20120827, 14:35  #1 
Feb 2011
Singapore
5×7 Posts 
Offset of primes from factorials
So i was doing some experiments with primes P of the form,
P = n! + k, for nonnegative integer n and smallest nonnegative integer k such that P is a prime. The first thing i saw was that except for 0 and 1s, k is always a prime! But i was able to derive why after sometime. My spirits went down but then, after looking again at the sequence of k, i came up with a guess : Let k be the set of numbers, {k0,k1,k2,k3,...,kn}, for which kn is the smallest nonnegative integer such that n! + k is a prime for all corresponding nonnegative integers n. Then, there exists an arbitrary large constant c such that any element in the set k except for 0 and 1 will appear not more than c times in the set. Does anyone know of a way to proof/disproof this? Last fiddled with by Lee Yiyuan on 20120827 at 14:36 
20120827, 16:32  #2 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
somewhat makes sense to me that k might be prime, Pk = n! for n>2 n! is 0 mod 6 so k= 1 or 5 mod 6 ( the same as the primes greater than 3, so some intersection is possible), also note that if gcd(k,n!)>1 that P has a factor x such that 1<x<n! making P not prime, so k must be coprime to n! a big part of this set of coprime numbers as n grows should be prime themselves hence k has high odds of being prime.
Last fiddled with by science_man_88 on 20120827 at 16:37 
20120827, 22:12  #3 
Aug 2006
3·1,993 Posts 

20120828, 17:44  #4 
Aug 2002
Buenos Aires, Argentina
2×3×241 Posts 
If P = n!+k with k composite, the prime factors of k must be greater than n, otherwise P would not be prime. Thus the range n! to n! + (n+1)^{2} must be devoid of primes.
Since log (n!) is approximately n log n, the conjecture that the lowest k must be 1 or prime is nearly equivalent to the Cramer conjecture which says that there is at least a prime between k and k + (log(k))^{2} Last fiddled with by alpertron on 20120828 at 17:50 
20120828, 18:11  #5  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Last fiddled with by science_man_88 on 20120828 at 18:12 

20120828, 18:20  #6  
Aug 2002
Buenos Aires, Argentina
5A6_{16} Posts 
Quote:


20120828, 18:25  #7 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
yes well I was just trying to strengthen what was said before, because (nextprime(n)^2)> ((n+1)^2) for all values of n such that isprime(n+1)==0 so for the majority of n values this is a higher value and hence a longer devoid is needed. if we can find one counterexample and it takes it all out I'm just trying to give a way of making it less likely that a composite k is a first value that it can take on.
Last fiddled with by science_man_88 on 20120828 at 18:36 
20120828, 18:40  #8 
Aug 2002
Buenos Aires, Argentina
2×3×241 Posts 
Asymptotically (nextprime(n)^2) / ((n+1)^2) = 1, so there is no difference at all.
Last fiddled with by alpertron on 20120828 at 18:40 
20120828, 18:47  #9 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
if n=8 nextprime(n) =11 is 11==9 ? no. and I can back this up with pari.
Last fiddled with by science_man_88 on 20120828 at 18:48 
20120828, 19:04  #10 
Aug 2002
Buenos Aires, Argentina
2646_{8} Posts 
Asymptotically means for big values of n (n > inf). For small values, I ran this program in UBASIC in order to test it up to n=300:
Code:
10 for T=2 to 300 20 print T;""; 30 K=!(T) 40 N=1 50 M=K+N 60 if modpow(3,M1,M)=1 then 140 70 N=T+1:if N\2*2=N then N=N+1 80 if gcd(N,T)>1 then 110 90 M=K+N 100 if modpow(3,M1,M)=1 then 130 110 N=N+2 120 goto 80 130 if nxtprm(N1)<>N then print T,N:stop 140 next T Last fiddled with by alpertron on 20120828 at 19:05 
20120828, 19:15  #11  
"Forget I exist"
Jul 2009
Dumbassville
8384_{10} Posts 
Quote:


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