%I #6 Sep 17 2014 20:48:21

%S 1,21,27,34,54,63,81,171,205,212,214,237,243,272,291,315,324,333,342,

%T 351,356,358,394,402,405,424,432,441,459,493,502,504,513,538,540,544,

%U 565,585,624,630,663,702,712,714,716,718,723,729,745,804,810,831,834,835

%N Numbers x such that the sum of all their cyclic permutations is equal to that of all cyclic permutations of their Euler totient functions phi(x).

%C The minimum number with all its cyclic permutations belonging to the sequence is 243: 243, 324, 432. But if a "0" is prepended to 54 then it could be considered the minimum one: 054, 405, 540.

%H Paolo P. Lava, <a href="/A247316/b247316.txt">Table of n, a(n) for n = 1..5000</a>

%e The sum of the cyclic permutations of 171 is 171 + 117 + 711 = 999; phi(171) = 108 and the sum of its cyclic permutations is 108 + 810 + 81 = 999.

%e The sum of the cyclic permutations of 1863 is 1863 + 3186 + 6318 + 8631 = 19998; phi(1863) = 1188 and the sum of its cyclic permutations is 1188 + 8118 + 8811 + 1881 = 19998.

%p with(numtheory):P:=proc(q) local a,b,c,d,k,n;

%p for n from 1 to q do a:=n; b:=a; c:=ilog10(a);

%p for k from 1 to c do a:=(a mod 10)*10^c+trunc(a/10); b:=b+a; od;

%p a:=phi(n); d:=a; c:=ilog10(a);

%p for k from 1 to c do a:=(a mod 10)*10^c+trunc(a/10); d:=d+a; od;

%p if d=b then print(n); fi; od; end: P(10^9);

%Y Cf. A000010, A247315, A247317.

%K nonn,base,easy

%O 1,2

%A _Paolo P. Lava_, Sep 12 2014