%I #29 Mar 02 2023 13:01:36
%S 0,1,2,3,2,2,3,4,3,2,3,6,4,2,2,4,4,2,2,5,5,5,4,4,4,4,4,5,4,5,4,4,3,4,
%T 3,4,4,5,6,5,5,5,4,7,3,3,4,6,4,2,3,5,6,3,4,5,6,5,2,5,4,5,3,2,4,5,4,3,
%U 3,3,6,7,5,5,6,10,6,3,4,8
%N Number of ordered pairs (k, m) with 0 <= k <= m such that n - binomial(2*k,k) - binomial(2*m,m) can be written as the sum of two squares.
%C Conjecture: a(n) > 0 for all n > 1.
%C a(n) > 0 for all n = 2..10^10.
%C See also A303540 and A303541 for related sequences.
%H Zhi-Wei Sun, <a href="/A303539/b303539.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.11.008">Refining Lagrange's four-square theorem</a>, J. Number Theory 175(2017), 167-190.
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/179b.pdf">New conjectures on representations of integers (I)</a>, Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1701.05868">Restricted sums of four squares</a>, arXiv:1701.05868 [math.NT], 2017-2018.
%e a(2) = 1 with 2 - binomial(2*0,0) - binomial(2*0,0) = 0^2 + 0^2.
%e a(3) = 2 with 3 - binomial(2*0,0) - binomial(2*0,0) = 0^2 + 1^2 and 3 - binomial(2*0,0) - binomial(2*1,1) = 0^2 + 0^2.
%e a(5) = 2 with 5 - binomial(2*0,0) - binomial(2*1,1) = 1^2 + 1^2 and 5 - binomial(2*1,1) - binomial(2*1,1) = 0^2 + 1^2.
%t c[n_]:=c[n]=Binomial[2n,n];
%t f[n_]:=f[n]=FactorInteger[n];
%t g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
%t QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
%t tab={};Do[r=0;k=0;Label[bb];If[c[k]>n,Goto[aa]];Do[If[QQ[n-c[k]-c[j]],r=r+1],{j,0,k}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]
%Y Cf. A000290, A000984, A001481, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303540, A303541, A303543.
%K nonn
%O 1,3
%A _Zhi-Wei Sun_, Apr 25 2018