OFFSET
1,5
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 3.
(ii) For any integer n > 10, prime(j)^3 + 2*prime(n-j)^2 is prime for some 0 < j < n, and prime(k)^3 + 2*prime(n-k)^3 is prime for some 0 < k < n.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Z.-W. Sun, On a^n+ bn modulo m, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014-2016.
EXAMPLE
a(4) = 1 since prime(3)^2 + 4*prime(1)^2 = 5^2 + 4*2^2 = 41 is prime.
a(6) = 1 since prime(5)^2 + 4*prime(1)^2 = 11^2 + 4*2^2 = 137 is prime.
a(8) = 1 since prime(3)^2 + 4*prime(5)^2 = 5^2 + 4*11^2 = 509 is prime.
a(16) = 1 since prime(6)^2 + 4*prime(10)^2 = 13^2 + 4*29^2 = 3533 is prime.
MATHEMATICA
a[n_]:=Sum[If[PrimeQ[Prime[k]^2+4*Prime[n-k]^2], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 09 2013
STATUS
approved