20190410, 14:17  #1 
Mar 2018
3^{2}·59 Posts 
for which values of p this expression has the form s*2^k?
let be p an arbitrary prime...
can you find values of p such that (10^p1)/9+1 has the form s*2^k? where s is a prime and k an integer >0 example p=3 does the job infact (10^31)/9+1 has the form 7*2^4 p=5 does not the job because (10^51)/9+1 has the form 2^3*3*463... 
20190410, 14:57  #2 
Mar 2018
3^{2}×59 Posts 
Primes that do the job up to 103
I found that the primes that do the job up to p=103 are: 3, 7, 43.
Is there some relation with the fact that 10^r+333667 is prime for r=3,7 and 43 and no other prime r up to r=26.000? Last fiddled with by enzocreti on 20190410 at 15:06 
20190411, 01:14  #3 
"William"
May 2003
New Haven
941_{16} Posts 

20190411, 14:01  #4 
Feb 2017
Nowhere
2·2,687 Posts 
I note that
if n > 3 then N = ((10^n  1)/9 + 1)/8 is an odd integer. Also, n need not be prime for N to be prime. Code:
? for(i=4,1000,n=1+(10^i1)/9;n=n/8;if(ispseudoprime(n),print(i))) 4 7 16 43 58 106 160 229 628 
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