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Old 2012-08-27, 14:35   #1
Lee Yiyuan
 
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Default 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 2012-08-27 at 14:36
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Old 2012-08-27, 16:32   #2
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Quote:
Originally Posted by Lee Yiyuan View Post
P = n! + k, for nonnegative integer n and smallest nonnegative integer k such that P is a prime.
somewhat makes sense to me that k might be prime, P-k = 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.

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Old 2012-08-27, 22:12   #3
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Quote:
Originally Posted by Lee Yiyuan View Post
Does anyone know of a way to proof/disproof this?
Clearly a nonnegative integer other than 1 can appear only finitely many times. Reo Fortune conjectured that the only numbers which appear are 1 and primes.
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Old 2012-08-28, 17:44   #4
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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

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Old 2012-08-28, 18:11   #5
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Originally Posted by alpertron View Post
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
since any composite k has to have at least 2 not necessarily distinct prime divisors and all primes below n are wiped out the lowest possible prime divisor jumps to nextprime(n) and hence k should if composite be no less than nextprime(n)^2 correct, using that and replacing the (n+1)^2 in the equation for this new value shows n!+nextprime(n)^2 is the next value using composite k=nextprime(n)^2 can n! to n!+nextprime(n)^2 be devoid if not should this not prove the lowest k value has to be prime ?

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Old 2012-08-28, 18:20   #6
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Quote:
Originally Posted by science_man_88 View Post
since any composite k has to have at least 2 not necessarily distinct prime divisors and all primes below n are wiped out the lowest possible prime divisor jumps to nextprime(n) and hence k should if composite be no less than nextprime(n)^2 correct, using that and replacing the (n+1)^2 in the equation for this new value shows n!+nextprime(n)^2 is the next value using composite k=nextprime(n)^2 can n! to n!+nextprime(n)^2 be devoid if not should this not prove the lowest k value has to be prime ?
Yes, but the Cramer conjecture is just that, it is not a proof. So we do not know if really the OP question is right or not.
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Old 2012-08-28, 18:25   #7
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Quote:
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Yes, but the Cramer conjecture is just that, it is not a proof. So we do not know if really the OP question is right or not.
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.

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Old 2012-08-28, 18:40   #8
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Asymptotically (nextprime(n)^2) / ((n+1)^2) = 1, so there is no difference at all.

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Old 2012-08-28, 18:47   #9
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Quote:
Originally Posted by alpertron View Post
Asymptotically (nextprime(n)^2) / ((n+1)^2) = 1, so there is no difference at all.
if n=8 nextprime(n) =11 is 11==9 ? no. and I can back this up with pari.

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Old 2012-08-28, 19:04   #10
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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,M-1,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,M-1,M)=1 then 130
 110   N=N+2
 120   goto 80
 130   if nxtprm(N-1)<>N then print T,N:stop
 140   next T
In this range the lowest value of k is always 1 or a prime.

Last fiddled with by alpertron on 2012-08-28 at 19:05
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Old 2012-08-28, 19:15   #11
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Quote:
Originally Posted by alpertron View Post
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,M-1,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,M-1,M)=1 then 130
 110   N=N+2
 120   goto 80
 130   if nxtprm(N-1)<>N then print T,N:stop
 140   next T
In this range the lowest value of k is always 1 or a prime.
yeah I just realized that pari's nextprime looks for pseudoprimes, I meant the next proven prime this doesn't always equal n+1 for all large n.
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