A247317 Numbers x such that the sum of all their cyclic permutations is equal to that of all cyclic permutations of sigma(x) and all cyclic permutations of Euler totient function phi(x).

1, 2907, 3339, 3726, 4293, 4371, 4614, 5049, 5319, 5607, 5751, 6291, 17901, 18009, 18441, 19413, 20349, 20655, 20943, 21219, 21267, 21573, 21627, 22137, 22191, 23355, 24831, 25647, 25731, 26019, 26145, 26163, 27405, 27537, 28035, 28215, 28227, 28305, 29601, 30429

Offset

1,2

Comments

Intersection of A247315 and A247316.

All numbers appear to be multiples of 3.

Big steps between a(135) and a(136), a(1387) and a(1388),...

Links

Paolo P. Lava, Table of n, a(n) for n = 1..1500

Paolo P. Lava, Plot of the first 1500 terms of the sequence

Example

The sum of the cyclic permutations of 4371 is 4371 + 1437 + 7143 + 3714 = 16667; sigma(4371) = 6144 and the sum of its cyclic permutations is 6144 + 4614 + 4461 + 1446 = 16667; phi(4371) = 2760 and the sum of its cyclic permutations is2760+276+6027+7602 = 16667.
The sum of the cyclic permutations of 24831 is 24831 + 12483 + 31248 + 83124 + 48312 = 199998; sigma(24831) = 37440 and the sum of its cyclic permutations is 37440 + 3744 + 40374 + 44037 + 74403 = 199998; phi(24831) = 15840 and the sum of its cyclic permutations is 15840 + 1584 + 40158 + 84015 + 58401 = 199998.

Maple

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

Crossrefs

A000010, A000203, A247315, A247316.

Sequence in context: A054828 A153412 A186819 * A236742 A254183 A254190

Adjacent sequences: A247314 A247315 A247316 * A247318 A247319 A247320

Keyword

nonn,base,easy

Author

Paolo P. Lava, Sep 12 2014

Status

approved