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a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/3 + 1, q = prime(p) - p + 1 and r = prime(q) - q + 1 are all prime}|, where phi(.) is Euler's totient function.
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%I #7 Apr 06 2014 04:09:21

%S 0,0,0,0,0,0,0,0,0,1,1,1,2,0,1,0,1,1,0,1,1,1,0,1,0,1,0,0,0,1,0,2,0,3,

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

%U 6,4,4,5,7,3,6,5,1,4,6,6,2,3,4,5,4,11,4,5,4,7,2,5,5,5,2,6,2,5,5,7

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

%C Conjecture: a(n) > 0 for all n > 37.

%C This implies that there are infinitely many primes p with q = prime(p) - p + 1 and r = prime(q) - q + 1 both prime.

%H Zhi-Wei Sun, <a href="/A235924/b235924.txt">Table of n, a(n) for n = 1..10000</a>

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014

%e a(20) = 1 since phi(6) + phi(14)/3 + 1 = 5, prime(5) - 4 = 11 - 4 = 7 and prime(7) - 6 = 17 - 6 = 11 are all prime.

%e a(77) = 1 since phi(59) + phi(18)/3 + 1 = 61, prime(61) - 60 = 283 - 60 = 223 and prime(223) - 222 = 1409 - 222 = 1187 are all prime.

%e a(1471) = 1 since phi(25) + phi(1446)/3 + 1 = 181, prime(181) - 180 = 1087 - 180 = 907 and prime(907) - 906 = 7057 - 906 = 6151 are all prime.

%t q[n_]:=Prime[n]-n+1

%t f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/3+1

%t p[n_,k_]:=PrimeQ[f[n,k]]&&PrimeQ[q[f[n,k]]]&&PrimeQ[q[q[f[n,k]]]]

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

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

%Y Cf. A000010, A000040, A234694, A234695.

%K nonn

%O 1,13

%A _Zhi-Wei Sun_, Jan 17 2014