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

1,4

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.

Links

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

Example

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.

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}]

Crossrefs

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

Sequence in context: A325252 A243334 A355531 * A135591 A114897 A360614

Adjacent sequences: A234806 A234807 A234808 * A234810 A234811 A234812

Keyword

nonn

Author

Zhi-Wei Sun, Dec 30 2013

Status

approved