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A236619
a(n) = |{0 < k < n: prime(m)^3 + 2*m^3 and m^3 + 2*prime(m)^3 are both prime with m = 3*phi(k) + phi(n-k) - 1}|, where phi(.) is Euler's totient function.
1
0, 1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 1, 2, 0, 0, 2, 2, 1, 1, 0, 1, 1, 2, 2, 1, 2, 4, 0, 1, 3, 0, 2, 3, 3, 2, 3, 1, 3, 2, 3, 3, 2, 4, 3, 4, 2, 0, 2, 5, 4, 2, 4, 2, 2, 3, 5, 5, 6
OFFSET
1,3
COMMENTS
Conjecture: a(n) > 0 for every n = 90, 91, ....
We have verified this for n up to 100000.
The conjecture implies that there are infinitely many positive integers m with prime(m)^3 + 2*m^3 and m^3 + 2*prime(m)^3 both prime.
EXAMPLE
a(51) = 1 since 3*phi(35) + phi(51-35) - 1 = 3*24 + 8 - 1 = 79 with prime(79)^3 + 2*79^3 = 401^3 + 2*79^3 = 65467279 and 79^3 + 2*prime(79)^3 = 79^3 + 2*401^3 = 129455441 both prime.
MATHEMATICA
p[n_]:=PrimeQ[Prime[n]^3+2*n^3]&&PrimeQ[n^3+2*Prime[n]^3]
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}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 29 2014
STATUS
approved