A236138 a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/3 - 1, prime(p-1) - (p-1) and prime(p-1) - 2*prime((p-1)/2) are all prime}|, where phi(.) is Euler's totient function.

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Offset: 1

Author

Zhi-Wei Sun, Jan 19 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 73.

(ii) For any integer n > 472, there is a positive integer k < n such that p = phi(k) + phi(n-k)/4 - 1, prime(p) - 2*prime((p-1)/2) and prime(p) - 2*prime((p+1)/2) are all prime.

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

Examples

			a(20) = 1 since phi(11) + phi(9)/3 - 1 = 11, prime(10) - 10 = 29 - 10 = 19 and prime(10) - 2*prime(5) = 29 - 2*11 = 7 are all prime.
a(293) = 1 since phi(267) + phi(26)/3 - 1 = 176 + 12/3 - 1 = 179, prime(178) - 178 = 1061 - 178 = 883 and prime(178) - 2*prime(89) = 1061 - 2*461 = 139 are all prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A234694, A235924, A236074, A236097.

Programs

Mathematica

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