A112726 - OEIS

%I #4 Mar 30 2012 17:37:43

%S 1,3,9,999,999999999,4899999987,19899999972,28999899936,49989892689,

%T 49999917897,68899199886,68899199886,68899199886,2678052898989,

%U 17902896898419,137530987695297,189281899170567,368055404997498

%N First positive multiple of 3^n whose reverse is also a multiple of 3^n.

%C a(0)=1; a(1)=3 and it is easily shown that for n>1, 10^3^(n-2)-1 is a multiple of 3^n whose reverse is also a multiple of 3^n (see comments line of A062567), so for each n, a(n) exists and for n>1, a(n)<=10^3^(n-2)-1. This sequence is a subsequence of A062567, a(n)=A062567(3^n). Jud McCranie conjectures that for n>1, a(n)=10^3^(n-2)-1 (see comments line of A062567), but we see that for n>4, a(n) is much smaller than 10^3^(n-2)-1, so his conjecture is rejected. It seems that only for n=2,3 & 4 we have, a(n)=10^3^(n-2)-1.

%e a(20)=218264275944702783 because 218264275944702783=3^20*62597583

%e 387207449572462812=3^20*111050012 & 218264275944702783 is the

%e smallest positive multiple of 3^20 whose reverse is also amultiple

%e of 3^20. I found a(n) for n<21, a(18) & a(19) are respectively

%e 14048104419899757 & 171101619858478932.

%t b[n_]:=(For[m=1, !IntegerQ[FromDigits[Reverse[IntegerDigits[m*n]]]/n], m++ ]; m*n);Do[Print[b[3^n]], {n, 0, 18}]

%Y Cf. A062567, A112725.

%K base,nonn

%O 0,2

%A _Farideh Firoozbakht_, Nov 13 2005