1, 60_7, 70_1, 10_21, 105_1, 420_1. The prime number 421 is counted twice since it is obtained as 4_21 and 42_1.
For each n there may be several numbers m for which the divisors create n prime numbers, we take a(n) to be smallest such m.
For n = 1 we can take the prime numbers m = p = 9*s 1 (see A061242). By concatenating the divisors we obtain the numbers 1_1, 1_p, p_1, p_p. It is easy to see that only 1_1 = 11 is prime. Thus there are infinitely many numbers m for which the concatenation produces a single prime number.
For n = 2 get the numbers m = 4, 8, 10, 16, 19, 22, 25, 32, 37, 38, 43, 58, 67, 68, 73, 74, 79, .... . This includes the primes 19, 43, 67, 73, 79, ... of the form p = 6s + 1. For these only one of the numbers 1_p or p_1 is prime.
For n = 3 we get the numbers m = 3, 7, 13, 14, 20, 26, 31, 35, 44, 46, 50, 62, 64, 65, 77, 82, 97, 103, 109, .... The prime numbers 3, 7, 13, 31, 97, 103, 109, ... have the property that 1_p and p_1 are simultaneously prime.
Conjecture 1: a(n) always exists.
Conjecture 2: For any n there are infinitely many choices for m.


EXAMPLE

For k = 1, Divisors(1) = {1} and 1_1 = 11, so a(1) = 1.
For k = 4, Divisors(4) = {1, 2, 4} and 1_1 = 11, 4_1 = 41, so a(2) = 4.
For k = 3, Divisors(3) = {1, 3} and 1_1 = 11, 1_3 = 13 , 3_1 = 31,so a(3) = 3.
For k = 9, Divisors(9) = {1, 3, 9} and 1_1 = 11, 1_3 = 13 , 1_9 = 19, 3_1 = 31,so a(4) = 9.
For k = 27, Divisors(27) = {1, 3, 9, 27} and 1_1 = 11, 1_3 = 13 , 1_9 = 19, 1_27 = 127, 3_1 = 31, 27_1 = 271 so a(6) = 27.
