A235912 a(n) = |{0 < k < n - 2: 2*m + 1, m*(m-1) - prime(m) and m*(m+1) - prime(m) are all prime with m = phi(k) + phi(n-k)/2}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 3, 2, 4, 2, 6, 5, 6, 7, 4, 8, 7, 8, 8, 11, 7, 12, 9, 9, 12, 5, 14, 10, 9, 9, 9, 9, 7, 8, 11, 9, 8, 7, 14, 8, 6, 9, 5, 5, 9, 11, 3, 9, 6, 13, 8, 8, 6, 7, 6, 5, 4, 3, 1, 5, 5, 5, 6, 5, 7, 7, 4, 7, 11, 8, 3, 5, 3, 10, 4, 4, 3, 9, 2, 4, 4, 5, 8, 12, 13, 4, 9, 5, 11, 5, 12, 7, 4, 4

Offset: 1

Author

Zhi-Wei Sun, Jan 16 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 11.

This implies that there are infinitely many odd primes p = 2*m + 1 with q = m*(m-1) - prime(m) and r = m*(m+1) - prime(m) both prime. Note that r - q = 2*m.

Examples

			 a(10) = 1 since phi(5) + phi(5)/2 = 6 with 2*6 + 1 = 13, 5*6 - prime(6) = 30 - 13 = 17 and 6*7 - prime(6) = 42 - 13 = 29 all prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A235592, A235728.

Programs

Mathematica

PQ[n_]:=n>0&&PrimeQ[n]
p[n_]:=PrimeQ[2n+1]&&PQ[n(n-1)-Prime[n]]&&PQ[n(n+1)-Prime[n]]
f[n_,k_]:=EulerPhi[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}]