A235919 a(n) = |{0 < k < n - 2: p = prime(k) + phi(n-k)/2, prime(p) - p + 1 and (p^2 - 1)/4 - prime(p) are all prime}|, where phi(.) is Euler's totient function.

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

Offset

1,18

Comments

Conjecture: a(n) > 0 for all n > 24.

This implies that there are infinitely many primes p with prime(p) - p + 1 and (p^2 - 1)/4 - prime(p) both prime.

Links

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

Example

 a(30) = 1 since prime(6) + phi(24)/2 = 13 + 4 = 17, prime(17) - 16 = 59 - 16 = 43 and (17^2 - 1)/4 - prime(17) = 72 - 59 = 13 are all prime.
  a(35) = 1 since prime(19) + phi(16)/2 = 67 + 4 = 71, prime(71) - 70 = 353 - 70 = 283 and (71^2 - 1)/4 - prime(71) = 1260 - 353 = 907 are all prime.

Mathematica

PQ[n_]:=PQ[n]=n>0&&PrimeQ[n]
p[n_]:=p[n]=PrimeQ[n]&&PrimeQ[Prime[n]-n+1]&&PQ[(n^2-1)/4-Prime[n]]
f[n_, k_]:=f[n, k]=Prime[k]+EulerPhi[n-k]/2
a[n_]:=a[n]=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-3}]
Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A000010, A000040, A234695, A235917.

Sequence in context: A231188 A249695 A136748 * A229654 A306288 A272188

Adjacent sequences: A235916 A235917 A235918 * A235920 A235921 A235922

Keyword

nonn

Author

Zhi-Wei Sun, Jan 17 2014

Status

approved