%I #6 Mar 30 2012 18:38:58
%S 1,5,17,479,71879
%N Smallest k such that k==i (mod(i!+1)) i=1,2,3,...,n.
%C There are no more terms. Any subsequent term would need to satisfy both k = 3 mod 7 and k = 6 mod 721 (6!+1). But, since 721 = 7*103, that implies that the next term would have to equal 3 mod 721 and 6 mod 721.  Larry Reeves (larryr(AT)acm.org), Sep 27 2002
%o (PARI) for(n=1,5,m=1; while(sum(i=1,n,abs(m%(1+i!)i))>0,m++); print1(m,","))
%Y Cf. A053664.
%K easy,nonn,fini,full
%O 1,2
%A _Benoit Cloitre_, May 12 2002
