OFFSET
1,14
COMMENTS
If p is a prime and p != 3, then a(p) = 0. Proof: For p = 2, phi(1) + phi(1) = 2 > phi(2) = 1. For p > 3, phi(x) + phi(p-x) <= x - 1 + p - x - 1 = p - 2 < p - 1 = phi(p).
If a(2*i) = 0, then i is a positive odd number. Proof: If i is a positive even number, then 2*i = 2^k*(2*m-1) with integers k, m where k > 1 and m > 0. Since phi(2^k*(2*m-1)) = phi(2^k)*phi(2*m-1) = 2^(k-1)*phi(2*m-1) = 2*2^(k-2)*phi(2*m-1) = 2*phi(2^(k-1)*(2*m-1)), it follows that x = 2^(k-1)*(2*m-1) is a solution to phi(x) + phi(2^k*(2*m-1)-x) = phi(2^k*(2*m-1)).
LINKS
Felix Huber, Table of n, a(n) for n = 1..10000
FORMULA
a(p) = 0 for primes p != 3.
a(2^k*(2*m-1)) > 0 for integers k, m where k > 1 and m > 0.
Conjecture: a(2*A065381(n)) = 0 for n > 1.
EXAMPLE
a(20) = 3 because phi(x) + phi(20-x) = phi(20) has 3 solutions for 0 <= x <= 10:
x = 6: phi(6) + phi(14) = 2 + 6 = 8 = phi(20).
x = 8: phi(8) + phi(12) = 4 + 4 = 8 = phi(20).
x = 10: phi(10) + phi(10) = 4 + 4 = 8 = phi(20).
MAPLE
PROG
(PARI) a(n) = my(e=eulerphi(n)); sum(x=1, n\2, eulerphi(x) + eulerphi(n-x) == e); \\ Michel Marcus, Mar 22 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Felix Huber, Mar 22 2025
STATUS
approved
