A281707 Even integers k such that phi(sum of even divisors of k) = sum of odd divisors of k.

2, 6, 14, 42, 62, 186, 254, 434, 762, 1302, 1778, 5334, 7874, 16382, 23622, 49146, 55118, 114674, 165354, 262142, 344022, 507842, 786426, 1048574, 1523526, 1834994, 2080514, 3145722, 3554894, 5504982, 6241542, 7340018, 8126402, 10664682, 14563598, 22020054

Offset: 1

Author

Michel Lagneau, Jan 28 2017

Keywords

• nonn

Comments

The number of divisors of a(n) is a power of 2, and sum of even divisors = 2^(m+1), sum of odd divisors = 2^m for some m.

a(n) == 2, 6 (mod 8) or a(n) == 2, 6 (mod 12).

a(n) is of the form 2*p1*p2*...pk where p1, p2, ..., pk are Mersenne primes = 3, 7, 31, 127, 8191, ... (see A000668).

Examples

			62 is a term because its divisors are 1, 2, 31 and 62, the sum of the even divisors of 62 = 62 + 2 = 2^6, the sum of odd divisors = 1 + 31 = 2^5, and phi(2^6) = 2^5.
		

Crossrefs

Cf. A000010, A000593, A000668, A146076.

Programs

Maple

with(numtheory):
for n from 2 by 2 to 10^6 do:
x:=divisors(n):n1:=nops(x):s0:=0:s1:=0:
   for k from 1 to n1 do:
    if irem(x[k],2)=0
     then
     s0:=s0+ x[k]:
     else
     s1:=s1+ x[k]:
    fi:
  od:
    if s1=phi(s0)
     then
     print(n):
     else
    fi:
od:

Mathematica

Select[2 * Range[10^6], (sodd = (s = DivisorSigma[1, #])/(2^(IntegerExponent[#, 2]+1) - 1)) == EulerPhi[s - sodd] &] (* Amiram Eldar, Aug 12 2023 *)

PARI

isok(n) = eulerphi(sumdiv(n, d, d*((d % 2)==0))) == sumdiv(n, d, d*(d%2)); \\ Michel Marcus, Jan 28 2017

Extensions

a(1) inserted by Amiram Eldar, Aug 12 2023