cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A299794 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x or 2*y is a power of 4 (including 4^0 = 1) and x + 15*y is also a power of 4.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 2, 1, 1, 2, 3, 1, 1, 2, 2, 1, 1, 6, 3, 1, 3, 3, 2, 2, 1, 3, 4, 2, 1, 5, 4, 4, 5, 1, 2, 3, 2, 5, 5, 2, 2, 8, 2, 2, 1, 5, 2, 4, 3, 4, 4, 4, 3, 6, 3, 2, 3, 3, 4, 3, 1, 3, 6, 4, 3, 11, 2, 2, 2, 4, 5, 1, 2, 3, 5, 3, 1
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 04 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0. Also, for any integer n > 1 we can write n^2 as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that 2*x or y is a power of 4 and also x + 15*y = 2^(2k+1) for some k = 0,1,2,....
Conjecture 2: Let d be 2 or 8, and let r be 0 or 1. Then any positive square n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x or y is a power of 2 and x + d*y = 2^(2k+r) for some k = 0,1,2,....
We have verified Conjecture 1 for n up to 10^7.
See also A299537, A300219 and A300396 for similar conjectures.

Examples

			a(2) = 1 since 2^2 = 1^2 + 1^2 + 1^2 + 1^2 with 1 = 4^0 and 1 + 15*1 = 4^2.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4 = 4^1 and 4 + 15*0 = 4^1.
a(19) = 1 since 19^2 = 1^2 + 0^2 + 6^2 + 18^2 with 1 = 4^0 and 1 + 15*0 = 4^0.
a(159) = 1 since 159^2 = 34^2 + 2^2 + 75^2 + 136^2 with 2*2 = 4^1 and 34 + 15*2 = 4^3.
a(1998) = 1 since 1998^2 = 256^2 + 256^2 + 286^2 + 1944^2 with 256 = 4^4 and 256 + 15*256 = 4^6.
a(3742) = 1 since 3742^2 = 2176^2 + 128^2 + 98^2 + 3040^2 with 2*128 = 4^4 and 2176 + 15*128 = 4^6.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]];
    tab={};Do[r=0;Do[If[Pow[2y]||Pow[4^k-15y],Do[If[SQ[n^2-y^2-(4^k-15y)^2-z^2],r=r+1],{z,0,Sqrt[Max[0,(n^2-y^2-(4^k-15y)^2)/2]]}]],
    {k,0,Log[4,Sqrt[226]*n]},{y,0,Min[n,4^(k-2)]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300362 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 2*y and (z + 2*w)/3 are squares and w is even.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 4, 1, 4, 2, 4, 1, 2, 2, 2, 3, 2, 1, 5, 1, 5, 2, 3, 3, 3, 1, 1, 2, 3, 1, 4, 3, 5, 1, 6, 6, 6, 1, 4, 6, 8, 2, 4, 4, 3, 1, 5, 3, 9, 1, 4, 4, 5, 3, 10, 4, 7, 3, 9, 2, 14, 2, 6, 2, 6, 1
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 04 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 7, 9, 14, 19, 22, 26, 34, 41, 4^k*m (k = 0,1,... and m = 1, 2, 3, 5, 10, 11, 13, 15).

Examples

			 a(9) = 1 since 9^2 = 9^2 + 0^2 + 0^2 + 0^2 with 9 + 2*0 = 3^2 and 0 + 2*0 = 3*0^2.
a(13) = 1 since 13^2 = 4^2 + 0^2 + 3^2 + 12^2 with 4 + 2*0 = 2^2 and 3 + 2*12 = 3*3^2.
a(14) = 1 since 14^2 = 4^2 + 6^2 + 12^2 + 0^2 with 4 + 2*6 = 4^2 and 12 + 2*0 = 3*2^2.
a(15) = 1 since 15^2 = 9^2 + 0^2 + 12^2 + 0^2 with 9 + 2*0 = 3^2 and 12 + 2*0 = 3*2^2.
a(41) = 1 since 41 = 38^2 + 13^2 + 8^2 + 2^2 with 38 + 2*13 = 8^2 and 8 + 2*2 = 3*2^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[x+2y]&&SQ[(n^2-x^2-y^2-z^2)/4]&&SQ[(z+2*Sqrt[n^2-x^2-y^2-z^2])/3],r=r+1],{x,0,n},{y,0,Sqrt[n^2-x^2]},{z,0,Sqrt[n^2-x^2-y^2]}];tab=Append[tab,r],{n,0,80}];Print[tab]

A302982 Number of ways to write n as x^2 + 5*y^2 + 2^z + 3*2^w with x,y,z,w nonnegative integers.

Original entry on oeis.org

0, 0, 0, 1, 2, 1, 2, 4, 3, 3, 5, 4, 6, 7, 4, 7, 5, 4, 7, 8, 5, 5, 8, 5, 9, 7, 6, 13, 10, 7, 9, 10, 7, 12, 11, 8, 11, 7, 7, 11, 11, 6, 11, 13, 6, 10, 7, 7, 17, 13, 6, 13, 14, 9, 11, 18, 10, 13, 14, 11
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 16 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
Clearly, a(4*n) > 0 if a(n) > 0. We have verified a(n) > 0 for all n = 4..2*10^8.
See also A302983 and A302984 for similar conjectures.

Examples

			a(4) = 1 with 4 = 0^2 + 5*0^2 + 2^0 + 3*2^0.
a(5) = 2 with 5 =  1^2 + 5*0^2 + 2^0 + 3*2^0 = 0^2 + 5*0^2 + 2^1 + 3*2^0.
a(6) = 1 with 6 = 1^2 + 3*0^2 + 2^1 + 3*2^0.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[n-3*2^k-2^j-5x^2],r=r+1],{k,0,Log[2,n/3]},{j,0,If[n==3*2^k,-1,Log[2,n-3*2^k]]},{x,0,Sqrt[(n-3*2^k-2^j)/5]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A302984 Number of ways to write n as x^2 + 2*y^2 + 2^z + 5*2^w with x,y,z,w nonnegative integers.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 2, 3, 3, 3, 4, 5, 5, 8, 5, 5, 7, 4, 6, 7, 9, 9, 10, 10, 7, 9, 8, 10, 15, 10, 9, 10, 8, 6, 10, 10, 11, 14, 14, 8, 12, 13, 13, 20, 15, 12, 16, 10, 15, 12, 10, 15, 17, 16, 12, 16, 14, 14, 21
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 16 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5.
Clearly, a(2*n) > 0 if a(n) > 0. We have verified a(n) > 0 for all n = 6...10^9.
See also A302982 and A302983 for similar conjectures.

Examples

			a(6) = 1 with 6 = 0^2 + 2*0^2 + 2^0 + 5*2^0.
a(7) = 2 with 7 = 1^2 + 2*0^2 + 2^0 + 5*2^0 = 0^2 + 2*0^2 + 2^1 + 5*2^0.
a(8) = 2 with 8 = 0^2 + 2*1^2 + 2^0 + 5*2^0 = 1^2 + 2*0^2 + 2^1 + 5*2^0.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[MemberQ[{5,7},Mod[Part[Part[f[n],i],1],8]]&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[n-5*2^k-2^j],Do[If[SQ[n-5*2^k-2^j-2x^2],r=r+1],{x,0,Sqrt[(n-5*2^k-2^j)/2]}]],{k,0,Log[2,n/5]},{j,0,Log[2,Max[1,n-5*2^k]]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A303234 Numbers of the form x*(x+1)/2 + 2^y with x and y nonnegative integers.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 14, 16, 17, 18, 19, 22, 23, 25, 26, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 42, 44, 46, 47, 49, 52, 53, 56, 57, 59, 60, 61, 63, 64, 65, 67, 68, 70, 71, 74, 77, 79, 80
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 20 2018

Keywords

Comments

Conjecture: Any integer n > 1 can be written as the sum of two terms of the current sequence.
This is equivalent to the author's conjecture in A303233.

Examples

			a(1) = 1 with 1 = 0*(0+1)/2 + 2^0.
a(2) = 2 with 2 = 1*(1+1)/2 + 2^0 = 0*(0+1)/2 + 2^1.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[Do[If[SQ[8(n-2^k)+1],tab=Append[tab,n];Goto[aa]],{k,0,Log[2,n]}];Label[aa],{n,1,80}];Print[tab]

A303389 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 5^c + 5^d, where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 3, 2, 2, 2, 4, 3, 2, 2, 3, 3, 3, 2, 2, 2, 4, 3, 2, 1, 5, 4, 3, 2, 5, 5, 5, 5, 3, 3, 5, 5, 4, 4, 4, 5, 5, 2, 5, 3, 5, 4, 7, 2, 4, 6, 6, 5, 4, 4, 5, 8, 4, 4, 4, 7, 6, 4, 3, 4, 8, 4, 7, 3, 3, 6, 8, 2, 5, 6, 5, 4, 6, 4, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integers n > 1 can be written as the sum of two triangular numbers and two powers of 5.
This has been verified for all n = 2..10^10.
See A303393 for the numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.
See also A303401, A303432 and A303540 for similar conjectures.

Examples

			a(4) = 1 with 4 = 1*(1+1)/2 + 1*(1+1)/2 + 5^0 + 5^0.
a(5) = 1 with 5 = 0*(0+1)/2 + 2*(2+1)/2 + 5^0 + 5^0.
a(7) = 1 with 7 = 0*(0+1)/2 + 1*(1+1)/2 + 5^0 + 5^1.
a(25) = 1 with 25 = 0*(0+1)/2 + 5*(5+1)/2 + 5^1 + 5^1.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    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;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[4(n-5^j-5^k)+1],Do[If[SQ[8(n-5^j-5^k-x(x+1)/2)+1],r=r+1],{x,0,(Sqrt[4(n-5^j-5^k)+1]-1)/2}]],{j,0,Log[5,n/2]},{k,j,Log[5,n-5^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300396 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that 2*x or y is a power of 4 (including 4^0 = 1) and x + 63*y = 2^(2k+1) for some k = 0,1,2,....

Original entry on oeis.org

0, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 3, 1, 3, 4, 2, 2, 4, 1, 2, 1, 1, 2, 4, 3, 1, 1, 2, 1, 6, 2, 2, 2, 5, 1, 4, 1, 2, 6, 3, 3, 3, 1, 2, 3, 4, 3, 3, 2, 4, 2, 2, 1, 7, 3, 1, 4, 1, 2, 8, 1, 3, 7, 3, 4, 6, 3, 4, 4, 6, 4, 3, 2, 4, 3, 1, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 05 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 5, 13, 25, 29, 59, 61, 79, 91, 95, 101, 103, 1315, 2^k (k = 1,2,3,...), 2^(2k+1)*m (k = 0,1,2,... and m = 3, 5, 7, 11, 15, 19, 23, 887).
This is stronger than the conjecture that A300360(n) > 0 for all n > 1. Note that a(387) = 3 < A300360(387) = 4 and a(1774) = 1 < A300360(1774) = 2.
We have verified that a(n) > 0 for all n = 2..10^7.
See also A299537, A299794 and A300219 for similar conjectures.

Examples

			a(29) = 1 since 29^2 = 2^2 + 2^2 + 7^2 + 28^2 with 2*2 = 4^1 and 2 + 63*2 = 2^7.
a(86) = 2 since 65^2 + 1^2 + 19^2 + 53^2 = 65^2 + 1^2 + 31^2 + 47^2 with 1 = 4^0 and 65 + 63*1 = 2^7.
a(1774) = 1 since 1774^2 = 8^2 + 520^2 + 14^2 + 1696^2 with 2*8 = 4^2 and 8 + 63*520 = 2^15.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]];
    tab={};Do[r=0;Do[If[Pow[y]||Pow[(2*4^k-63y)/2],Do[If[SQ[n^2-y^2-(2*4^k-63y)^2-z^2],r=r+1],{z,0,Sqrt[Max[0,(n^2-y^2-(2*4^k-63y)^2)/2]]}]],{k,0,Log[4,Sqrt[63^2+1]*n/2]},{y,0,Min[n,2*4^k/63]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A303393 Numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.

Original entry on oeis.org

1, 2, 4, 5, 6, 7, 8, 11, 15, 16, 20, 22, 25, 26, 28, 29, 31, 33, 35, 37, 40, 41, 46, 50, 53, 56, 60, 61, 67, 70, 71, 79, 80, 83, 91, 92, 96, 103, 106, 110, 116, 121, 125, 126, 128, 130, 131, 135, 137, 140, 141, 145, 146, 153, 154, 158, 161, 170, 172, 176
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

The author's conjecture in A303389 has the following equivalent version: Each integer n > 1 can be expressed as the sum of two terms of the current sequence.
This has been verified for all n = 2..2*10^8.

Examples

			a(1) = 1 with 1 = 0*(0+1)/2 + 5^0.
a(2) = 2 with 2 = 1*(1+1)/2 + 5^0.
a(3) = 4 with 4 = 2*(2+1)/2 + 5^0.
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
    tab={};Do[Do[If[TQ[m-5^k],tab=Append[tab,m];Goto[aa]],{k,0,Log[5,m]}];Label[aa],{m,1,176}];Print[tab]

A272084 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 4*x^2 + 5*y^2 + 20*z*w a square, where x,y,z,w are nonnegative integers with z < w.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 19 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k*q (k = 0,1,2,... and q = 1, 3, 7), 4^k*m (k = 0,1,2,... and m = 21, 30, 38, 62, 70, 77, 142, 217, 237, 302, 382, 406, 453, 670).
(ii) For each triple (a,b,c) = (1,8,8), (7,9,-12), (9,40,-24), (9,40,-60), any positive integer can be written as x^2 + y^2 + z^2 + w^2 with a*x^2 + b*y^2 + c*z*w a square, where w is a positive integer and x,y,z are nonnegative integers.
(iii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with (3*x+5*y)^2 -24*z*w a square, where x,y,z,w are nonnegative integers. Also, for each ordered pair (a,b) = (1,4), (1,8), (1,12), (1,24), (1,32), (1,48), (25,24), (1,-4), (9,-4), (121,-20), every natural number can be written as x^2 + y^2 + z^2 + w^2 with a*x^2 + b*y*z a square, where x,y,z,w are nonnegative integers.
(iv) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with (x^2-y^2)*(w^2-2*z^2) (or (x^2-y^2)*(2*w^2-z^2) or (x^2-y^2)*(w^2-5*z^2)) a square, where w,x,y,z are integers.
See also A262357, A268507, A269400, A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775, A271778 and A271824 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 < 1 and 4*0^2 + 5*0^2 + 20*0*1 = 0^2.
a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 0 < 1 and 4*1^2 + 5*0^2 + 20*0*1 = 2^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 0 < 1 and 4*1^2 + 5*1^2 + 20*0*1 = 3^2.
a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 0 < 2 and 4*1^2 + 5*1^2 + 20*0*2 = 3^2.
a(14) = 1 since 14 = 1^2 + 3^2 + 0^2 + 2^2 with 0 < 2 and 4*1^2 + 5*3^2 + 20*0*2 = 7^2.
a(21) = 1 since 21 = 0^2 + 2^2 + 1^2 + 4^2 with 1 < 4 and 4*0^2 + 5*2^2 + 20*1*4 = 10^2.
a(30) = 1 since 30 = 4^2 + 2^2 + 1^2 + 3^2 with 1 < 3 and 4*4^2 + 5*2^2 + 20*1*3 = 12^2.
a(38) = 1 since 38 = 1^2 + 1^2 + 0^2 + 6^2 with 0 < 6 and 4*1^2 + 5*1^2 + 20*0*6 = 3^2.
a(62) = 1 since 62 = 1^2 + 3^2 + 4^2 + 6^2 with 4 < 6 and 4*1^2 + 5*3^2 + 20*4*6 = 23^2.
a(70) = 1 since 70 = 7^2 + 1^2 + 2^2 + 4^2 with 2 < 4 and 4*7^2 + 5*1^2 + 20*2*4 = 19^2.
a(77) = 1 since 77 = 4^2 + 6^2 + 3^2 + 4^2 with 3 < 4 and 4*4^2 + 5*6^2 + 20*3*4 = 22^2.
a(142) = 1 since 142 = 4^2 + 6^2 + 3^2 + 9^2 with 3 < 9 and 4*4^2 + 5*6^2 + 20*3*9 = 28^2.
a(217) = 1 since 217 = 6^2 + 6^2 + 8^2 + 9^2 with 8 < 9 and 4*6^2 + 5*6^2 + 20*8*9 = 42^2.
a(237) = 1 since 237 = 5^2 + 8^2 + 2^2 + 12^2 with 2 < 12 and 4*5^2 + 5*8^2 + 20*2*12 = 30^2.
a(302) = 1 since 302 = 11^2 + 9^2 + 6^2 + 8^2 with 6 < 8 and 4*11^2 + 5*9^2 + 20*6*8 = 43^2.
a(382) = 1 since 382 = 11^2 + 7^2 + 4^2 + 14^2 with 4 < 14 and 4*11^2 + 5*4^2 + 20*4*14 = 43^2.
a(406) = 1 since 406 = 8^2 + 6^2 + 9^2 + 15^2 with 9 < 15 and 4*8^2 + 5*6^2 + 20*9*15 = 56^2.
a(453) = 1 since 453 = 8^2 + 14^2 + 7^2 + 12^2 with 7 < 12 and 4*8^2 + 5*14^2 + 20*7*12 = 54^2.
a(670) = 1 since 670 = 17^2 + 11^2 + 2^2 + 16^2 with 2 < 16 and 4*17^2 + 5*11^2 + 20*2*16 = 49^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[4x^2+5y^2+20*z*Sqrt[n-x^2-y^2-z^2]],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[(n-1-x^2-y^2)/2]}];Print[n," ",r];Continue,{n,1,100}]

A272332 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 36*x^2*y + 12*y^2*z + z^2*x a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 2, 2, 6, 4, 3, 3, 3, 8, 5, 2, 6, 6, 4, 1, 7, 10, 6, 8, 8, 5, 2, 2, 7, 16, 8, 3, 12, 6, 4, 3, 6, 13, 8, 8, 8, 6, 5, 7, 15, 14, 4, 2, 12, 7, 3, 2, 5, 18, 8, 12, 14, 8, 7, 4, 6, 8, 7, 5, 14, 8, 5, 2, 12, 18, 8, 12, 10, 6, 3, 5, 10, 19, 10, 3, 8, 3, 1, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 26 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m (k = 0,1,2,... and m = 1, 79, 591, 599, 1752, 1839, 10264).
We have verified that a(n) > 0 for all n = 1,...,400000.
For more refinements of Lagrange's four-square theorem, see arXiv:1604.06723.

Examples

			a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 > 0 and 36*0^2*0 + 12*0^2*0 + 0^2*0 = 0^2.
a(79) = 1 since 79 = 7^2 + 1^2 + 5^2 + 2^2 with 7 > 0 and 36*1^2*5 + 12*5^2*2 + 2^2*1 = 28^2.
a(591) = 1 since 591 = 23^2 + 1^2 + 6^2 + 5^2 with 23 > 0 and 36*1^2*6 + 12*6^2*5 + 5^2*1 = 49^2.
a(599) = 1 since 599 = 6^2 + 1^2 + 11^2 + 21^2 with 6 > 0 and 36*1^2*11 + 12*11^2*21 + 21^2*1 = 177^2.
a(1752) = 1 since 1752 = 10^2 + 4^2 + 40^2 + 6^2 with 10 > 0 and 36*4^2*40 + 12*40^2*6 + 6^2*10 = 372^2.
a(1839) = 1 since 1839 = 17^2 + 37^2 + 9^2 + 10^2 with 17 > 0 and 36*37^2*9 + 12*9^2*10 + 10^2*37 = 676^2.
a(10264) = 1 since 10264 = 96^2 + 30^2 + 2^2 + 12^2 with 96 > 0 and 36*30^2*2 + 12*2^2*12 + 12^2*30 = 264^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[36*x^2*y+12*y^2*z+z^2*x],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]
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