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A070294
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Smallest k such that k==i (mod(i!+1)) i=1,2,3,...,n.
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0
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OFFSET
| 1,2
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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
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PROG
| (PARI) for(n=1, 5, m=1; while(sum(i=1, n, abs(m%(1+i!)-i))>0, m++); print1(m, ", "))
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CROSSREFS
| Cf. A053664.
Sequence in context: A093428 A081479 A096996 * A161469 A119531 A172996
Adjacent sequences: A070291 A070292 A070293 * A070295 A070296 A070297
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KEYWORD
| easy,nonn,fini,full
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), May 12 2002
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