OFFSET
1,14
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 100.
(ii) For any integer n > 4, there is a positive integer k < n with phi(k)^2 + prime(n-k)^2 prime.
Note that phi(k^2) (k = 1, 2, 3, ...) are pairwise distinct and this can be easily proved by induction.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(16) = 1 since 16 = 3^2 + 7 with phi(3^2) + prime(7) = 6 + 17 = 23 prime.
a(611) = 1 since 611 = 22^2 + 127 with phi(22^2) + prime(127) = 220 + 709 = 929 prime.
MATHEMATICA
p[n_, k_]:=PrimeQ[EulerPhi[k^2]+Prime[n-k^2]]
a[n_]:=Sum[If[p[n, k], 1, 0], {k, 1, Sqrt[n-1]}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 28 2014
STATUS
approved