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%I #8 Nov 18 2018 00:13:35
%S 1,13,15,17,106,107,108,109,1089
%N a(n) = smallest k such that floor(reverse(k)/k) >= n.
%C 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
%e a(3)= 15 as floor(51/15) = 3, and 15 is the smallest such number.
%t 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 *)
%Y Cf. A079830.
%K base,fini,full,nonn
%O 1,2
%A _Amarnath Murthy_, Feb 11 2003
%E More terms from _Ryan Propper_, Aug 27 2005