Jeppe, I tried the same, but my code for creating the number is a bit more complicated... :)

m=2000;p=0;k=0;for (i=1,m,k+=ceil(log(i)/log(10));p+=i*10^k;if(ispseudoprime(p),print("Prime: ",p)))



And the other suggested format of increasing and then decreasing values like 12345678910987654321:

updown(r,s) = {
b=s;
for(a=r,s-1,
p=0;k=0;for (i=a,b,p+=i*10^k;k+=(floor(log(i)/log(10))+1));
q=0;k=0;for (i=1,b-a,j=b-i;q+=j*10^k;k+=(floor(log(j)/log(10))+1));
p+=q*10^(floor(log(p)/log(10))+1);
\\ print(p);
if(ispseudoprime(p),print("yay: ",p," is prime!"))
)
}

And to run it: for(i=2,100,updown(1,i)) tests all number with 2 ... 100 in the middle and beginning at 1 to the middle value. Such as 121, 232, 12321, 343, 23432, 1234321, ...

Your slim code could be used there as well though. I got to understand what's going on there at first though.
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1281979 * 2^485014 + 1 is prime ... no further hits up to: n = 5,700,000

Ah, I had read ?logint but didn't see why it could be used at first.

So the output of logint(a,b) is always identical to floor(log_b(a)) it just saves computing time by not going through the actual calculation of the log?

Interesting take on the & instead of if. Now I realized I actually used that a couple of times before just not in such simple constructs but only within an if clause to indicate progress by throwing an &!(print(...)) in there. Btw, is it convention that print(a) returns 0 upon successfully printing?
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1281979 * 2^485014 + 1 is prime ... no further hits up to: n = 5,700,000