A233918 a(n) = |{0 < k <= n/2: (phi(k) + phi(n-k))/2 is prime}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 1, 1, 2, 2, 1, 2, 3, 1, 3, 2, 4, 3, 2, 7, 1, 3, 3, 4, 7, 2, 4, 5, 5, 5, 5, 6, 6, 4, 7, 5, 6, 4, 4, 11, 5, 5, 5, 11, 4, 3, 5, 7, 12, 4, 6, 11, 3, 6, 7, 8, 6, 7, 8, 11, 10, 5, 9, 7, 9, 5, 4, 14, 8, 9, 6, 10, 7, 6, 10, 9, 10, 7, 10, 11, 7, 7, 13, 11, 13, 5, 8, 11, 9, 9, 3, 12, 4, 11, 13, 11, 19, 8, 12, 11, 7

Offset: 1

Author

Zhi-Wei Sun, Dec 21 2013

Keywords

• nonn

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.

Examples

			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.
		

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.

Crossrefs

Cf. A000010, A000040, A233542, A233544, A233547, A233566, A233567, A233867, A234200.

Programs

Mathematica

a[n_]:=Sum[If[PrimeQ[(EulerPhi[k]+EulerPhi[n-k])/2],1,0],{k,1,n/2}]
Table[a[n],{n,1,100}]