0,2

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.

a(18)=18771463736417781 because 18771463736417781=3^18*48452429 is the smallest positive palindromic multiple of 3^18.

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

Cf. A020485, A062567, A112726.

Sequence in context: A069028 A137043 A112726 this_sequence A060712 A122463 A072005

Adjacent sequences: A112722 A112723 A112724 this_sequence A112726 A112727 A112728

base,nonn

Farideh Firoozbakht (mymontain(AT)yahoo.com), Nov 12 2005