OFFSET
1,8
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 5.
(ii) If n > 5 is not equal to 19, then phi(k) + phi(n-k) - 1 and phi(k) + phi(n-k) + 1 are both prime for some 0 < k < n.
(iii) If n > 5, then (phi(k)/2)^2 + (phi(n-k)/2)^2 is prime for some 0 < k < n.
(iv) If n > 8, then (sigma(k) + phi(n-k))/2 is prime for some 0 < k < n, where sigma(k) is the sum of all positive divisors of k.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, New representation problems involving Euler's totient function, a message to Number Theory List, Dec. 18, 2013.
EXAMPLE
a(6) = 1 since (phi(3) + phi(3))/2 = 2 is prime.
a(7) = 1 since (phi(3) + phi(4))/2 = 2 is prime.
a(10) = 1 since (phi(4) + phi (6))/2 = 2 is prime.
a(13) = 1 since (phi(3) + phi(10))/2 = 3 is prime.
a(20) = 1 since (phi(4) + phi(16))/2 = 5 is prime.
MATHEMATICA
a[n_]:=Sum[If[PrimeQ[(EulerPhi[k]+EulerPhi[n-k])/2], 1, 0], {k, 1, n/2}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 21 2013
STATUS
approved