A308621 -id:A308621 - cp's OEIS frontend

A308623 Number of ways to write n as a(a+1)/2 + b(b+1)/2 + 2^c*10^(2d) with a,b,c,d nonnegative integers.

1, 2, 2, 3, 3, 2, 3, 5, 2, 4, 4, 3, 3, 5, 3, 3, 6, 4, 4, 5, 2, 6, 5, 4, 4, 5, 2, 4, 6, 3, 4, 8, 5, 3, 5, 4, 5, 8, 5, 5, 4, 3, 5, 7, 4, 5, 8, 4, 2, 8, 2, 6, 7, 4, 3, 4, 6, 5, 8, 4, 4, 6, 5, 5, 5, 5, 6, 8, 4, 6, 7, 4, 6, 10, 4, 4, 7, 5, 2, 10, 4, 7, 7, 4, 8, 4, 4, 7, 8, 2, 4, 9, 5, 5, 9, 5, 5, 7, 5, 6

Offset: 1

Zhi-Wei Sun, Jun 11 2019

nonn

Conjecture: a(n) > 0 for all n > 0. Equivalently, any positive integer n can be written as a*(a+1)/2 + b*(b+1)/2 + 2^c*10^(2d) with a,b,c,d nonnegative integers.
This was motivated by A308566, and we verified a(n) > 0 for all n = 1..2*10^8. Then, on the author's request, Giovanni Resta verified the above conjecture for n up to 10^10. G. Resta also noted that 729546026 cannot be written as a*(a+1)/2 + b*(b+1)/2 + 2^c*3^d with a,b,c,d nonnegative integers.
See also A308566, A308594 and A308621 for similar conjectures.

			a(1) = 1 with 1 = 0*1/2 + 0*1/2 + 2^0*10^(2*0).
a(10107) = 1 with 10107 = 82*83/2 + 96*97/2 + 2^11*10^(2*0).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000079, A000217, A011557, A308566, A308584, A308621.

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[r=0;Do[If[TQ[n-10^(2k)*2^m-x(x+1)/2],r=r+1],{k,0,Log[10,n]/2},{m,0,Log[2,n/10^(2k)]},{x,0,(Sqrt[4(n-10^(2k)*2^m)+1]-1)/2}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308640 Number of ways to write n as (2^a3^b)^2 + c(2c+1) + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 1, 2, 4, 1, 4, 3, 3, 4, 1, 7, 2, 2, 7, 2, 5, 2, 4, 5, 1, 8, 5, 2, 3, 4, 6, 2, 3, 4, 2, 3, 7, 6, 5, 4, 7, 6, 1, 7, 5, 4, 6, 4, 4, 1, 6, 9, 2, 5, 3, 3, 5, 6, 7, 4, 7, 5, 4, 6, 6, 6, 4, 4, 5, 3, 9, 7, 4, 8, 2, 8, 5, 4, 10, 3, 9, 6, 5, 6, 4, 11, 7, 5, 8, 4, 7, 7, 8, 8, 2, 14, 6, 3, 8, 4

Offset: 1

Zhi-Wei Sun, Jun 12 2019

nonn

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: Let k be 1 or 2. Then, any positive integer n can be written as (2^a*3^b)^2 + k*c^2 + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.
Conjecture 3: Let k be 1 or -1. Then, any positive integer n can be written as (2^a*3^b)^2 + c*(5c+3k)/2 + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.
We have verified Conjectures 1-3 for all n = 1..10^6.
See also A308641 for similar conjectures.

			a(230) = 1 with 230 =(2^3*3^0)^2 + 3*(2*3+1) + 10*(3*10+1)/2.
a(2058) = 1 with 2058 = (2^0*3^0)^2 + 25*(2*25+1) + (-23)*(3*(-23)+1)/2.
a(26550) = 1 with 26550 = (2^0*3^3)^2 + 14*(2*14+1) + 130*(3*130+1)/2.
a(39433) = 1 with 39433 = (2*3^3)^2 + 135*(2*135+1) + 17*(3*17+1)/2.
a(505330) = 1 with 505330 = (2*3^2)^2 + 198*(2*198+1) + 533*(3*533+1)/2.
a(537830) = 1 with 537830 = (2^5*3^2)^2 + 402*(2*402+1) + (-296)*(3*(-296)+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), No. 7, 1367-1396.

Cf. A000079, A000244, A000290, A000566, A001318, A014105, A147875, A308566, A308584, A308621, A308623, A308641.

PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={};Do[r=0;Do[If[PenQ[n-4^a*9^b-c(2c+1)],r=r+1],{a,0,Log[4,n]},{b,0,Log[9,n/4^a]},{c,0,(Sqrt[8(n-4^a*9^b)+1]-1)/4}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308641 Number of ways to write n as (2^a5^b)^2 + c(3c+1)/2 + d(3d+1)/2, where a and b are nonnegative integers, and c and d are integers with c(3c+1)/2 <= d*(3d+1)/2.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 3, 2, 2, 3, 1, 3, 2, 2, 4, 3, 6, 2, 3, 4, 1, 5, 3, 4, 4, 4, 8, 3, 5, 6, 5, 4, 3, 4, 2, 4, 6, 5, 4, 5, 6, 5, 4, 5, 4, 3, 4, 5, 1, 5, 6, 6, 4, 2, 7, 4, 5, 4, 4, 4, 5, 6, 3, 5, 7, 7, 5, 3, 5, 5, 5, 7, 6, 3, 7, 6, 8, 5, 5, 7, 5, 7, 4, 2, 8, 6, 6, 3, 3, 7, 2, 9, 4, 7, 6, 5, 7, 2, 8

Offset: 1

Zhi-Wei Sun, Jun 13 2019

nonn

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: If f(x) is one of the polynomials x^2, x*(5x+1), x*(5x+1)/2, x*(7x+1)/2, x*(7x+5)/2, then any positive integers n can be written as (2^a*5^b)^2 + f(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
Conjecture 3: Any positive integers n can be written as (2^a*7^b)^2 + c*(7c+5)/2 + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
See also A308640 for similar conjectures.

			a(12) = 1 with 12 = (2^1*5^0)^2 + (-1)*(3*(-1)+1)/2 + 2*(3*2+1)/2.
a(22) = 1 with 22 = (2^2*5^0)^2 + (-1)*(3*(-1)+1)/2 + (-2)*(3*(-2)+1)/2.
a(50) = 1 with 50 = (2^2*5^0)^2 + (-3)*(3*(-3)+1)/2 + (-4)*(3*(-4)+1)/2.
a(330) = 1 with 330 = (2^2*5^0)^2 + (-8)*(3*(-8)+1)/2 + 12*(3*12+1)/2.
a(8650) = 1 with 8650 = (2^5*5^0)^2 + 8*(3*8+1)/2 + (-71)*(3*(-71)+1)/2.
a(29440) = 1 with 29440 = (2*5)^2 + (-80)*(3*(-80)+1)/2 + (-115)*(3*(-115)+1)/2.
a(48459) = 1 with 48459 = (2^7*5^0)^2 + 20*(3*20+1)/2 + (-145)*(3*(-145)+1)/2.
a(153035) = 1 with 153035 = (2*5^2)^2 + 35*(3*35+1)/2 + (-315)*(3*(-315)+1)/2.
a(164043) = 1 with 164043 = (2^2*5^2)^2 + (-46)*(3*(-46)+1)/2 + 317*(3*317+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), No. 7, 1367-1396.

Cf. A000079, A000290, A000351, A001318, A308566, A308584, A308621, A308623, A308640.

PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={};Do[r=0;Do[If[PenQ[n-4^a*25^b-c(3c+1)/2],r=r+1],{a,0,Log[4,n]},{b,0,Log[25,n/4^a]},{c,-Floor[(Sqrt[12(n-4^a*25^b)+1]+1)/6],(Sqrt[12(n-4^a*25^b)+1]-1)/6}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308644 Number of ways to write n as (3^a5^b)^2 + c(3c+1)/2 + d*(7d+1)/2, where a and b are nonnegative integers, and c and d are integers.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 1, 2, 2, 2, 2, 3, 4, 2, 5, 3, 2, 3, 2, 3, 1, 3, 4, 3, 4, 5, 3, 5, 4, 6, 2, 2, 4, 4, 6, 2, 4, 6, 7, 5, 3, 4, 6, 3, 4, 4, 2, 4, 3, 4, 3, 3, 4, 5, 5, 5, 2, 3, 8, 3, 5, 4, 7, 5, 4, 4, 4, 4, 5, 4, 1, 4, 5, 4, 1, 3, 3, 6, 4, 7, 7, 3, 5, 7, 8, 2, 4, 5, 6, 7, 3, 8, 5, 7, 8, 4, 7, 8, 2

Offset: 1

Zhi-Wei Sun, Jun 13 2019

nonn

Conjecture 1: a(n) > 0 for all n > 0. Also, any positive integer n can be written as (3^a*5^b)^2 + c*(3c+1)/2 + d*(7d+3)/2, where a and b are nonnegative integers, and c and d are integers.
Conjecture 2: Let r be 1 or 3. Then, any positive integer n can be written as (3^a*4^b)^2 + c*(3c+1)/2 + d*(7d+r)/2, where a and b are nonnegative integers, and c and d are integers.
We have verified Conjectures 1-2 for all n = 1..10^6.
See also A308640, A308641 and A308656 for similar conjectures.

			a(152) = 1 with 152 = (3^0*5^0)^2 + (-4)*(3*(-4)+1)/2 + 6*(7*6+1)/2.
a(129894) = 1 with 129894 = (3^0*5^1)^2 + 154*(3*154+1)/2 + 164*(7*164+1)/2.
a(200963) = 1 with 200963 = (3^1*5^0)^2 + 364*(3*364+1)/2 + 24*(7*24+1)/2.
a(371278) = 1 with 371278 = (3^3*5^1)^2 + (-382)*(3*(-382)+1)/2 + (-196)*(7*(-196)+1)/2.
a(534699) = 1 with 534699 = (3^2*5^2)^2 + 543*(3*543+1)/2 + (-109)*(3*(-109)+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Universal sums of three quadratic polynomials, Sci. China Math., in press.

Cf. A000244, A000290, A000351, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308656.

PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={};Do[r=0;Do[If[PenQ[n-9^a*25^b-x(7x+1)/2],r=r+1],{a,0,Log[9,n]},{b,0,Log[25,n/9^a]},{x,-Floor[(Sqrt[56(n-9^a*25^b)+1]+1)/14],(Sqrt[56(n-9^a*25^b)+1]-1)/14}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308656 Number of ways to write n as (2^a9^b)^2 + c(2c+1) + d*(3d+1), where a and b are nonnegative integers, and c and d are integers.

Original entry on oeis.org

1, 1, 1, 3, 2, 3, 3, 2, 3, 1, 4, 2, 1, 4, 3, 4, 3, 5, 4, 3, 6, 2, 2, 4, 3, 6, 2, 4, 5, 3, 6, 4, 4, 4, 4, 4, 4, 1, 4, 5, 5, 2, 3, 3, 2, 8, 3, 4, 5, 3, 5, 3, 3, 5, 3, 7, 1, 3, 5, 4, 6, 3, 6, 2, 2, 6, 5, 4, 6, 6, 7, 3, 4, 9, 5, 4, 5, 3, 4, 4, 11, 5, 5, 12, 5, 7, 5, 4, 10, 2, 7, 8, 4, 8, 7, 12, 5, 5, 5, 5

Offset: 1

Zhi-Wei Sun, Jun 14 2019

nonn

Note that {x*(2x+1): x is an integer} = {n*(n+1)/2: n = 0,1,2,...}.
Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: If f(x) is one of the polynomials x*(4x+1), x*(5x+2), x*(5x+4), x*(7x+3)/2 and x(7x+5)/2, then any positive integer n can be written as (2^a*9^b)^2 + f(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
Conjecture 3: Let r be 1 or 2. Then any positive integer n can be written as (2^a*7^b)^2 + c*(2c+1) + d*(3d+r), where a and b are nonnegative integers, and c and d are integers.
Conjecture 4: If g(x) is one of the polynomials x*(x+1), x*(4x+3), x*(7x+1)/2, x*(7x+3)/2 and x*(7x+5)/2, then any positive integer n can be written as (2^a*7^b)^2 + g(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.
We have verified a(n) > 0 for all n = 1..10^8, and Conjectures 2-4 for all n = 1..10^6.
See also A308640, A308641, and A308644 for similar conjectures.
Jiao-Min Lin (a student at Nanjing University) has found a counterexample to Conjecture 1: a(2109982225) = 0. - Zhi-Wei Sun, Jul 30 2022

			a(13) = 1 with 13 = (2^0*9^0)^2 + 2*(2*2+1) + (-1)*(3*(-1)+1).
a(3515) = 1 with 3515 = (2^0*9^1)^2 + 0*(2*0+1) + (-34)*(3*(-34)+1).
a(124076) = 1 with 124076 = (2^3*9^1)^2 + 206*(2*206+1) + 106*(3*106+1).
a(141518) = 1 with 141518 = (2^1*9^2)^2 + (-188)*(2*(-188)+1) + 122*(3*122+1).
a(345402) = 1 with 345402 = (2^7*9^0)^2 + 18*(2*18+1) + (-331)*(3*(-331)+1).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

A000079, A000217, A000420, A001019, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308644.

PQ[n_]:=PQ[n]=IntegerQ[Sqrt[12n+1]];
tab={};Do[r=0;Do[If[PQ[n-81^a*4^b-x(2x+1)],r=r+1],{a,0,Log[81,n]},{b,0,Log[4,n/81^a]},{x,-Floor[(Sqrt[8(n-81^a*4^b)+1]+1)/4],(Sqrt[8(n-81^a*4^b)+1]-1)/4}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308661 Number of ways to write 12n+5 as (2^a5^b)^2 + c^2 + d^2, where a,b,c,d are nonnegative integers with a > 0 and c <= d.

Original entry on oeis.org

1, 2, 3, 3, 2, 3, 3, 5, 5, 4, 5, 3, 5, 5, 5, 6, 3, 6, 4, 3, 5, 4, 7, 6, 6, 6, 2, 8, 8, 5, 5, 5, 6, 5, 6, 10, 6, 6, 8, 4, 6, 8, 8, 7, 3, 10, 5, 7, 9, 6, 7, 3, 9, 7, 2, 7, 6, 9, 8, 6, 8, 6, 8, 9, 5, 4, 7, 6, 4, 5, 7, 8, 5, 8, 7, 6, 4, 8, 10, 6, 10, 3, 6, 9, 6, 11, 5, 9, 4, 4, 8, 8, 10, 9, 7, 4, 5, 11, 7, 9, 10

Offset: 0

Zhi-Wei Sun, Jun 15 2019

nonn

Conjecture 1: a(n) > 0 for all n = 0,1,2,.... Equivalently, for each nonnegative integer n we can write 3*n+1 as a*(a+1)/2 + b*(b+1)/2 + (2^c*5^d)^2 with a,b,c,d nonnegative integers.
Conjecture 2: For each n = 0,1,2,... we can write 24*n+10 as a^2 + b^2 + (2^c*3^d)^2 with a,b,c,d nonnegative integers and d > 0.
We have verified Conjectures 1 and 2 for n up to 2*10^8 and 10^8 respectively.
By the Gauss-Legendre theorem on sums of three squares, for each n = 0,1,... we can write 4*n+1 (or 4*n+2, or 8*n+3) as the sum of three squares.
Conjecture 1 holds for n < 8.33*10^9. - Giovanni Resta, Jun 19 2019

			a(0) = 1 with 12*0 + 5 = (2^1*5^0)^2 + 0^2 + 1^2.
a(4) = 2 with 12*4 + 5 = 53 = (2^1*5^0)^2 + 0^2 + 7^2 = (2^2*5^0)^2 + 1^2 + 6^2.
a(441019) = 2 with 12*441019 + 5 = 5292233 = (2^1*5^2)^2 + 513^2 + 2242^2 = (2^3*5^1)^2 + 757^2 + 2172^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000079, A000217, A000244, A000290, A000351, A000378, A001481, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308662.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[12n+5-4^a*25^b-x^2],r=r+1],{a,1,Log[4,12n+5]},{b,0,Log[25,(12n+5)/4^a]},{x,0,Sqrt[(12n+5-4^a*25^b)/2]}];tab=Append[tab,r],{n,0,100}];Print[tab]

A308662 Number of ways to write n as (2^a5^b)^2 + c(3c+1) + d*(3d+2), where a and b are nonnegative integers, and c and d are integers.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 1, 2, 3, 1, 3, 2, 2, 2, 2, 4, 2, 2, 5, 3, 3, 3, 3, 3, 4, 6, 4, 3, 3, 5, 4, 4, 3, 6, 5, 6, 3, 2, 6, 3, 6, 2, 3, 4, 4, 6, 5, 5, 4, 4, 6, 1, 4, 4, 4, 6, 3, 5, 2, 6, 7, 3, 2, 5, 5, 4, 5, 6, 8, 5, 6, 5, 4, 8, 3, 7, 3, 3, 7, 3, 6, 7, 4, 4, 7, 7, 4, 4, 8, 7, 4, 3, 6, 4, 7, 7, 4, 1, 6, 7

Offset: 1

Zhi-Wei Sun, Jun 15 2019

nonn

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: Let r be 1 or 2. Then, any positive integer n can be written as (2^a*5^b)^2 + c*(2c+1) + d*(3d+r), where a and b are nonnegative integers, and c and d are integers.
We have verified Conjectures 1 and 2 for all n = 1..10^8.

			a(3) = 1 with 3 = (2^0*5^0)^2 + (-1)*(3*(-1)+1) + 0*(3*0+2).
a(7) = 1 with 7 = (2^1*5^0)^2 + (-1)*(3*(-1)+1) + (-1)*(3*(-1)+2).
a(10) = 1 with 10 = (2^0*5^0)^2 + 1*(3*1+1) + 1*(3*1+2).
a(52) = 1 with 52 = (2^0*5^0)^2 + 3*(3*3+1) + (-3)*(3*(-3)+2).
a(98) = 1 with 98 = (2^0*5^1)^2 + 4*(3*4+1) + (-3)*(3*(-3)+2).
a(14596) = 1 with 14596 = (2^3*5^0)^2 + (-36)*(3*(-36)+1) + (-60)*(3*(-60)+2).
a(22163) = 1 with 22163 = (2^3*5^0)^2 + 66*(3*66+1) + (-55)*(3*(-55)+2).
a(150689) = 1 with 150689 = (2^6*5^1)^2 + 117*(3*117+1) + (-49)*(3*(-49)+2).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), No. 7, 1367-1396.

Cf. A000079, A000351, A001082, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308661.

OctQ[n_]:=OctQ[n]=IntegerQ[Sqrt[3n+1]];
tab={};Do[r=0;Do[If[OctQ[n-4^a*25^b-x(3x+1)],r=r+1],{a,0,Log[4,n]},{b,0,Log[25,n/4^a]},{x,-Floor[(Sqrt[12(n-4^a*25^b)+1]+1)/6],(Sqrt[12(n-4^a*25^b)+1]-1)/6}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308734 Number of ordered ways to write n as (2^a3^b)^2 + (2^c5^d)^2 + x^2 + y^2, where a,b,c,d,x,y are nonnegative integers with x <= y.

Original entry on oeis.org

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

Zhi-Wei Sun, Jun 21 2019

nonn

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

			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.

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.)

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

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]

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A308623 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 2^c*10^(2d) with a,b,c,d nonnegative integers.

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A308640 Number of ways to write n as (2^a*3^b)^2 + c*(2c+1) + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.

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A308641 Number of ways to write n as (2^a*5^b)^2 + c*(3c+1)/2 + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers with c*(3c+1)/2 <= d*(3d+1)/2.

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A308644 Number of ways to write n as (3^a*5^b)^2 + c*(3c+1)/2 + d*(7d+1)/2, where a and b are nonnegative integers, and c and d are integers.

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A308656 Number of ways to write n as (2^a*9^b)^2 + c*(2c+1) + d*(3d+1), where a and b are nonnegative integers, and c and d are integers.

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A308661 Number of ways to write 12*n+5 as (2^a*5^b)^2 + c^2 + d^2, where a,b,c,d are nonnegative integers with a > 0 and c <= d.

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A308662 Number of ways to write n as (2^a*5^b)^2 + c*(3c+1) + d*(3d+2), where a and b are nonnegative integers, and c and d are integers.

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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.

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A308623 Number of ways to write n as a(a+1)/2 + b(b+1)/2 + 2^c*10^(2d) with a,b,c,d nonnegative integers.

A308640 Number of ways to write n as (2^a3^b)^2 + c(2c+1) + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.

A308641 Number of ways to write n as (2^a5^b)^2 + c(3c+1)/2 + d(3d+1)/2, where a and b are nonnegative integers, and c and d are integers with c(3c+1)/2 <= d*(3d+1)/2.

A308644 Number of ways to write n as (3^a5^b)^2 + c(3c+1)/2 + d*(7d+1)/2, where a and b are nonnegative integers, and c and d are integers.

A308656 Number of ways to write n as (2^a9^b)^2 + c(2c+1) + d*(3d+1), where a and b are nonnegative integers, and c and d are integers.

A308661 Number of ways to write 12n+5 as (2^a5^b)^2 + c^2 + d^2, where a,b,c,d are nonnegative integers with a > 0 and c <= d.

A308662 Number of ways to write n as (2^a5^b)^2 + c(3c+1) + d*(3d+2), where a and b are nonnegative integers, and c and d are integers.

A308734 Number of ordered ways to write n as (2^a3^b)^2 + (2^c5^d)^2 + x^2 + y^2, where a,b,c,d,x,y are nonnegative integers with x <= y.