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A242567 Least number k >= 0 such that (n!+k)/(n+k) is an integer. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,5

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

LINKS

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

EXAMPLE

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

PROG

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)

CROSSREFS

Cf. A006093.

Sequence in context: A089275 A182052 A223520 * A040319 A040318 A040320

Adjacent sequences:  A242564 A242565 A242566 * A242568 A242569 A242570

KEYWORD

nonn

AUTHOR

Derek Orr, May 17 2014

EXTENSIONS

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

STATUS

approved

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Last modified June 24 04:02 EDT 2021. Contains 345416 sequences. (Running on oeis4.)