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A233918 a(n) = |{0 < k <= n/2: (phi(k) + phi(n-k))/2 is prime}|, where phi(.) is Euler's totient function. 8
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 (list; graph; refs; listen; history; text; internal format)
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 A072781 A236566

Adjacent sequences:  A233915 A233916 A233917 * A233919 A233920 A233921

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 21 2013

STATUS

approved

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Last modified September 17 13:00 EDT 2019. Contains 327131 sequences. (Running on oeis4.)