%I #13 Feb 26 2021 22:53:12
%S 1,2,2,3,4,4,4,6,4,6,6,7,6,7,5,6,10,5,9,10,7,7,9,9,4,12,10,9,8,7,10,9,
%T 10,7,15,10,6,13,10,9,10,16,10,10,9,8,15,9,8,15,12,12,7,12,11,14,12,8,
%U 16,6,10,11,14,8,11,17,10,16,9,13,16,15,8,18,13,10,14,10,12,16,12,13,18,11,9,17,17,9,15,16,15,9,12,12,17,12,9,21,10,11
%N Number of ordered ways to write n as 2^x + y^2 + pi(z^2) with x >= 0, y >= 0 and z > 0, where pi(m) denotes the number of primes not exceeding m.
%C Conjecture: (i) a(n) > 0 for all n > 0.
%C (ii) Each positive integer can be written as 2^x + pi(y^2) + pi(z^2) with x >= 0, y > 0 and z > 0.
%D Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.
%H Zhi-Wei Sun, <a href="/A262976/b262976.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641 [math.NT], 2014.
%e a(1) = 1 since 1 = 2^0 + 0^2 + pi(1^2).
%e a(2) = 2 since 2 = 2^0 + 1^2 + pi(1^2) = 2 + 0^2 + pi(1^2).
%e a(3) = 2 since 3 = 2^0 + 0^2 + pi(2^2) = 2 + 1^2 + pi(1^2).
%t SQ[n_]:=IntegerQ[Sqrt[n]]
%t f[n_]:=PrimePi[n^2]
%t Do[r=0;Do[If[f[x]>=n,Goto[aa]];Do[If[2^y>n-f[x],Goto[bb]];If[SQ[n-f[x]-2^y],r=r+1],{y,0,Log[2,n-f[x]]}];Label[bb];Continue,{x,1,n}];Label[aa];Print[n," ",r];Continue,{n,1,100}]
%Y Cf. A000079, A000290, A000720, A038107, A262746, A262887.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Oct 05 2015