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A178028 Numbers n dividing every cyclic permutation of n^2. 4

%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

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Last modified March 29 02:23 EDT 2024. Contains 371264 sequences. (Running on oeis4.)