A245697 Least number k such that (n!+k)/n and (n!-k)/n are both prime.

0, 4, 25, 42, 133, 152, 279, 170, 121, 204, 1079, 938, 5295, 3632, 2771, 1062, 1159, 2260, 7413, 682, 33281, 13704, 9725, 4966, 9099, 24724, 2929, 54690, 20429, 22688, 5379, 46274, 15365, 11052, 40441, 65854, 97149, 42520, 44731, 83958, 61877, 4796, 123885, 27922, 122999, 12912, 5047

Offset: 3

Author

Derek Orr, Jul 29 2014

Keywords

• nonn

Comments

a(n) < n! for all n > 2.

It is believed that a(n) exists for all n > 2.

a(n) = n times (least m such that (n-1)!+m and (n-1)!-m are both prime) = n*A075409(n-1). - Jens Kruse Andersen, Jul 30 2014 [Goldbach's conjecture would then imply that a(n) always exists.]

Examples

			(4!+4)/4 = 7 is prime and (4!-4)/4 = 5 is prime. Thus a(4) = 4.
		

Links

• Jens Kruse Andersen, Table of n, a(n) for n = 3..101

Crossrefs

Cf. A075409, A245695, A245696.

Programs

PARI

a(n)=for(k=0,10^7,s1=(n!-k)/n;s2=(n!+k)/n;if(floor(s1)==s1&&floor(s2)==s2,if(ispseudoprime(s1)&&ispseudoprime(s2),return(k))))
n=3;while(n<100,print1(a(n),", ");n++)