%I #51 Feb 22 2024 02:17:57
%S 0,1,1,1,2,3,3,1,3,5,2,3,4,4,5,1,4,8,4,4,8,8,4,3,8,7,7,6,5,13,6,1,10,
%T 11,7,7,10,9,9,5,7,18,7,5,14,11,6,3,10,11,9,8,7,15,9,4,14,12,5,10,9,
%U 10,11,1,11,19,10,6,17,21,6,8,14,12,13,7,14,21,7,4
%N Number of ordered ways to write n as (2^a*3^b)^2 + (2^c*5^d)^2 + x^2 + y^2, where a,b,c,d,x,y are nonnegative integers with x <= y.
%C Four-square Conjecture: a(n) > 0 for all n > 1.
%C This is much stronger than Lagrange's four-square theorem. We have verified a(n) > 0 for all n = 2..10^9.
%C Note that 16265031 cannot be written as (2^a*3^b)^2 + (2^c*3^d)^2 + x^2 + y^2 with a,b,c,d,x,y nonnegative integers.
%C a(n) > 0 for 1 < n <= 10^10. - _Giovanni Resta_, Jun 28 2019
%C I promise to offer 2500 US dollars as the prize for the first correct proof of the Four-square Conjecture. - _Zhi-Wei Sun_, Jul 09 2019
%C Jiao-Min Lin (a student at Nanjing University) has verified a(n) > 0 for all 1 < n <= 1.6*10^11. - _Zhi-Wei Sun_, Jul 30 2022
%H Zhi-Wei Sun, <a href="/A308734/b308734.txt">Table of n, a(n) for n = 1..10000</a>
%H Soumyarup Banerjee, <a href="https://doi.org/10.1016/j.jnt.2023.09.004">On a conjecture of Sun about sums of restricted squares</a>, J. Number Theory 256 (2024), 253-289.
%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="https://doi.org/10.1142/S1793042119501045">Restricted sums of four squares</a>, Int. J. Number Theory 15 (2019), 1863-1893.
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/135-solution.pdf">Various Refinements of Lagrange's Four-Square Theorem</a>, Westlake Number Theory Symposium (Nanjing University, China, 2020).
%H Zhi-Wei Sun, <a href="http://hitpress.hit.edu.cn/2021/1015/c12593a261001/page.htm">New Conjectures in Number Theory and Combinatorics</a> (in Chinese), Harbin Institute of Technology Press, 2021. (See Conjecture 5.16.)
%e a(2^(2k+1)) = 1 with 2^(2k+1) = (2^k*3^0)^2 + (2^k*5^0)^2 + 0^2 + 0^2.
%e a(2^(2k+2)) = 1 with 2^(2k+2) = (2^k*3^0)^2 + (2^k*5^0)^2 + (2^k)^2 + (2^k)^2.
%e a(3) = 1 with 3 = (2^0*3^0)^2 + (2^0*5^0)^2 + 0^2 + 1^2.
%e a(5) = 2 with 5 = (2^0*3^0)^2 + (2^1*5^0)^2 + 0^2 + 0^2 = (2^1*3^0)^2 + (2^0*5^0)^2 + 0^2 + 0^2.
%e a(11) = 2 with 11 = (2^0*3^0)^2 + (2^0*5^0)^2 + 0^2 + 3^2 = (2^0*3^1)^2 + (2^0*5^0)^2 + 0^2 + 1^2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t tab={};Do[r=0;Do[If[SQ[n-4^a*9^b-4^c*25^d-x^2],r=r+1],{a,0,Log[4,n]},{b,0,Ceiling[Log[9,n/4^a]]-1},
%t {c,0,Log[4,n-4^a*9^b]},{d,0,Log[25,(n-4^a*9^b)/4^c]},{x,0,Sqrt[(n-4^a*9^b-4^c*25^d)/2]}];tab=Append[tab,r],{n,1,80}];Print[tab]
%Y Cf. A000079, A000118, A000290, A000244, A000351, A271518, A281976, A303656, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308661, A308662.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Jun 21 2019