
COMMENTS

In A075053(m), the primes obtained as permutations of digits of m are counted several times if they can be obtained in several different ways. See sequence A076730 which uses A039993 instead, i.e., counting only different primes.  M. F. Hasler, Mar 11 2014
The original data given for n = 3, 4, 5 was erroneously A007526(n).  Up to n = 6, a(n) = A076730(n), but the two will differ not later than for n = 10, where A134596(10) = 1123456789 gives a(10) >= 398100 = A075053(1123456789) > A039993(1123456789) = 362451 = A076730(10). The difference arises because each prime containing a single '1' will be counted twice by A075053, but only once by A039993.  M. F. Hasler, Oct 14 2019


EXAMPLE

From M. F. Hasler, Oct 14 2019: (Start)
a(2) = 4 = A075053(37), because from 37 one can obtain the primes {3, 7, 37, 73}, and there is obviously no 2digit number which could give more primes.
a(3) = 11 = A075053(137), because from 137 one can obtain the primes {3, 7, 13, 17, 31, 37, 71, 73, 137, 173, 317}, and no 3digit number yields more.
a(4) = 31 = A075053(1379), because from 1379 one can obtain the 31 primes {3, 7, 13, 17, 19, 31, 37, 71, 73, 79, 97, 137, 139, 173, 179, 193, 197, 317, 379, 397, 719, 739, 937, 971, 1973, 3719, 3917, 7193, 9137, 9173, 9371}, and no 4digit number yields more.
a(5) = 106 = A075053(13679). a(6) = 402 = A075053(123479).
a(7) = 1953 = A075053(1234679). (End)
