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A079829
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a(n) = smallest k such that floor(reverse(k)/k) >= n.
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0
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OFFSET
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1,2
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COMMENTS
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The complete sequence is now given. Proof (that there is no k such that reverse(k)/k >= 10): Since reverse(k) and k have the same number of digits, we see that reverse(k)/k < 10, otherwise reverse(k) would have at least one more base-10 digit. - Ryan Propper, Aug 27 2005
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LINKS
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EXAMPLE
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a(3)= 15 as floor(51/15) = 3, and 15 is the smallest such number.
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MATHEMATICA
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r[n_] := FromDigits[Reverse[IntegerDigits[n]]]; Do[k = 1; While[Floor[r[k]/k] < n, k++ ]; Print[k], {n, 1, 9}] (* Ryan Propper, Aug 27 2005 *)
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CROSSREFS
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KEYWORD
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base,fini,full,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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