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

1,8

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

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

Sequence in context: A284155 A002345 A321325 * A071694 A365393 A366286

Adjacent sequences: A233915 A233916 A233917 * A233919 A233920 A233921

Keyword

nonn

Author

Zhi-Wei Sun, Dec 21 2013

Status

approved