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Numbers k dividing every cyclic permutation of k^3.
3

%I #13 Jun 04 2019 21:39:07

%S 1,2,3,9,11,41,63,77,91,99,219,303,411,999,1353,5291,6363,6993,7777,

%T 8547,9009,9191,9901,9999,12561,23661,41841,47027,75609,90243,99999,

%U 110011,122859,124533,125341,152207,169983,170017,473211,487179,513513,575757,578369,626373,683527,703703,740259,904761,999001,999999,2463661,2709729,2754573

%N Numbers k dividing every cyclic permutation of k^3.

%C Includes k if 10^(d-1) <= k^3 < 10^d and k | 10^d-1. Is 2 the only member of the sequence that is not of this form? - _Robert Israel_, Jun 04 2019

%H Robert Israel, <a href="/A242680/b242680.txt">Table of n, a(n) for n = 1..130</a>

%e 41 is a term as the cyclic permutations of 41^3 = 68921 are {68921, 89216, 92168, 21689, 16892}

%e and

%e 68921 = 41*1681;

%e 89216 = 41*2176;

%e 92168 = 41*2248;

%e 21689 = 41*529;

%e 16892 = 41*412.

%p filter:= proc(n) local d,t,r,i;

%p d:= ilog10(n^3);

%p t:= n^3;

%p for i from 1 to d do

%p r:= t mod 10;

%p t:= 10^d*r + (t-r)/10;

%p if not (t/n)::integer then return false fi;

%p od;

%p true

%p end proc:

%p select(filter, [$1..10^7]); # _Robert Israel_, Jun 04 2019

%t Select[Range[300000], And@@Divisible[FromDigits/@Table[ RotateRight[ IntegerDigits[ #^3], n], {n, IntegerLength[#^3]}], #]&]

%Y Cf. A178028.

%K nonn,base

%O 1,2

%A _Michel Lagneau_, May 20 2014

%E More terms from _Robert Israel_, Jun 04 2019