OFFSET
1,6
COMMENTS
Conjecture: a(n) > 0 for all n > 5.
This implies that there are infinitely many odd primes p = 2*m + 1 with m*(m+1) - prime(m) and m*(m+1)- prime(m+1) both prime.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(8) = 2 since phi(4) + phi(4)/2 = 3 with 2*3 + 1 = 7, 3*4 - prime(3) = 7 and 3*4 - prime(4) = 5 all prime, and phi(5) + phi(3)/2 = 5 with 2*5 + 1 = 11, 5*6 - prime(5) = 19 and 5*6 - prime(6) = 17 all prime.
MATHEMATICA
q[n_]:=PrimeQ[2n+1]&&PrimeQ[n(n+1)-Prime[n]]&&PrimeQ[n(n+1)-Prime[n+1]]
f[n_, k_]:=EulerPhi[k]+EulerPhi[n-k]/2
a[n_]:=Sum[If[q[f[n, k]], 1, 0], {k, 1, n-3}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 16 2014
STATUS
approved