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Old 2019-04-10, 14:17   #1
enzocreti
 
Mar 2018

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Default 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^p-1)/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 (10^3-1)/9+1 has the form 7*2^4


p=5 does not the job because


(10^5-1)/9+1 has the form 2^3*3*463...
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Old 2019-04-10, 14:57   #2
enzocreti
 
Mar 2018

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Default 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 2019-04-10 at 15:06
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Old 2019-04-11, 01:14   #3
wblipp
 
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"William"
May 2003
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Quote:
Originally Posted by enzocreti View Post
let be p an arbitrary prime...
can you find values of p such that


(10^p-1)/9+1 has the form s*2^k?
2, 3, 7, 229, 1579
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Old 2019-04-11, 14:01   #4
Dr Sardonicus
 
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Feb 2017
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Default

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^i-1)/9;n=n/8;if(ispseudoprime(n),print(i)))
4
7
16
43
58
106
160
229
628
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