Search: id:A062567
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%I A062567
%S A062567 1,2,3,4,5,6,7,8,9,0,11,48,494,252,510,272,272,216,171,0,168,22,161,
%T A062567 696,525,494,999,252,232,0,434,2112,33,272,525,216,111,494,585,0,656,
%U A062567 252,989,44,540,414,141,2112,343,0,969,676,212,4698,55,616,171,232,767
%N A062567 First multiple of n whose reverse is also divisible by n, or 0 if no 
               such multiple exists.
%C A062567 a(81) = 999999999. 10^27-1 is a solution for a(3^5), but it may not be 
               the smallest one. However, it seems likely (and perhaps easy to prove) 
               that a(3^i) is 3^(i-2) "9"s, for i > 1. - Jud McCranie (j.mccranie(AT)comcast.net), 
               Aug 07 2001
%C A062567 a(3^5)=4899999987<10^27-1 so Jud McCranie's conjecture "for n>1, a(3^n)=10^3^(n-2)-1 
               " is incorrect. I found a(3^n) for n<21; A112726 gives this subsequence. 
               From the terms of A112726 we see that for n>4, a(3^n) is much smaller 
               than 10^3^(n-2)-1. It seems that only for n=2,3 & 4 we have a(3^n)=10^3^(n-2)-1. 
               - Farideh Firoozbakht (mymontain(AT)yahoo.com), Nov 13 2005
%e A062567 48 and 84 are both divisible by 12
%t A062567 Block[{k = 1}, While[ !IntegerQ[k/n] || !IntegerQ[ FromDigits[ Reverse[ 
               IntegerDigits[k]]]/n] && k < 10^5, k++ ]; If[k != 10^5, k, 0]]; Table[ 
               a[n], {n, 1, 60}] (from Robert G. Wilson v)
%t A062567 a[n_]:=(For[m=1, !IntegerQ[FromDigits[Reverse[IntegerDigits[m*n]]]/n], 
               m++ ]; m*n);Do[Print[a[n]], {n, 60}] (Firoozbakht)
%Y A062567 Cf. A112725, A112726.
%Y A062567 Sequence in context: A056967 A067079 A080434 this_sequence A069554 A020485 
               A083116
%Y A062567 Adjacent sequences: A062564 A062565 A062566 this_sequence A062568 A062569 
               A062570
%K A062567 base,nonn
%O A062567 0,2
%A A062567 Erich Friedman (efriedma(AT)stetson.edu), Jul 03 2001

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