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

3,2

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.]

Links

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

Example

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

Prog

(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++)

Crossrefs

Cf. A075409, A245695, A245696.

Sequence in context: A231176 A335350 A199772 * A089767 A337173 A135784

Adjacent sequences: A245694 A245695 A245696 * A245698 A245699 A245700

Keyword

nonn

Author

Derek Orr, Jul 29 2014

Status

approved