A070294 Smallest k such that k==i (mod(i!+1)) i=1,2,3,...,n.

1, 5, 17, 479, 71879

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

Table of n, a(n) for n=1..5.

Prog

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

Crossrefs

Cf. A053664.

Sequence in context: A318751 A096996 A256236 * A377779 A245568 A218378

Adjacent sequences: A070291 A070292 A070293 * A070295 A070296 A070297

Keyword

easy,nonn,fini,full

Author

Benoit Cloitre, May 12 2002

Status

approved