A079829 a(n) = smallest k such that floor(reverse(k)/k) >= n.

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

Links

Table of n, a(n) for n=1..9.

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

Adjacent sequences: A079826 A079827 A079828 * A079830 A079831 A079832

Keyword

base,fini,full,nonn

Author

Amarnath Murthy, Feb 11 2003

Extensions

More terms from Ryan Propper, Aug 27 2005

Status

approved