|
| |
|
|
A079829
|
|
a(n) = smallest k such that floor(reverse(k)/k) >= n.
|
|
0
|
| |
|
|
|
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
|
|
|
|
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
|
|
|
|
KEYWORD
|
base,fini,full,nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
