Here is the exact code being used and its current output:

{

i=1;n=0;e=10;

while(i,

p=prime(i);if(p>e,e*=10);

for(j=1,i^2,

n*=e;n+=p;if(ispseudoprime(n),print1("("i","j") "))

);

i++;next()

)

}

-----------------------------------------------------------------

(1,1) (2,1) (2,2) (2,3) (2,4)

Jim

P.S. I answered the question more along the lines requested in a response to Mr. Hasler already, not realizing it wasn't going to the group. If anybody else really didn't understand, after the given value of 23333, nine 5s are appended and then sixteen 7s, twenty-five 11s, and so on; and the current question on the appearance of the next prime doesn't look at numbers ending after the first 1 of an 11 (and the like).

On Sun Jul 31st, 2011 5:20 AM EDT Maximilian Hasler wrote:

>On Sun, Jul 31, 2011 at 7:08 AM, James Merickel <merk7777777@...> wrote:

>> Assuming that my PARI/GP did not crash, after 2, 23, 233, 2333 and 23333 are all prime, concatenation of n^2 copies of each prime prime(n) does not produce another prime (after adjoining any single copy of a prime) until some very large number.

>

>

>could you explain to the simple minded as me, what kind of numbers you

>test exactly ?

>I can't see what the n^2, nor the prime(n), refer to w.r.t. the

>example 2,23,233,...

>

>

>>. Unless somebody knows how to break the program,

>

>on PARI versions not older than 2 years or so (since v.2.4.3 IIRC)

>you can simply hit Ctrl-C, which interrupts the computation,

>gives you a command prompt to let you examine variables and do

>whatever nasty thing you want, and continue the computation at the

>very point it stopped.

>

>

>>the time with the CPU overtaxed, as information on where exactly in the search I am.

>

>so you know where you are ? can you give this information ?

>

>M.