%I #12 Apr 25 2018 14:01:40
%S 0,1,1,1,1,2,1,2,2,2,2,3,3,2,2,2,3,2,2,2,2,3,2,2,1,4,3,3,2,5,4,4,4,3,
%T 3,4,4,3,4,4,4,3,2,4,3,3,3,5,2,4,5,4,4,4,4,3,5,3,4,4,4,4,4,3,3,5,4,5,
%U 3,3,5,5,2,4,6,3,3,4,4,3
%N Number of ordered pairs (a, b) with 0 <= a <= b such that n - 5^a - 5^b can be written as the sum of two triangular numbers.
%C Conjecture: a(n) > 0 for all n > 1.
%C This is equivalent to the author's conjecture in A303389. It has been verified that a(n) > 0 for all n = 2..6*10^9.
%C Note that a nonnegative integer m is the sum of two triangular numbers if and only if 4*m + 1 can be written as the sum of two squares.
%H Zhi-Wei Sun, <a href="/A303399/b303399.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(6) = 2 with 6 - 5^0 - 5^0 = 1*(1+1)/2 + 2*(2+1)/2 and 6 - 5^0 - 5^1 = 0*(0+1)/2 + 0*(0+1)/2.
%e a(7) = 1 with 7 - 5^0 - 5^1 = 0*(0+1)/2 + 1*(1+1)/2.
%e a(25) = 1 with 25 - 5^1 - 5^1 = 0*(0+1)/2 + 5*(5+1)/2.
%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;Do[If[QQ[4(n-5^j-5^k)+1],r=r+1],{j,0,Log[5,n/2]},{k,j,Log[5,n-5^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]
%Y Cf. A000217, A000351, A271518, A273812, A281976, A299924, A300219, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234, A303235, A303338, A303363, A303389, A303393.
%K nonn
%O 1,6
%A _Zhi-Wei Sun_, Apr 23 2018