OFFSET
1,2
COMMENTS
a(81) = 999999999. 10^27-1 is a solution for a(3^5), but it may not be the smallest one. However, it seems likely (and perhaps easy to prove) that a(3^i) is 3^(i-2) "9"s, for i > 1. - Jud McCranie, Aug 07 2001
a(3^5)=4899999987<10^27-1 so Jud McCranie's conjecture "for n>1, a(3^n)=10^3^(n-2)-1 " is incorrect. I found a(3^n) for n<21; A112726 gives this subsequence. From the terms of A112726 we see that for n>4, a(3^n) is much smaller than 10^3^(n-2)-1. It seems that only for n=2,3 & 4 we have a(3^n)=10^3^(n-2)-1. - Farideh Firoozbakht, Nov 13 2005
EXAMPLE
48 and 84 are both divisible by 12.
MATHEMATICA
Block[{k = 1}, While[ !IntegerQ[k/n] || !IntegerQ[ FromDigits[ Reverse[ IntegerDigits[k]]]/n] && k < 10^5, k++ ]; If[k != 10^5, k, 0]]; Table[ a[n], {n, 1, 60}] (* Robert G. Wilson v *)
a[n_]:=(For[m=1, !IntegerQ[FromDigits[Reverse[IntegerDigits[m*n]]]/n], m++ ]; m*n); Do[Print[a[n]], {n, 60}] (* Farideh Firoozbakht *)
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Erich Friedman, Jul 03 2001
EXTENSIONS
Offset corrected by Sean A. Irvine, Apr 03 2023
STATUS
approved