|
Search: id:A070294
|
|
|
| A070294 |
|
Smallest k such that k==i (mod(i!+1)) i=1,2,3,...,n. |
|
+0
1
|
| |
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
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
|
|
PROGRAM
|
(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: A093428 A081479 A096996 this_sequence A119531 A085832 A062223
Adjacent sequences: A070291 A070292 A070293 this_sequence A070295 A070296 A070297
|
|
KEYWORD
|
easy,nonn,fini,full
|
|
AUTHOR
|
Benoit Cloitre (benoit7848c(AT)orange.fr), May 12 2002
|
|
|
Search completed in 0.002 seconds
|
