20080617, 13:30  #1 
Feb 2008
20_{16} Posts 
Subfactorial primes
!3 = 2 is the only subfactorial prime
!2+1 = 2 !3+1 = 3 !51 = 43 Any other !N +1 or !N  1 that are prime? 
20080617, 14:41  #2 
Nov 2003
7460_{10} Posts 

20080617, 17:00  #3 
"Mark"
Apr 2003
Between here and the
2^{2}·1,607 Posts 

20080617, 17:09  #4 
Feb 2008
2^{5} Posts 
N subfactorial, designated !N is calculted as follows:
N! * (11/1!+1/2!1/3!+1/4!.....1/n!) !1 = 0 !2 = 1 !3 = 2 !4 = 9 !5 = 44 !6 = 265 
20080617, 18:00  #5 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
25306_{8} Posts 

20080617, 18:15  #6  
(loop (#_fork))
Feb 2006
Cambridge, England
3·19·113 Posts 
Quote:
Bring out gp: Code:
for(t=10,264,s=floor(exp(1)*factorial(t)+0.5);for(j=3,3,if(isprime(s+j),print([t,j])))) Code:
[11, 3] [15, 1] [17, 1] [21, 3] [149, 3] [173, 3] [185, 3] [202,2] [264,2] 

20080617, 19:57  #7 
Einyen
Dec 2003
Denmark
5^{2}·127 Posts 
Another formula:
!n = Floor((n!+1)/e) for n>=1 Last fiddled with by ATH on 20080617 at 20:02 
20080618, 12:15  #8 
Feb 2006
Denmark
E6_{16} Posts 
If N is even then N divides !N1, and !N+1 is even.
If N is odd then N divides !N+1. That leaves !N1 for odd N. The only primes for N<=10000: !51 = 43 !151 = 481066515733 !171 = 130850092279663 These were already computed by http://www.research.att.com/~njas/sequences/A100015 and fivemack. I'm continuing to 20000. Candidates are computed by PARI/GP and tested by PrimeForm/GW. Like R.D. Silverman, I expect infinitely many but rare primes. 
20080618, 15:03  #9 
Aug 2002
Buenos Aires, Argentina
2^{2}×3×5×23 Posts 
I think it is best to use a sieve to discard many candidates.
I will prove that . Let N = mp+n (0<=n<p) Let Let When computing u, if k<mp, then that term will have more p's on the numerator than on the denominator, so it will be congruent to zero. The same occurs when computing v for k<mp+p. So discarding those terms we have: From Wilson's theorem: Using induction: Replacing the last formula on the formula for v: So the terms of u are negatives of those of v. Thus Last fiddled with by alpertron on 20080618 at 15:05 
20080618, 15:41  #10 
Aug 2002
Buenos Aires, Argentina
2^{2}·3·5·23 Posts 
For example, in the case !N1 we have the following zeros:
p = 2 > N = 0 (mod 2) p = 3 > N = 0, 2 (mod 6) p = 5 > N = 0, 2, 9 (mod 10) p = 7 > N = 0, 2, 13 (mod 14) p = 11 > N = 0, 2, 6, 10 (mod 22) etc. So for these congruences of N, !N1 is composite. Of course the even values are not needed because they are already covered by the case p=2. Other values of N for which !N1 is composite: p=5 > N = 9 (mod 10) p=7 > N = 13 (mod 14) p=17 > N = 7 (mod 34) p=19 > N = 13, 25 (mod 38) p=23 > N = 19 (mod 46) p=29 > N = 49 (mod 58) p=43 > N = 5 (mod 86) (of course when N=5, !N1 is prime). p=47 > N = 27 (mod 94) p=59 > N = 11 (mod 118) p=67 > N = 63, 81 (mod 134) p=73 > N = 121 (mod 146) p=79 > N = 109, 115 (mod 158) p=89 > N = 103 (mod 178) p=97 > N = 179 (mod 194) Last fiddled with by alpertron on 20080618 at 15:55 
20080618, 21:26  #11 
Feb 2006
Denmark
E6_{16} Posts 
I'm stopping when 20000 is reached. This is a one day search and I'm not implementing a sieve. I also think the advantage would be quite small. The primes which divide more than one value up to !20000 are so small that only little trial factoring would be spared. A trial factoring program using !n = n * !(n1) + (1)^n would probably be a better use of programming time but I'm not doing that either.

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