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A079829
a(n) = smallest k such that floor(reverse(k)/k) >= n.
0
1, 13, 15, 17, 106, 107, 108, 109, 1089
OFFSET
1,2
COMMENTS
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
EXAMPLE
a(3)= 15 as floor(51/15) = 3, and 15 is the smallest such number.
MATHEMATICA
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 *)
CROSSREFS
Cf. A079830.
Sequence in context: A341329 A109019 A068893 * A096090 A037286 A166533
KEYWORD
base,fini,full,nonn
AUTHOR
Amarnath Murthy, Feb 11 2003
EXTENSIONS
More terms from Ryan Propper, Aug 27 2005
STATUS
approved