A235917 a(n) = |{0 < k < n - 2: p = prime(k) + phi(n-k)/2, p^2 - 1 - prime(p) and (p^2 - 1)/2 - prime(p) are all prime}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 3, 1, 4, 2, 4, 3, 1, 2, 2, 3, 3, 1, 2, 2, 1, 1, 2, 4, 1, 5, 2, 2, 3, 2, 6, 2, 1, 3, 3, 2, 4, 5, 4, 2, 5, 3, 4, 2, 3, 4, 4, 3, 3, 2, 1, 4, 3, 2, 3, 4, 5, 7, 3, 5, 1, 6, 1, 7, 3, 6, 5, 3, 5, 2, 3, 4, 5, 3, 8, 6, 4, 2, 6, 4, 8, 3, 7, 5, 6, 6, 4, 3, 5, 6, 4, 3

Offset: 1

Author

Zhi-Wei Sun, Jan 17 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 14.

(ii) For any integer n > 12, there is a positive integer k < n such that p = prime(k) + phi(n-k), (p^2 - 1)/2 - prime(p) and (p^2 - 1)/4 - prime(p) are all prime.

Clearly, part (i) of the conjecture implies that there are infinitely many primes p with p^2 - 1 - prime(p) and (p^2 - 1)/2 - prime(p) both prime.

Examples

			a(10) = 1 since prime(5) + phi(5)/2 = 11 + 2 = 13, 13^2 - 1 - prime(13) = 168 - 41 = 127 and (13^2 - 1)/2 - prime(13) = 84 - 41 = 43 are all prime.
a(71) = 1 since prime(19) + phi(52)/2 = 67 + 12 = 79, 79^2 - 1 - prime(79) = 6240 - 401 = 5839 and (79^2 - 1)/2 - prime(79) = 3120 - 401 = 2719 are all prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A234694, A235912.

Programs

Mathematica

PQ[n_]:=n>0&&PrimeQ[n]
p[n_]:=PrimeQ[n]&&PQ[n^2-1-Prime[n]]&&PQ[(n^2-1)/2-Prime[n]]
f[n_,k_]:=Prime[k]+EulerPhi[n-k]/2
a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-3}]
Table[a[n],{n,1,100}]