A112388 a(n) is the smallest prime p such that p^n contains every digit.

10123457689, 101723, 5437, 2339, 1009, 257, 139, 173, 83, 67, 31, 29, 37, 17, 17, 47, 19, 7, 5, 23, 23, 5, 11, 11, 17, 5, 5, 5, 5, 11, 5, 11, 11, 5, 5, 7, 5, 7, 3, 5, 5, 7, 7, 7, 3, 7, 3, 3, 5, 5, 5, 5, 3, 7, 7, 5, 3, 7, 5, 3, 3, 3, 3, 3, 3, 5, 3, 2, 3, 2, 3, 3, 3, 3, 5, 3, 3, 3, 2, 3, 5, 2

Offset: 1

Author

Tanya Khovanova, Dec 05 2005

Keywords

• base
• nonn

Comments

Conjecture: a(n)=2 for all n>168. Checked up to n = 20000. - Robert Israel, Aug 28 2020

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A020666, A020665, A007377, A137214.

Programs

Maple

f:= proc(n) local k;
  k:= 1:
  do k:= nextprime(k);
    if convert(convert(k^n,base,10),set) = {$0..9} then return k fi
  od
end proc:
f(1):= 10123457689:
map(f, [$1..100]); # Robert Israel, Aug 28 2020

Mathematica

f[n_] := Block[{k = 1}, While[ Union@IntegerDigits[ Prime[k]^n] != {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, k++ ]; Prime[k]]; Array[f, 82] (* Robert G. Wilson v, Dec 06 2005 *)

Python

from sympy import nextprime
def a(n):
    if n == 1: return 10123457689
    p = 2
    while not(len(set(str(p**n))) == 10): p = nextprime(p)
    return p
print([a(n) for n in range(1, 83)]) # Michael S. Branicky, Jul 04 2021

Extensions

More terms from Robert G. Wilson v, Dec 06 2005