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

1,6

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.

Links

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

Example

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.

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}]

Crossrefs

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

Sequence in context: A318048 A235805 A168231 * A205546 A322348 A081315

Adjacent sequences: A235725 A235726 A235727 * A235729 A235730 A235731

Keyword

nonn

Author

Zhi-Wei Sun, Jan 15 2014

Status

approved