A236074 - OEIS

%I #10 Jan 19 2014 04:44:00

%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,1,1,1,0,0,1,1,0,1,1,2,1,1,2,0,1,

%T 0,0,0,0,2,2,0,1,0,0,1,1,0,1,0,0,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0,0,0,

%U 0,1,0,0,0,2,0,0,0,1,0,4,0,0,1,2,0,1,2,1,0,3,4,0,0,0,2,1,3,1,2,0

%N a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/6 + 1, prime(2*p) - 2*prime(p) and prime(p) - 2*prime((p-1)/2) are all prime}|, where phi(.) is Euler's totient function.

%C Conjecture: (i) a(n) > 0 for all n > 116.

%C (ii) For any integer n > 196, there is a positive integer k < n such that p = phi(k) + phi(n-k)/6 + 1, prime(2*p) - 2*prime(p) and prime(p-1) - 2*prime((p-1)/2) are all prime.

%C Clearly, part (i) (or part (ii))  implies that there are infinitely many odd primes p with prime(2*p) - 2*prime(p) and prime(p) - 2*prime((p-1)/2) (or prime(p-1) - 2*prime((p-1)/2), resp.) both prime.

%H Zhi-Wei Sun, <a href="/A236074/b236074.txt">Table of n, a(n) for n = 1..10000</a>

%e a(30) = 1 since phi(4) + phi(26)/6 + 1 = 5, prime(2*5) - 2*prime(5) = 29 - 2*11 = 7 and prime(5) - 2*prime((5-1)/2) = 11 - 2*3 = 5 are all prime.

%e a(204) = 1 since phi(159) + phi(45)/6 + 1 = 109, prime(2*109) - 2*prime(109) = 1361 - 2*599 = 163, and prime(109) - 2*prime((109-1)/2) = 599 - 2*251 = 97 are all prime.

%t PQ[n_]:=PQ[n]=n>0&&PrimeQ[n]

%t p[n_]:=PQ[n]&&PQ[Prime[2n]-2Prime[n]]&&PQ[Prime[n]-2*Prime[(n-1)/2]]

%t f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/6+1

%t a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-1}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000010, A000040, A234694, A235924, A236075.

%K nonn

%O 1,29

%A _Zhi-Wei Sun_, Jan 19 2014