A235728 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, 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, 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, 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

Offset: 1

Author

Zhi-Wei Sun, Jan 15 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 5, and a(n) = 1 only for n = 191.

(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.

(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).

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.

Examples

			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.
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.
		

Links

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

Crossrefs

Cf. A000010, A000040, A235508, A235592, A235682, A235703, A235727.

Programs

Mathematica

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