OFFSET
1,12
COMMENTS
Conjecture: a(n) > 0 for all n > 11.
This implies that there are infinitely many odd primes p = 2*m + 1 with q = m*(m-1) - prime(m) and r = m*(m+1) - prime(m) both prime. Note that r - q = 2*m.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(10) = 1 since phi(5) + phi(5)/2 = 6 with 2*6 + 1 = 13, 5*6 - prime(6) = 30 - 13 = 17 and 6*7 - prime(6) = 42 - 13 = 29 all prime.
MATHEMATICA
PQ[n_]:=n>0&&PrimeQ[n]
p[n_]:=PrimeQ[2n+1]&&PQ[n(n-1)-Prime[n]]&&PQ[n(n+1)-Prime[n]]
f[n_, k_]:=EulerPhi[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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 16 2014
STATUS
approved