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A147553
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Numbers k such that k^2 divides k.k where dot "." means concatenation.
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5
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OFFSET
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1,2
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COMMENTS
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I proved that for n > 0, a(n+1) = (10^(6n-3) + 1)/7. Namely for n > 1, a(n) is of the form 142857.142857. ... .142857.143. Except for a(1), 11 divides all terms, so there is no prime p such that p^2 divides p.p. For n > 1, a(n).a(n)/(a(n)*a(n))=7.
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LINKS
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EXAMPLE
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143*143|143.143 (143143/(143*143)=7) so 143 is in the sequence.
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MAPLE
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MATHEMATICA
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a[0]=1; a[n_]:=(10^(6n-3)+1)/7; Table[a[k], {k, 0, 8}]
Do[d=Divisors[10^i+1]; s=Select[d, Length[IntegerDigits[#]]==i&]; If[Length[s]>0, Do[Print[s[[j]]], {j, Length[s]}]], {i, 69}] (* Hans Havermann, May 31 2014 *)
LinearRecurrence[{1000001, -1000000}, {1, 143, 142857143}, 20] (* Harvey P. Dale, Apr 02 2018 *)
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CROSSREFS
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KEYWORD
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base,easy,nice,nonn
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AUTHOR
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STATUS
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approved
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