

A112725


Smallest positive palindromic multiple of 3^n.


2



1, 3, 9, 999, 999999999, 29799999792, 39789998793, 39989598993, 68899199886, 68899199886, 68899199886, 68899199886, 68899199886, 2699657569962, 146189959981641, 191388777883191, 191388777883191, 18641845754814681
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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 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^(n2)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^(n2)1, if his conjecture were true then from the above facts we conclude that for n>1 a(n)=10^3^(n2)1, but we see that for 4<n<=18 a(n) is much smaller than 10^3^(n2)1, so his conjecture will be rejected.


LINKS

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


EXAMPLE

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


MATHEMATICA

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


CROSSREFS

Cf. A020485, A062567, A112726.
Sequence in context: A318246 A137043 A112726 * A060712 A122463 A072005
Adjacent sequences: A112722 A112723 A112724 * A112726 A112727 A112728


KEYWORD

base,nonn


AUTHOR

Farideh Firoozbakht, Nov 12 2005


STATUS

approved



