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%I #16 Nov 23 2013 23:44:24
%S 0,0,0,1,1,2,1,2,2,2,1,2,3,4,2,2,3,3,3,3,3,3,3,4,4,3,3,2,2,5,4,5,5,3,
%T 3,4,4,4,5,4,2,6,5,5,5,3,4,8,5,5,6,2,4,7,6,6,5,3,5,7,6,7,6,4,6,6,5,7,
%U 5,6,6,7,7,7,6,4,5,7,8,8,7,7,7,7,8,9,5,6,9,9,7,6,7,6,7,8,3,8,9,5
%N Number of ways to write n = p + q - pi(q), where p and q are odd primes, and pi(q) is the number of primes not exceeding q.
%C Conjecture: (i) a(n) > 0 for all n > 3. Moreover, every n = 6, 7, ... can be written as p + q - pi(q) with p, p + 6 and q all prime.
%C (ii) For each integer n > 7, there is a prime p < n with n + p - pi(p) prime.
%C (iii) Any integer n > 4 not equal to 9 or 17 can be written as p + q + pi(q) with p and q both prime.
%C (iv) Each integer n > 7 can be written as p + q + pi(p) + pi(q) with p and q both prime.
%H Zhi-Wei Sun, <a href="/A232443/b232443.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1211.1588">Conjectures involving primes and quadratic forms</a>, preprint, arXiv:1211.1588.
%e a(4) = 1 since 4 = 3 + 3 - pi(3) with 3 prime.
%e a(5) = 1 since 5 = 3 + 5 - pi(5) with 3 and 5 prime.
%e a(6) = 2 since 6 = 3 + 7 - pi(7) = 5 + 3 - pi(3) with 3, 5, 7 all prime.
%e a(7) = 1 since 7 = 5 + 5 - pi(5) with 5 prime.
%e a(11) = 1 since 11 = 5 + 11 - pi(11) with 5 and 11 both prime.
%t PQ[n_]:=PQ[n]=n>2&&PrimeQ[n]
%t a[n_]:=Sum[If[PQ[n-Prime[k]+k],1,0],{k,2,PrimePi[2n-2]}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000040, A000720, A002375, A232398.
%K nonn
%O 1,6
%A _Zhi-Wei Sun_, Nov 23 2013
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