A112726 First positive multiple of 3^n whose reverse is also a multiple of 3^n.

1, 3, 9, 999, 999999999, 4899999987, 19899999972, 28999899936, 49989892689, 49999917897, 68899199886, 68899199886, 68899199886, 2678052898989, 17902896898419, 137530987695297, 189281899170567, 368055404997498

Offset: 0

Author

Farideh Firoozbakht, Nov 13 2005

Keywords

• base
• nonn

Comments

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.

Examples

			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.
		

Crossrefs

Cf. A062567, A112725.

Programs

Mathematica

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