A242680 -id:A242680 - cp's OEIS frontend

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

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

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

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