A234514 Number of ways to write n = k + m with k > 0 and m > 0 such that p = k + phi(m)/2 and q(p) + 1 are both prime, where phi(.) is Euler's totient function, and q(.) is the strict partition function (A000009).

0, 0, 0, 1, 2, 2, 1, 1, 1, 0, 1, 0, 2, 2, 2, 3, 4, 2, 4, 2, 3, 3, 3, 2, 2, 3, 1, 4, 2, 1, 4, 2, 4, 2, 5, 3, 4, 1, 5, 6, 4, 2, 5, 5, 5, 3, 5, 4, 6, 3, 5, 7, 10, 2, 4, 5, 6, 5, 5, 2, 3, 5, 6, 6, 4, 2, 5, 3, 7, 4, 5, 3, 8, 7, 2, 5, 9, 3, 3, 2, 9, 9, 6, 6, 7, 6, 9, 4, 7, 4, 10, 8, 6, 11, 11, 4, 6, 4, 9, 7

Offset: 1

Author

Zhi-Wei Sun, Dec 27 2013

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 12.

(ii) For any integer n > 4, there is a prime p < n - 2 such that q(p + phi(n-p)/2) + 1 is prime.

Clearly, part (i) of the conjecture implies that there are infinitely many primes p with q(p) + 1 prime (cf. A234530).

We have verified part (i) for n up to 10^5.

Examples

			a(11) = 1 since 11 = 1 + 10 with 1 + phi(10)/2 = 3 and q(3) + 1 = 3 both prime.
a(27) = 1 since 27 = 7 + 20 with 7 + phi(20)/2 = 11 and q(11) + 1 = 13 both prime.
a(30) = 1 since 30 = 8 + 22 with 8 + phi(22)/2 = 13 and q(13) + 1 = 19 both prime.
a(38) = 1 since 38 = 21 + 17 with 21 + phi(17)/2 = 29 and q(29) + 1 = 257 both prime.
a(572) = 1 since 572 = 77 + 495 with 77 + phi(495)/2 = 197 and q(197) + 1 = 406072423 both prime.
a(860) = 1 since 860 = 523 + 337 with 523 + phi(337)/2 = 691 and q(691) + 1 = 712827068077888961 both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000009, A000010, A000040, A229835, A233307, A233390, A233417, A234451, A234470, A234475, A234503, A234504, A234530

Programs

Mathematica

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