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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.)