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%I #4 Mar 30 2012 17:37:43
%S 1,3,9,999,999999999,29799999792,39789998793,39989598993,68899199886,
%T 68899199886,68899199886,68899199886,68899199886,2699657569962,
%U 146189959981641,191388777883191,191388777883191,18641845754814681
%N Smallest positive palindromic 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 palindromic 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 A020485(a(n)=A020485(3^n)) and for all n, A062567(3^n)<=a(n) because for all n, A062567(n)<= A020485(n). Jud McCranie conjectures that for n>1 A062567(3^n) =10^3^(n-2)-1, if his conjecture were true then from the above facts we conclude that for n>1 a(n)=10^3^(n-2)-1, but we see that for 4<n<=18 a(n) is much smaller than 10^3^(n-2)-1, so his conjecture will be rejected.
%e a(18)=18771463736417781 because 18771463736417781=3^18*48452429 is the smallest positive palindromic multiple of 3^18.
%t b[n_]:=(For[m=1, !FromDigits[Reverse[IntegerDigits[m*n]]]==m*n, m++ ]; m*n);Do[Print[b[3^n]], {n, 0, 18}]
%Y Cf. A020485, A062567, A112726.
%K base,nonn
%O 0,2
%A _Farideh Firoozbakht_, Nov 12 2005
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