A235917 a(n) = |{0 < k < n - 2: p = prime(k) + phi(n-k)/2, p^2 - 1 - prime(p) and (p^2 - 1)/2 - prime(p) are all prime}|, where phi(.) is Euler's totient function.

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

Offset

1,12

Comments

Conjecture: (i) a(n) > 0 for all n > 14.

(ii) For any integer n > 12, there is a positive integer k < n such that p = prime(k) + phi(n-k), (p^2 - 1)/2 - prime(p) and (p^2 - 1)/4 - prime(p) are all prime.

Clearly, part (i) of the conjecture implies that there are infinitely many primes p with p^2 - 1 - prime(p) and (p^2 - 1)/2 - prime(p) both prime.

Links

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

Example

a(10) = 1 since prime(5) + phi(5)/2 = 11 + 2 = 13, 13^2 - 1 - prime(13) = 168 - 41 = 127 and (13^2 - 1)/2 - prime(13) = 84 - 41 = 43 are all prime.
a(71) = 1 since prime(19) + phi(52)/2 = 67 + 12 = 79, 79^2 - 1 - prime(79) = 6240 - 401 = 5839 and (79^2 - 1)/2 - prime(79) = 3120 - 401 = 2719 are all prime.

Mathematica

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

Crossrefs

Cf. A000010, A000040, A234694, A235912.

Sequence in context: A089990 A071427 A248813 * A093949 A324518 A108807

Adjacent sequences: A235914 A235915 A235916 * A235918 A235919 A235920

Keyword

nonn

Author

Zhi-Wei Sun, Jan 17 2014

Status

approved