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 2008-06-17, 13:30 #1 Housemouse     Feb 2008 25 Posts Subfactorial primes !3 = 2 is the only subfactorial prime !2+1 = 2 !3+1 = 3 !5-1 = 43 Any other !N +1 or !N - 1 that are prime?
2008-06-17, 14:41   #2
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by Housemouse !3 = 2 is the only subfactorial prime !2+1 = 2 !3+1 = 3 !5-1 = 43 Any other !N +1 or !N - 1 that are prime?
They should form a very thin infinite set.

2008-06-17, 17:00   #3
rogue

"Mark"
Apr 2003
Between here and the

146138 Posts

Quote:
 Originally Posted by Housemouse !3 = 2 is the only subfactorial prime !2+1 = 2 !3+1 = 3 !5-1 = 43 Any other !N +1 or !N - 1 that are prime?
Based upon this post, I'm not clear what a "subfactorial" prime is.

 2008-06-17, 17:09 #4 Housemouse     Feb 2008 25 Posts N subfactorial, designated !N is calculted as follows: N! * (1-1/1!+1/2!-1/3!+1/4!.....1/n!) !1 = 0 !2 = 1 !3 = 2 !4 = 9 !5 = 44 !6 = 265
2008-06-17, 18:00   #5
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

71×157 Posts

Quote:
 Originally Posted by rogue Based upon this post, I'm not clear what a "subfactorial" prime is.
Neither was I, but Google quickly told me.

Paul

2008-06-17, 18:15   #6
fivemack
(loop (#_fork))

Feb 2006
Cambridge, England

22·1,613 Posts

Quote:
 Originally Posted by Housemouse N subfactorial, designated !N is calculated as follows: N! * (1-1/1!+1/2!-1/3!+1/4!.....1/n!) !1 = 0 !2 = 1 !3 = 2 !4 = 9 !5 = 44 !6 = 265
OK, by that definition and standard properties of alternating sequences it's also nearest_integer(exp(-1) * factorial(N))

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]))))
says
Code:
[11, -3]
[15, -1]
[17, -1]
[21, 3]
[149, -3]
[173, -3]
[185, 3]
[202,-2]
[264,-2]

 2008-06-17, 19:57 #7 ATH Einyen     Dec 2003 Denmark 3,257 Posts Another formula: !n = Floor((n!+1)/e) for n>=1 Last fiddled with by ATH on 2008-06-17 at 20:02
 2008-06-18, 12:15 #8 Jens K Andersen     Feb 2006 Denmark 2·5·23 Posts If N is even then N divides !N-1, and !N+1 is even. If N is odd then N divides !N+1. That leaves !N-1 for odd N. The only primes for N<=10000: !5-1 = 43 !15-1 = 481066515733 !17-1 = 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.
 2008-06-18, 15:03 #9 alpertron     Aug 2002 Buenos Aires, Argentina 1,423 Posts I think it is best to use a sieve to discard many candidates. I will prove that $!N \equiv -!(N+p) \pmod p$. Let N = mp+n (0<=n
 2008-06-18, 15:41 #10 alpertron     Aug 2002 Buenos Aires, Argentina 142310 Posts For example, in the case !N-1 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, !N-1 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 !N-1 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, !N-1 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 2008-06-18 at 15:55
 2008-06-18, 21:26 #11 Jens K Andersen     Feb 2006 Denmark 2×5×23 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 * !(n-1) + (-1)^n would probably be a better use of programming time but I'm not doing that either.

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