A242567 Least number k >= 0 such that (n!+k)/(n+k) is an integer.

1, 1, 0, 1, 18, 1, 712, 5031, 14, 1, 18, 1, 479001586, 1719, 87178291184, 1, 3024, 1, 40, 633, 124748, 1, 86, 51847, 625793187628, 123, 20404, 1, 210, 1, 265252859812191058636308479999968, 755, 263130836933693530167218012159999966

Offset: 1

Author

Derek Orr, May 17 2014

Keywords

• nonn

Comments

a(n) = 1 iff n+1 is prime.

For n > 2, in order for (n!+k)/(n+k) to be an integer, the smallest integer possible is 2. Thus, a(n) <= n!-2n for all n > 2.

Let q = (n!+k)/(n+k). Then, k = (n!-n)/(q-1)-n. So, a(n) = d-n, where d is the smallest divisor (> n) of n!-n. (for all n >= 4) - Hiroaki Yamanouchi, Sep 29 2014

Examples

			(6!+1)/(6+1) = 103 is an integer. Thus a(6) = 1.
		

Links

• Hiroaki Yamanouchi, Table of n, a(n) for n = 1..79

Crossrefs

Cf. A006093.

Programs

PARI

a(n)=for(k=1,10^5,s=(n!+k)/(n+k);if(floor(s)==s,return(k)));
n=1;while(n<100,print(a(n));n+=1)

Extensions

a(13), a(15), a(21), a(25), a(31) and a(33) from Hiroaki Yamanouchi, Sep 29 2014