

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.


0



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

2,1


COMMENTS

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


LINKS

Table of n, a(n) for n=2..50.
Carlos Rivera, Puzzle 892. Primes as a concatenation of a series of powers of a prime, The Prime Puzzles and Problems Connection.


EXAMPLE

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.


MAPLE

g:= proc(p, n) local i, t;
t:= p^(n1):
for i from n2 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


PROG

(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; }


CROSSREFS

Cf. A047253.
Sequence in context: A113466 A007301 A009715 * A242886 A221947 A138662
Adjacent sequences: A292160 A292161 A292162 * A292164 A292165 A292166


KEYWORD

nonn,base


AUTHOR

Michel Marcus, Sep 10 2017


EXTENSIONS

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


STATUS

approved



