A234567 Number of ways to write n = k + m with k > 0 and m > 0 such that p = phi(k) + phi(m)/2 + 1 and P(p-1) are both prime, where phi(.) is Euler's totient function and P(.) is the partition function (A000041).

0, 0, 0, 1, 2, 1, 1, 3, 2, 2, 3, 2, 4, 2, 4, 4, 2, 4, 3, 5, 1, 3, 2, 3, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 2, 0, 1, 2, 1, 1, 2, 1, 2, 3, 2, 8, 2, 1, 2, 2, 3, 3, 1, 2, 7, 0, 2, 3, 3, 4, 5, 7, 3, 4, 1, 9, 1, 4, 3, 1, 2

Offset: 1

Author

Zhi-Wei Sun, Dec 28 2013

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 727.

(ii) For the strict partition function q(.) (cf. A000009), any n > 93 can be written as k + m with k > 0 and m > 0 such that p = phi(k) + phi(m)/2 + 1 and q(p-1) - 1 are both prime.

(iii) If n > 75 is not equal to 391, then n can be written as k + m with k > 0 and m > 0 such that f(k,m) - 1, f(k,m) + 1 and q(f(k,m)) + 1 are all prime, where f(k,m) = phi(k) + phi(m)/2.

Part (i) of the conjecture implies that there are infinitely many primes p with P(p-1) prime.

Examples

			a(21) = 1 since 21 = 6 + 15 with  phi(6) + phi(15)/2 + 1 = 7 and P(6) = 11 both prime.
a(700) = 1 since 700 = 247 + 453 with phi(247) + phi(453)/2 + 1 = 367 and P(366) = 790738119649411319 both prime.
a(945) = 1 since 945 = 687 + 258 with phi(687) + phi(258)/2 + 1 = 499 and P(498) = 2058791472042884901563 both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Crossrefs

Cf. A000009, A000010, A000040, A000041, A234470, A234475, A234514, A234530.

Programs

Mathematica

f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/2
q[n_,k_]:=PrimeQ[f[n,k]+1]&&PrimeQ[PartitionsP[f[n,k]]]
a[n_]:=Sum[If[q[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]