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%I #14 Jan 15 2014 22:23:26
%S 0,0,0,0,0,2,3,2,4,4,4,4,4,5,3,6,4,7,4,3,6,5,6,4,7,4,7,3,5,5,5,6,6,6,
%T 3,6,3,4,2,2,4,3,4,5,4,3,6,4,2,4,2,4,3,3,6,4,2,6,8,6,10,4,6,7,4,6,6,8,
%U 6,6,2,9,5,9,10,12,4,10,6,10,6,9,5,11,10,7,10,10,6,9,11,7,8,8,13,6,5,5,6,9
%N 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.
%C Conjecture: (i) a(n) > 0 for all n > 5, and a(n) = 1 only for n = 191.
%C (ii) If n > 8 is not equal to 32, then there is a positive integer k < n - 2 such that 2*m + 1, m*(m+1) - prime(m) and m*(m+1) + prime(m) are all prime, where m = sigma(k) + phi(n-k)/2, and sigma(k) is the sum of all positive divisors of k.
%C (iii) If n > 444, then there is a positive integer k < n such that 2*m + 1, m^2 - prime(m) and m^2 + prime(m) are all prime, where m = sigma(k) + phi(n-k).
%C Clearly, part (i) of the conjecture implies that there are infinitely many odd primes p = 2*m + 1 with m*(m+1) - prime(m) = (p^2-1)/4 - prime((p-1)/2) and m*(m+1) + prime(m) = (p^2-1)/4 + prime((p-1)/2) both prime.
%H Zhi-Wei Sun, <a href="/A235728/b235728.txt">Table of n, a(n) for n = 1..10000</a>
%e a(6) = 2 since phi(1) + phi(5)/2 = phi(3) + phi(3)/2 = 3 with 2*3 + 1 = 7, 3*4 - prime(3) = 7 and 3*4 + prime(3) = 17 all prime.
%e a(191) = 1 since phi(153) + phi(38)/2 = 105 with 2*105 + 1 = 211, 105*106 - prime(105) = 11130 - 571 = 10559 and 105*106 + prime(105) = 11130 + 571 = 11701 all prime.
%t q[n_]:=PrimeQ[2n+1]&&PrimeQ[n(n+1)-Prime[n]]&&PrimeQ[n(n+1)+Prime[n]]
%t f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/2
%t a[n_]:=Sum[If[q[f[n,k]],1,0],{k,1,n-3}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000010, A000040, A235508, A235592, A235682, A235703, A235727.
%K nonn
%O 1,6
%A _Zhi-Wei Sun_, Jan 15 2014
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