

A112726


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


2



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

0,2


COMMENTS

a(0)=1; a(1)=3 and it is easily shown that for n>1, 10^3^(n2)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^(n2)1. This sequence is a subsequence of A062567, a(n)=A062567(3^n). Jud McCranie conjectures that for n>1, a(n)=10^3^(n2)1 (see comments line of A062567), but we see that for n>4, a(n) is much smaller than 10^3^(n2)1, so his conjecture is rejected. It seems that only for n=2,3 & 4 we have, a(n)=10^3^(n2)1.


LINKS

Table of n, a(n) for n=0..17.


EXAMPLE

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.


MATHEMATICA

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


CROSSREFS

Cf. A062567, A112725.
Sequence in context: A069028 A318246 A137043 * A112725 A060712 A122463
Adjacent sequences: A112723 A112724 A112725 * A112727 A112728 A112729


KEYWORD

base,nonn


AUTHOR

Farideh Firoozbakht, Nov 13 2005


STATUS

approved



