A292163 a(n) is the least prime p such that the orderly concatenation of the n successive powers of p yields a prime number; a(n)=0 if n is a multiple of 6.

3, 3, 337, 23, 0, 373, 37, 839, 421, 7, 0, 1447, 2113, 29, 43, 17, 0, 1789, 523, 84737, 7669, 397, 0, 3851, 3583, 99149, 146023, 157, 0, 14173, 38329, 1229, 8017, 1021, 0, 18979, 5437, 17207, 6571, 47, 0, 347, 43669, 25847, 257353, 2887, 0, 33889, 71287

Offset: 2

Author

Michel Marcus, Sep 10 2017

Keywords

• nonn
• base

Comments

See in the Prime Puzzle link the discussion for when n is a multiple of 6.

Examples

			For n=2, the concatenation of 3^0 and 3^1 is 13 which is prime (while 12 was not prime); so a(2) = 3.
For n=3, the concatenation of 3^0, 3^1 and 3^2 is 139 which is prime (while 124 was not prime); so a(3) = 3.
		

Links

• Carlos Rivera, Puzzle 892. Primes as a concatenation of a series of powers of a prime, The Prime Puzzles and Problems Connection.

Crossrefs

Cf. A047253.

Programs

Maple

g:= proc(p,n) local i,t;
  t:= p^(n-1):
  for i from n-2 to 0 by -1 do
    t:= t + 10^(1+ilog10(t))*p^i
  od;
  t
end proc:
f:= proc(n)
  local p;
  if n mod 6 = 0 then return 0 fi;
  p:= 3;
  while not isprime(g(p,n)) do
    p:= nextprime(p);
    if n mod 3 = 0 then while p mod 3 = 1 do p:= nextprime(p) od fi:
  od;
  p
end proc:
map(f, [$2..30]); # Robert Israel, Sep 10 2017

PARI

pconc(p, n) = {my(s = "1"); for (k=1, n, s = concat(s, Str(p^k));); eval(s);}
a(n) = {if (!(n % 6), return (0)); n --; my(p = 2); while (! isprime(pconc(p, n)), p = nextprime(p+1)); p;}

Extensions

a(27)-a(50) from Robert Israel, Sep 10 2017