A235919 a(n) = |{0 < k < n - 2: p = prime(k) + phi(n-k)/2, prime(p) - p + 1 and (p^2 - 1)/4 - prime(p) are all prime}|, where phi(.) is Euler's totient function.

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

Offset: 1

Author

Zhi-Wei Sun, Jan 17 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 24.

This implies that there are infinitely many primes p with prime(p) - p + 1 and (p^2 - 1)/4 - prime(p) both prime.

Examples

			 a(30) = 1 since prime(6) + phi(24)/2 = 13 + 4 = 17, prime(17) - 16 = 59 - 16 = 43 and (17^2 - 1)/4 - prime(17) = 72 - 59 = 13 are all prime.
  a(35) = 1 since prime(19) + phi(16)/2 = 67 + 4 = 71, prime(71) - 70 = 353 - 70 = 283 and (71^2 - 1)/4 - prime(71) = 1260 - 353 = 907 are all prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A234695, A235917.

Programs

Mathematica

PQ[n_]:=PQ[n]=n>0&&PrimeQ[n]
p[n_]:=p[n]=PrimeQ[n]&&PrimeQ[Prime[n]-n+1]&&PQ[(n^2-1)/4-Prime[n]]
f[n_,k_]:=f[n,k]=Prime[k]+EulerPhi[n-k]/2
a[n_]:=a[n]=Sum[If[p[f[n,k]],1,0],{k,1,n-3}]
Table[a[n],{n,1,100}]