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A236508
a(n) = |{0 < k < n-2: p = 2*phi(k) + phi(n-k)/2 - 1, p + 2, p + 6 and prime(p) + 6 are all prime}|, where phi(.) is Euler's totient function.
3
0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 2, 1, 1, 2, 1, 3, 2, 2, 0, 2, 3, 1, 2, 1, 3, 3, 2, 2, 1, 1, 1, 3, 0, 2, 3, 2, 1, 3, 0, 2, 0, 1, 1, 1, 1, 2, 0, 0, 0, 0, 2, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 2, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1
OFFSET
1,11
COMMENTS
Conjecture: a(n) > 0 for all n > 146.
We have verified this for n up to 52000.
The conjecture implies that there are infinitely many prime triples {p, p + 2, p + 6} with {prime(p), prime(p) + 6} a sexy prime pair. See A236509 for such primes p.
EXAMPLE
a(13) = 1 since 2*phi(3) + phi(10)/2 - 1 = 5, 5 + 2 = 7, 5 + 6 = 11 and prime(5) + 6 = 11 + 6 = 17 are all prime.
a(244) = 1 since 2*phi(153) + phi(244-153)/2 - 1 = 2*96 + 72/2 - 1 = 227, 227 + 2 = 229, 227 + 6 = 233 and prime(227) + 6 = 1433 + 6 = 1439 are all prime.
MATHEMATICA
p[n_]:=PrimeQ[n]&&PrimeQ[n+2]&&PrimeQ[n+6]&&PrimeQ[Prime[n]+6]
f[n_, k_]:=2*EulerPhi[k]+EulerPhi[n-k]/2-1
a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-3}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 27 2014
STATUS
approved