%I
%S 0,0,0,0,3,2,2,1,1,4,2,2,1,1,2,2,1,1,2,2,1,1,2,1,1,4,3,3,1,1,2,2,1,1,
%T 2,2,1,1,2,1,1,1,1,2,2,1,1,1,1,5,3,3,3,3,3,3,3,2,3,3,2,3,3,2,2,2,3,4,
%U 3,2,2,2,3,3,2,3,4,3,2,2,3,3,2,2,3,3,2,2,2,3,3,3,2,2,2,2,4,2,2,3
%N Number of ways to write n = p^2 + q  pi(q) with p prime and q among 1, ..., n, where pi(.) is given by A000720.
%C For any positive integer q, clearly q + 1  pi(q+1)  (q  pi(q)) = 1 + pi(q)  pi(q+1) is 0 or 1. Thus {q  pi(q): q = 1, ..., n} = {1, 2, ..., npi(n)}. Note that n  pi(n) > n/2 for n > 8. If p is at least sqrt(n/2), then n  p^2 <= n/2 and hence n  p^2 = q  pi(q) for some q = 1, ..., n. So it can be proved that a(n) > 0 for all n > 4.  LiLu Zhao and _ZhiWei Sun_, Feb 26 2014
%H ZhiWei Sun, <a href="/A238402/b238402.txt">Table of n, a(n) for n = 1..10000</a>
%e a(9) = 1 since 9 = 2^2 + 9  pi(9) with 2 prime and pi(9) = 4.
%e a(40) = 1 since 40 = 5^2 + 24  pi(24) with 5 prime and pi(24) = 9.
%e a(120) = 1 since 120 = 7^2 + 95  pi(95) with 7 prime and pi(95) = 24.
%t SQ[n_]:=IntegerQ[Sqrt[n]]&&PrimeQ[Sqrt[n]]
%t a[n_]:=Sum[If[SQ[nq+PrimePi[q]],1,0],{q,1,n}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000040, A000290, A000720, A232443, A232463, A238386.
%K nonn
%O 1,5
%A _ZhiWei Sun_, Feb 26 2014
