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 A147553 Numbers n such that n^2 divides n.n where dot"." means concatenation. 1
 1, 143, 142857143, 142857142857143, 142857142857142857143, 142857142857142857142857143, 142857142857142857142857142857143, 142857142857142857142857142857142857143 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS I proved that for n>0, a(n)=(10^(6n-3)+1)/7. Namely for n>1, a(n) is of the form 142857.142857. ... .142857.143. Except the first term 11 divides all other terms, so there is no prime p such that p^2 divides p.p. For n>0 a(n).a(n)/(a(n)*a(n))=7. LINKS EXAMPLE 143*143|143.143 (143143/(143*143)=7) so 143 is in the sequence. MATHEMATICA a[0]=1; a[n_]:=(10^(6n-3)+1)/7; Table[a[k], {k, 0, 8}] CROSSREFS Cf. A147554. Sequence in context: A204683 A205159 A205308 * A030122 A057404 A174683 Adjacent sequences:  A147550 A147551 A147552 * A147554 A147555 A147556 KEYWORD base,easy,nice,nonn AUTHOR Farideh Firoozbakht, Dec 23 2008 STATUS approved

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Last modified June 17 23:56 EDT 2013. Contains 226327 sequences.