A178028 -id:A178028 - cp's OEIS frontend

A242680 Numbers k dividing every cyclic permutation of k^3.

1, 2, 3, 9, 11, 41, 63, 77, 91, 99, 219, 303, 411, 999, 1353, 5291, 6363, 6993, 7777, 8547, 9009, 9191, 9901, 9999, 12561, 23661, 41841, 47027, 75609, 90243, 99999, 110011, 122859, 124533, 125341, 152207, 169983, 170017, 473211, 487179, 513513, 575757, 578369, 626373, 683527, 703703, 740259, 904761, 999001, 999999, 2463661, 2709729, 2754573

Offset: 1

Michel Lagneau, May 20 2014

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

			41 is a term as the cyclic permutations of 41^3 = 68921 are {68921, 89216, 92168, 21689, 16892}
and
  68921 = 41*1681;
  89216 = 41*2176;
  92168 = 41*2248;
  21689 = 41*529;
  16892 = 41*412.

Robert Israel, Table of n, a(n) for n = 1..130

Cf. A178028.

filter:= proc(n) local d,t,r,i;
  d:= ilog10(n^3);
  t:= n^3;
  for i from 1 to d do
    r:= t mod 10;
    t:= 10^d*r + (t-r)/10;
    if not (t/n)::integer then return false fi;
  od;
  true
end proc:
select(filter, [$1..10^7]); # Robert Israel, Jun 04 2019

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

More terms from Robert Israel, Jun 04 2019

A177928 Let n be the number whose square n^2 has the decimal expansion { d(1) d(2) ... d(D) }, and let q be the corresponding number whose decimal expansion is { d(2) d(3) ... d(D) d(1)}. Sequence lists numbers n dividing q.

Original entry on oeis.org

1, 2, 3, 9, 27, 33, 66, 99, 123, 246, 271, 333, 351, 407, 429, 462, 481, 518, 546, 567, 666, 693, 702, 715, 777, 814, 819, 924, 936, 999, 1434, 2151, 2868, 3333, 4521, 4818, 6666, 7227, 7373, 7535, 8631, 9042, 9999, 33333, 53658, 54546, 66666, 80487, 81819

Offset: 1

Michel Lagneau, May 15 2010

A178028 is a subsequence of this sequence.

When n divides q, n divides d(D)*(10^D - 1) because q = 10*n^2 - d(D)*(10^D - 1). If n is prime, n divides (10^D - 1); for example, the prime term 271 divides 10^5 - 1 = 99999 = 271*369.

			429 is in the sequence because 429^2 = 184041 and 840411/429 = 1959.

Eric Weisstein's World of Mathematics, Repunits

Cf. A002275, A003020, A005422, A067063, A102380, A178028.

for n from 1 to 10^6 do: d:=convert(n^2, base, 10):n1:=nops(d):s:=sum('d[i]*10^i','i'=1..n1-1)+d[n1]:if irem(s,n)=0 then printf(`%d, `,n):else fi:od:

Select[Range[100000], Mod[FromDigits[RotateLeft[IntegerDigits[#^2]]], #] == 0 &] (* T. D. Noe, Jul 27 2012 *)

A242740 Numbers n dividing every cyclic permutation of n^4.

Original entry on oeis.org

1, 3, 9, 21, 27, 73, 99, 111, 271, 693, 707, 777, 819, 909, 999, 2151, 2629, 3441, 3813, 4551, 6987, 7227, 7373, 9999, 18981, 19019, 20007, 20979, 23199, 24453, 25641, 27027, 27417, 30303, 81819, 82113, 83883, 99999, 125523, 172013, 194841, 201917, 238139

Offset: 1

Michel Lagneau, May 21 2014

Property of the sequence :
Consider the sequence A178028 (Numbers n dividing every cyclic permutation of n^2), so
a(1) = A178028 (1) = 1;
a(5) = A178028 (5) = 27;
a(7) = A178028 (7) = 99;
a(9) = A178028 (9) = 271;
a(10) = A178028 (15) = 693;
a(13) = A178028 (17) = 819;
a(15) = A178028 (18) = 999;
a(16) = A178028 (19) = 2151;
a(22) = A178028 (22) = 7227;
...........................

			21 is a member as all the six cyclic permutations of 21^4 = 194481 are :
{194481, 944811, 448119, 481194, 811944, 119448} and :
194481 = 21*9261;
944811 = 21*44991;
448119 = 21*21339;
811944 = 21*38664;
119448 = 21*5688.

Cf. A178028, A242680.

Select[Range[300000], And@@Divisible[FromDigits/@Table[ RotateRight[ IntegerDigits[ #^4], n], {n, IntegerLength[#^4]}], #]&]

A262814 Numbers k dividing every cyclic permutation of k^k.

Original entry on oeis.org

1, 2, 3, 7, 9, 11, 27, 63, 99, 111, 129, 159, 231, 271, 273, 303, 333, 351, 357, 403, 457, 711, 991, 999, 1111, 1147, 1241, 2121, 2479, 4227, 4653, 5151, 5547, 5837, 6191, 6237, 6643, 6993, 7133, 8229, 8547, 8683, 8811, 8987, 9009, 9633, 9999, 11009, 13449, 13531

Offset: 1

Michel Lagneau, Oct 03 2015

Conjecture: 10^n-1 is a term of the sequence for all n > 0. - Chai Wah Wu, Nov 03 2015

			7 is a member as the six cyclic permutations of 7^7 = 823543 are {823543, 382354, 438235, 543823, 354382, 235438} and these 6 integers are divisible by 7.

Chai Wah Wu, Table of n, a(n) for n = 1..100

Cf. A178028, A242680, A242740.

Select[Range[1000], And@@Divisible[FromDigits/@Table[ RotateRight[ IntegerDigits[ #^#], n], {n, IntegerLength[#^#]}], #]&]

isok(n) = {my(nn = n^n); for (j=1, #Str(nn)-1, cp = eval(Str(nn%10^j, nn\10^j)); if (cp % n, return (0));); return (1);} \\ Michel Marcus, Oct 11 2015

A262814_list = []
for k in range(1,10**3):
    n = k**k
    if not n % k:
        s = str(n)
        for i in range(len(s)-1):
            s = s[1:]+s[0]
            if int(s) % k:
                break
        else:
            A262814_list.append(k) # Chai Wah Wu, Oct 26 2015

a(24)-a(27) from Michel Marcus, Oct 11 2015

a(28)-a(50) from Chai Wah Wu, Oct 26 2015

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A242680 Numbers k dividing every cyclic permutation of k^3.

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A177928 Let n be the number whose square n^2 has the decimal expansion { d(1) d(2) ... d(D) }, and let q be the corresponding number whose decimal expansion is { d(2) d(3) ... d(D) d(1)}. Sequence lists numbers n dividing q.

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Maple

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A242740 Numbers n dividing every cyclic permutation of n^4.

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Mathematica

A262814 Numbers k dividing every cyclic permutation of k^k.

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