0,2

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.

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

387207449572462812=3^20*111050012 & 218264275944702783 is the

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

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

14048104419899757 & 171101619858478932.

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

Cf. A062567, A112725.

Sequence in context: A088031 A069028 A137043 this_sequence A112725 A060712 A122463

Adjacent sequences: A112723 A112724 A112725 this_sequence A112727 A112728 A112729

base,nonn

Farideh Firoozbakht (mymontain(AT)yahoo.com), Nov 13 2005