cp's OEIS Frontend

A062567 First multiple of n whose reverse is also divisible by n, or 0 if no such multiple exists.

%I A062567 #17 Apr 03 2023 20:12:58
%S A062567 1,2,3,4,5,6,7,8,9,0,11,48,494,252,510,272,272,216,171,0,168,22,161,
%T A062567 696,525,494,999,252,232,0,434,2112,33,272,525,216,111,494,585,0,656,
%U A062567 252,989,44,540,414,141,2112,343,0,969,676,212,4698,55,616,171,232,767
%N A062567 First multiple of n whose reverse is also divisible by n, or 0 if no such multiple exists.
%C A062567 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
%C A062567 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
%e A062567 48 and 84 are both divisible by 12.
%t A062567 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_ *)
%t A062567 a[n_]:=(For[m=1, !IntegerQ[FromDigits[Reverse[IntegerDigits[m*n]]]/n], m++ ]; m*n);Do[Print[a[n]], {n, 60}] (* _Farideh Firoozbakht_ *)
%Y A062567 Cf. A112725, A112726.
%K A062567 base,nonn
%O A062567 1,2
%A A062567 _Erich Friedman_, Jul 03 2001
%E A062567 Offset corrected by _Sean A. Irvine_, Apr 03 2023