A247316 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).

1, 21, 27, 34, 54, 63, 81, 171, 205, 212, 214, 237, 243, 272, 291, 315, 324, 333, 342, 351, 356, 358, 394, 402, 405, 424, 432, 441, 459, 493, 502, 504, 513, 538, 540, 544, 565, 585, 624, 630, 663, 702, 712, 714, 716, 718, 723, 729, 745, 804, 810, 831, 834, 835

Offset: 1

Author

Paolo P. Lava, Sep 12 2014

Keywords

• nonn
• base
• easy

Comments

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.

Examples

			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.
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.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A000010, A247315, A247317.

Programs

Maple

with(numtheory):P:=proc(q) local a,b,c,d,k,n;
for n from 1 to q do a:=n; b:=a; c:=ilog10(a);
for k from 1 to c do a:=(a mod 10)*10^c+trunc(a/10); b:=b+a; od;
a:=phi(n); d:=a; c:=ilog10(a);
for k from 1 to c do a:=(a mod 10)*10^c+trunc(a/10); d:=d+a; od;
if d=b then print(n); fi; od; end: P(10^9);