cp's OEIS Frontend

Base-8 pandigital primes must have at least 9 octal digits, since sum(d_i 8^i) = sum(d_i) (mod 7), and 0+1+...+6+7 is divisible by 7. So the smallest ones should be of the form "10123...." in base 8, where "...." is a permutation of "4567". By chance, the identical permutation already yields a prime: a(1)="101234567" in base-8.

Terms in this sequence have at least 10 digits in base 9, i.e., are larger than 9^9, since sum(d_i 9^i) = sum(d_i) (mod 8), and 0+1+2+3+4+5+6+7+8 is divisible by 4. So there must be at least one repeated digit, which may not be even, else the resulting number is even. The smallest terms are therefore of the form "10123...." in base 9, where "...." is a permutation of "45678", cf. examples.

Terms in this sequence have at least 8 digits in base 7, i.e., are larger than 7^7, since sum(d_i 7^i) = sum(d_i) (mod 6), and 0+1+2+3+4+5+6 is divisible by 3. So there must be at least one repeated digit, which may not be 0 nor 6 neither odd (else the resulting number is even). The smallest terms are therefore of the form "1022...." in base 7, where "...." is a permutation of "3456", cf. examples.

These numbers need to have at least 13 digits in base 12 since any permutation of the digits 0,...,9,A,B will result in a number divisible by 11. For the same reason, it must be digit different from 0 which is repeated. Thus the smallest terms in this sequence are written "10123456....." in base 12, where ..... is a permutation of {7,8,9,A,B}.

Note: Due to the implementation of numtoperm(), the PARI script will not necessarily print the terms in the correct order. In some cases, more than the desired number of terms have to be calculated, and vecsort() to be used to get the correct sequence. - M. F. Hasler, May 27 2010

Base-16 (a.k.a. hexadecimal, sexadecimal, senidenary or hexadecadic) pandigital primes must have at least 17 hexadecimal digits (i.e. they are larger than 16^16 = 2^64 > 10^19), since sum(d_i 16^i) = sum(d_i) (mod 15), and 0+1+...+14+15 is divisible by 15. So the smallest ones should be of the form "101234567...." in base 16, where "...." is a permutation of "89ABCDEF".

The same reasoning shows that numbers of this form ("1012...") are congruent to 1 modulo 15 and thus modulo 30 (since also = 1 [mod 2]). This explains that all terms < 2*16^16 end in the (decimal!) digit 1.

a(n) == 1 (mod 30) for a(n) < 2^65 = 3.69*10^19.

Base-20 pandigital primes must have at least 21 base-20 digits (i.e. they are larger than 20^20 > 10^26), since sum(d_i 20^i) = sum(d_i) (mod 19), and 0+1+...+18+19 is divisible by 19. So the smallest ones should be of the form "10123456789ABCD..." in base 20, where "..." is a permutation of "EFHGIJ" (with A..J representing digits 10..19).

Terms in this sequence have at least 6 digits in base 5, i.e., are larger than 5^5, since sum(d_i 5^i) = sum(d_i) (mod 4), and 0+1+2+3+4 is divisible by 2. So the smallest ones should be of the form "10...." in base 5, where "...." is a permutation of "1234". By chance the identical permutation already yields a prime, i.e. a(1) = "101234" in base 5.

Terms in this sequence have at least 7 digits in base 6, i.e., are larger than 6^6, since sum(d_i 6^i) = sum(d_i) (mod 5), and 0+1+2+3+4+5 is divisible by 5. So the smallest ones should be of the form "101...." in base 6, where "...." is a permutation of "2345". Actually there is only one such prime, cf. examples.