A308734 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.

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, 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, 10, 11, 1, 11, 19, 10, 6, 17, 21, 6, 8, 14, 12, 13, 7, 14, 21, 7, 4

Offset: 1

Author

Zhi-Wei Sun, Jun 21 2019

Keywords

• nonn

Comments

Four-square Conjecture: a(n) > 0 for all n > 1.

This is much stronger than Lagrange's four-square theorem. We have verified a(n) > 0 for all n = 2..10^9.

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.

a(n) > 0 for 1 < n <= 10^10. - Giovanni Resta, Jun 28 2019

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

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

Examples

			a(2^(2k+1)) = 1 with 2^(2k+1) = (2^k*3^0)^2 + (2^k*5^0)^2 + 0^2 + 0^2.
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.
a(3) = 1 with 3 = (2^0*3^0)^2 + (2^0*5^0)^2 + 0^2 + 1^2.
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.
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.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Soumyarup Banerjee, On a conjecture of Sun about sums of restricted squares, J. Number Theory 256 (2024), 253-289.
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175 (2017), 167-190.
• Zhi-Wei Sun, Restricted sums of four squares, Int. J. Number Theory 15 (2019), 1863-1893.
• Zhi-Wei Sun, Various Refinements of Lagrange's Four-Square Theorem, Westlake Number Theory Symposium (Nanjing University, China, 2020).
• Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021. (See Conjecture 5.16.)

Crossrefs

Cf. A000079, A000118, A000290, A000244, A000351, A271518, A281976, A303656, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308661, A308662.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
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},
{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]