A234809 a(n) = |{0 < k < n: p = k + phi(n-k) and 2*(n-p) + 1 are both prime}|, where phi(.) is Euler's totient function.

0, 0, 1, 2, 1, 3, 1, 4, 1, 1, 1, 5, 3, 7, 3, 1, 1, 7, 5, 9, 4, 2, 1, 9, 5, 2, 4, 3, 1, 10, 5, 14, 2, 2, 2, 1, 6, 14, 5, 4, 1, 15, 5, 16, 5, 5, 3, 17, 8, 4, 5, 6, 3, 17, 7, 5, 2, 6, 6, 17, 11, 25, 3, 5, 3, 1, 11, 25, 4, 4, 4, 22, 10, 26, 6, 7, 8, 3, 9, 26, 7, 9, 6, 25, 8, 3, 7, 9, 10, 25, 15, 6, 2, 9, 9, 2, 13, 29, 3, 7

Offset: 1

Author

Zhi-Wei Sun, Dec 30 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 2.

Clearly, this implies Lemoine's conjecture which states that any odd number 2*n + 1 > 5 can be written as 2*p + q with p and q both prime.

See also A234808 for a similar conjecture.

Examples

			a(5) = 1 since 1 + phi(4) = 3 and 2*(5-3) + 1 = 5 are both prime.
a(16) = 1 since 7 + phi(9) = 13 and 2*(16-13) + 1 = 7 are both prime.
a(41) = 1 since 7 +phi(34) = 23 and 2*(41-23) + 1 = 37 are both prime.
a(156) = 1 since 131 + phi(25) = 151 and 2*(156-151) + 1 = 11 are both prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A046927, A234470, A234475, A234514, A234567, A234615, A234694, A234808

Programs

Mathematica

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