|
%I #20 Aug 05 2019 09:26:41
%S 0,0,0,1,2,1,2,1,3,4,4,8,4,6,3,1,7,3,8,5,2,9,2,11,8,7,5,4,8,7,8,8,8,7,
%T 5,9,5,10,9,7,13,9,11,10,14,5,11,10,10,11,12,7,13,10,10,8,15,11,12,11,
%U 13,14,6,12,11,22,21,5,15,7,13,15,17,15,10,16,11,13,14,12,17,12,16,16,19,22,17,12,19,17,19,17,16,17,18,20,19,17,10,16
%N a(n) = |{0 < k < n: prime(k)^2 + 4*prime(n-k)^2 is prime}|
%C Conjecture: (i) a(n) > 0 for all n > 3.
%C (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.
%C (iii) If n > 5, then prime(k)^3 + 2*p(n-k)^3 is prime for some 0 < k < n, where p(.) is the partition function (A000041). If n > 2, then prime(k)^3 + 2*q(n-k)^3 is prime for some 0 < k < n, where q(.) is the strict partition function (A000009).
%H Zhi-Wei Sun, <a href="/A233439/b233439.txt">Table of n, a(n) for n = 1..10000</a>
%H Z.-W. Sun, <a href="http://arxiv.org/abs/1312.1166">On a^n+ bn modulo m</a>, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.
%H Z.-W. Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641 [math.NT], 2014-2016.
%e a(4) = 1 since prime(3)^2 + 4*prime(1)^2 = 5^2 + 4*2^2 = 41 is prime.
%e a(6) = 1 since prime(5)^2 + 4*prime(1)^2 = 11^2 + 4*2^2 = 137 is prime.
%e a(8) = 1 since prime(3)^2 + 4*prime(5)^2 = 5^2 + 4*11^2 = 509 is prime.
%e a(16) = 1 since prime(6)^2 + 4*prime(10)^2 = 13^2 + 4*29^2 = 3533 is prime.
%t a[n_]:=Sum[If[PrimeQ[Prime[k]^2+4*Prime[n-k]^2],1,0],{k,1,n-1}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000040, A002313, A220413, A232174, A232269, A232465, A232502, A233150, A233204, A233206, A233296, A233307, A233346.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Dec 09 2013
|
