

A070294


Smallest k such that k==i (mod(i!+1)) i=1,2,3,...,n.


0




OFFSET

1,2


COMMENTS

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


LINKS



PROG

(PARI) for(n=1, 5, m=1; while(sum(i=1, n, abs(m%(1+i!)i))>0, m++); print1(m, ", "))


CROSSREFS



KEYWORD

easy,nonn,fini,full


AUTHOR



STATUS

approved



