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Let's call the numbers of the form $k\times p\# \mp1$, the Generalized Primorial Primes.

One can find many $k$ for a fixed $p$ such that $k\times p\# \mp1$ be prime. As an example for $p = 8933$ there are 13 $k$ below 20000.

I'd like to know, is there any way to prove (or even approach the proof) that there are infinite number of $k$ for each $p$?

  1. $3347\times 8933\#+1$
  2. $5570\times 8933\#+1$
  3. $6980\times 8933\#+1$
  4. $7227\times 8933\#+1$
  5. $10087\times 8933\#+1$
  6. $11731\times 8933\#+1$
  7. $14532\times 8933\#+1$
  8. $16957\times 8933\#+1$
  9. $17173\times 8933\#+1$
  10. $17575\times 8933\#+1$
  11. $18563\times 8933\#+1$
  12. $19518\times 8933\#+1$
  13. $19625\times 8933\#+1$
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1 Answer 1

up vote 2 down vote accepted

This is a corollary of Dirichlet's theorem on primes in arithmetic progressions, which states that if $\gcd(a,d)=1$, then the arithmetic progression $$a,\quad a+d,\quad a+2d,\quad\ldots$$ contains infinitely many prime numbers. For any prime number $p$, letting $d=p\#$ and $a=1$, we see that there are infinitely many prime numbers of the form $$k\cdot (p\#) +1.$$ For the minus case, just use $a=p\#-1$ instead.

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Thank you, brilliant explanation ... –  Mohsen Afshin Mar 6 '13 at 11:08
    
Glad to help!${}$ –  Zev Chonoles Mar 6 '13 at 11:09

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