%I #16 Feb 20 2023 06:13:43
%S 1,2,3,9,27,33,99,123,271,333,351,407,429,481,693,777,819,999,2151,
%T 3333,4521,7227,7373,9999,33333,81819,99999,194841,326733,333333,
%U 340067,366337,369963,386139,389961,437229,534391,623763,706293,762377,863247
%N Numbers n dividing every cyclic permutation of n^2.
%e 123 is a member as all the five cyclic permutations of 123^2 are :
%e {15129, 51291, 12915, 29151, 91512};
%e 15129 = 123*123 ;
%e 51291 = 123*417 ;
%e 12915 = 123*105 ;
%e 29151 = 123*237 ;
%e 91512 = 123*744.
%p with(numtheory):for n from 1 to 100000 do:n0:=n^2:l:=length(n0) :ind:=0:for j
%p from 1 to l do:s:=0:for m from 1 to l do:q:=n0:u:=irem(q, 10):v:=iquo(q, 10):n0:=v
%p :s:=s+ u*10^m:od:s:=floor(s-u*10^l+u):if irem(s, n)=0 then ind:=ind+1:n0:=s:else
%p fi: od:if ind=l then printf(`%d, `, n):else fi: od:
%t Select[Range[900000],And@@Divisible[FromDigits/@Table[ RotateRight[ IntegerDigits[ #^2], n],{n,IntegerLength[#^2]}],#]&] (* _Harvey P. Dale_, Jul 31 2013 *)
%o (Sage)
%o def cycle(x): return (cp(x) for cp in CyclicPermutationGroup(len(x)))
%o is_A178028 = lambda n: all(n.divides(Integer(cx,base=10)) for cx in cycle(str(n**2))) # _D. S. McNeil_, Jan 08 2011
%Y Cf. A177950, A177928.
%K nonn,base
%O 1,2
%A _Michel Lagneau_, May 17 2010