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A236511
a(n) = |{0 < k < n: p = 3*phi(k) + phi(n-k) - 1, p + 2, p + 6 and p + 8 are all prime}|, where phi(.) is Euler's totient function.
2
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 0, 0, 0, 0, 1, 1, 2, 1, 1, 1, 1, 0, 2, 2, 2, 2, 2, 2, 0, 2, 0, 4, 4, 2, 1, 3, 4, 2, 2, 3, 0, 1, 3, 2, 3, 1, 4, 4, 3, 1
OFFSET
1,13
COMMENTS
Conjecture: a(n) > 0 for all n > 1075.
We have verified this for n up to 50000.
The above conjecture implies the well-known conjecture that there are infinitely many prime quadruplets (p, p + 2, p + 6, p + 8).
EXAMPLE
a(10) = 1 since 3*phi(3) + phi(7) - 1 = 6 + 6 - 1 = 11, 11 + 2 = 13, 11 + 6 = 17 and 11 + 8 = 19 are all prime.
a(57) = 1 since 3*phi(31) + phi(26) - 1 = 90 + 12 - 1 = 101, 101 + 2 = 103, 101 + 6 = 107 and 101 + 8 = 109 are all prime.
MATHEMATICA
p[n_]:=PrimeQ[n]&&PrimeQ[n+2]&&PrimeQ[n+6]&&PrimeQ[n+8]
f[n_, k_]:=3*EulerPhi[k]+EulerPhi[n-k]-1
a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 27 2014
STATUS
approved