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

%I A262814 #21 Nov 05 2015 11:22:07
%S A262814 1,2,3,7,9,11,27,63,99,111,129,159,231,271,273,303,333,351,357,403,
%T A262814 457,711,991,999,1111,1147,1241,2121,2479,4227,4653,5151,5547,5837,
%U A262814 6191,6237,6643,6993,7133,8229,8547,8683,8811,8987,9009,9633,9999,11009,13449,13531
%N A262814 Numbers k dividing every cyclic permutation of k^k.
%C A262814 Conjecture: 10^n-1 is a term of the sequence for all n > 0. - _Chai Wah Wu_, Nov 03 2015
%H A262814 Chai Wah Wu, <a href="/A262814/b262814.txt">Table of n, a(n) for n = 1..100</a>
%e A262814 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.
%t A262814 Select[Range[1000], And@@Divisible[FromDigits/@Table[ RotateRight[ IntegerDigits[ #^#], n], {n, IntegerLength[#^#]}], #]&]
%o A262814 (PARI) 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
%o A262814 (Python)
%o A262814 A262814_list = []
%o A262814 for k in range(1,10**3):
%o A262814     n = k**k
%o A262814     if not n % k:
%o A262814         s = str(n)
%o A262814         for i in range(len(s)-1):
%o A262814             s = s[1:]+s[0]
%o A262814             if int(s) % k:
%o A262814                 break
%o A262814         else:
%o A262814             A262814_list.append(k) # _Chai Wah Wu_, Oct 26 2015
%Y A262814 Cf. A178028, A242680, A242740.
%K A262814 nonn,base
%O A262814 1,2
%A A262814 _Michel Lagneau_, Oct 03 2015
%E A262814 a(24)-a(27) from _Michel Marcus_, Oct 11 2015
%E A262814 a(28)-a(50) from _Chai Wah Wu_, Oct 26 2015