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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A300844 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x or 2*y or z is a square and (12*x)^2 + (21*y)^2 + (28*z)^2 is also a square.

Original entry on oeis.org

1, 4, 4, 2, 5, 6, 2, 2, 4, 5, 7, 1, 3, 7, 2, 3, 5, 7, 7, 2, 6, 1, 2, 2, 2, 11, 7, 3, 3, 8, 5, 1, 4, 5, 9, 4, 6, 8, 6, 4, 7, 9, 3, 3, 2, 9, 2, 1, 3, 6, 16, 5, 9, 7, 6, 5, 1, 5, 9, 4, 4, 7, 5, 5, 5, 17, 6, 4, 7, 3, 6, 3, 6, 11, 11, 4, 3, 1, 8, 2, 6
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 13 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n = 0,1,2,....
Conjecture 2: Any positive integer can be written as x^2 + y^2 + z^2 + w^2 with w a positive integer and x,y,z nonnegative integers such that x or y or z is a square and 144*x^2 + 505*y^2 + 720*z^2 is also a square.
By the author's 2017 JNT paper, each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x (or 2*x) is a square.
In 2016, the author conjectured in A271510 that each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y >= z such that (3*x)^2 + (4*y)^2 + (12*z)^2 is a square.
See also A300791 and A300792 for similar conjectures.

Examples

			a(11) = 1 since 11 = 0^2 + 1^2 + 1^2 + 3^2 with 0 = 0^2 and (12*0)^2 + (21*1)^2 + (28*1)^2 = 35^2.
a(56) = 1 since 56 = 4^2 + 6^2 + 2^2 + 0^2 with 4 = 2^2 and (12*4)^2 + (21*6)^2 + (28*2)^2 = 146^2.
a(77) = 1 since 77 = 4^2 + 0^2 + 5^2 + 6^2 with 4 = 2^2 and (12*4)^2 + (21*0)^2 + (28*5)^2 = 148^2.
a(184) = 1 since 184 = 12^2 + 2^2 + 0^2 + 6^2 with 0 = 0^2 and (12*12)^2 + (21*2)^2 + (28*0)^2 = 150^2.
a(599) = 1 since 599 = 21^2 + 11^2 + 1^2 + 6^2 with 1 = 1^2 and (12*21)^2 + (21*11)^2 + (28*1)^2 = 343^2.
a(7836) = 1 since 7836 = 38^2 + 18^2 + 68^2 + 38^2 with 2*18 = 6^2 and (12*38)^2 + (21*18)^2 + (28*68)^2 = 1994^2.
a(15096) = 1 since 15096 = 16^2 + 6^2 + 52^2 + 110^2 with 16 = 4^2 and (12*16)^2 + (21*6)^2 + (28*52)^2 = 1474^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[x]||SQ[2y]||SQ[z])&&SQ[(12x)^2+(21y)^2+(28z)^2]&&SQ[n-x^2-y^2-z^2],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];tab=Append[tab,r],{n,0,80}]

A300908 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square , where w is a positive integer and x,y,z are nonnegative integers such that z or 2*z or 3*z is a square.

Original entry on oeis.org

1, 3, 1, 2, 6, 1, 2, 3, 2, 7, 2, 2, 7, 1, 5, 2, 7, 9, 3, 6, 2, 3, 4, 1, 9, 7, 3, 4, 5, 7, 2, 3, 5, 6, 3, 4, 7, 3, 8, 6, 7, 6, 4, 3, 7, 3, 2, 2, 4, 13, 5, 8, 6, 5, 3, 1, 8, 8, 3, 6, 5, 1, 11, 2, 16, 5, 4, 8, 1, 8, 2, 7, 11, 7, 5, 4, 2, 13, 2, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 15 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0. Moreover, any positive integer can be written as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square, where w is a positive integer and x,y,z are nonnegative integers for which one of z, z/2, z/3 is a square and z/2 (or z/3) is a power of 4 (including 1).
Conjecture 2: Each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square, where x,y,z,w are nonnegative integers for which x or z is a square or y = 2^k for some k = 0,1,2,....
Conjecture 3. Let a,b,c be positive integers with a <= b <= c and gcd(a,b,c) = 1. If each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that (a*x)^2 + (b*y)^2 + (c*z)^2 is a square, then (a,b,c) must be among the three triples (3,4,12), (12,15,20) and (12,21,28).
We have verified Conjectures 1 and 2 for n up to 5*10^5 and 10^6 respectively.
See also A300791 and A300844 for similar conjectures.

Examples

			a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 0 = 0^2 and (3*1)^2 + (4*1)^2 + (12*0)^2 = 5^2.
a(14) = 1 since 14 = 3^2 + 0^2 + 1^2 + 2^2 with 1 = 1^2 and (3*3)^2 + (4*0)^2 + (12*1)^2 = 15^2.
a(69) = 1 since 69 = 0^2 + 8^2 + 2^2 + 1^2 with 2*2 = 2^2 and (3*0)^2 + (4*8)^2 + (12*2)^2 = 40^2.
a(671) = 1 since 671 = 18^2 + 17^2 + 3^2 + 7^2 with 3*3 = 3^2 and (3*18)^2 + (4*17)^2 + (12*3)^2 = 94^2.
a(1175) = 1 since 1175 = 30^2 + 5^2 + 9^2 + 13^2 with 9 = 3^2 and (3*30)^2 + (4*5)^2 + (12*9)^2 = 142^2.
a(12151) = 1 since 12151 = 50^2 + 71^2 + 49^2 + 47^2 with 49 = 7^2 and (3*50)^2 + (4*71)^2 + (12*49)^2 = 670^2.
a(16204) = 1 since 16204 = 90^2 + 90^2 + 0^2 + 2^2 with 0 = 0^2 and (3*90)^2 + (4*90)^2 + (12*0)^2 = 450^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[z]||SQ[2z]||SQ[3z], Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(3x)^2+(4y)^2+(12z)^2],r=r+1],{x,0,Sqrt[n-1-z^2]},{y,0,Sqrt[n-1-x^2-z^2]}]],{z,0,Sqrt[n-1]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A272336 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x^4 + y^3*z a square, where x,y,z,w are nonnegative integers with z > 0.

Original entry on oeis.org

1, 3, 2, 1, 4, 5, 1, 3, 5, 5, 5, 3, 4, 7, 3, 1, 10, 10, 4, 5, 8, 5, 1, 5, 6, 12, 7, 1, 10, 7, 1, 3, 11, 8, 6, 6, 4, 11, 2, 5, 15, 11, 4, 5, 12, 6, 2, 3, 8, 15, 8, 4, 13, 16, 2, 7, 11, 6, 10, 4, 11, 13, 4, 1, 15, 20, 5, 10, 15, 9, 1, 10, 10, 18, 11, 4, 15, 9, 1, 5
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 26 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 23, 31, 71, 191, 271, 391, 503, 943, 1591, 2351, 2791, 4791, 8863, 9983, 4^k*m (k = 0,1,2,... and m = 1, 7, 79).
(ii) For each ordered pair (a,b) = (1,1), (1,15), (1,20), (1,36), (1,60), (9,260), any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and w > 0 (or z > 0) such that a*x^4 + b*y^3*z is a square.
See also A280831 for a similar conjecture.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 1^2 + 0^2 with 0^4 + 0^3*1 = 0^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1^4 + 2^3*1 = 3^2.
a(23) = 1 since 23 = 1^2 + 2^2 + 3^2 + 3^2 with 1^4 + 2^3*3 = 5^2.
a(31) = 1 since 31 = 1^2 + 2^2 + 1^2 + 5^2 with 1^4 + 2^3*1 = 3^2.
a(71) = 1 since 71 = 5^2 + 6^2 + 1^2 + 3^2 with 5^4 + 6^3*1 = 29^2.
a(79) = 1 since 79 = 3^2 + 6^2 + 3^2 + 5^2 with 3^4 + 6^3*3 = 27^2.
a(191) = 1 since 191 = 7^2 + 6^2 + 5^2 + 9^2 with 7^4 + 6^3*5 = 59^2.
a(271) = 1 since 271 = 5^2 + 10^2 + 5^2 + 11^2 with 5^4 + 10^3*5 = 75^2.
a(391) = 1 since 391 = 9^2 + 6^2 + 15^2 + 7^2 with 9^4 + 6^3*15 = 99^2.
a(503) = 1 since 503 = 5^2 + 6^2 + 1^2 + 21^2 with
5^4 + 6^3*1 = 29^2.
a(943) = 1 since 943 = 6^2 + 3^2 + 27^2 + 13^2 with 6^4 + 3^3*27 = 45^2.
a(1591) = 1 since 1591 = 18^2 + 27^2 + 3^2 + 23^2 with 18^4 + 27^3*3 = 405^2.
a(2351) = 1 since 2351 = 6^2 + 45^2 + 13^2 + 11^2 with 6^4 + 45^3*13 = 1089^2.
a(2791) = 1 since 2791 = 19^2 + 38^2 + 19^2 + 25^2 with 19^4 + 38^3*19 = 1083^2.
a(4791) = 1 since 4791 = 9^2 + 2^2 + 41^2 + 55^2 with 9^4 + 2^3*41 = 83^2.
a(8863) = 1 since 8863 = 27^2 + 54^2 + 27^2 + 67^2 with 27^4 + 54^3*27 = 2187^2.
a(9983) = 1 since 9983 = 63^2 + 54^2 + 17^2 + 53^2 with 63^4 + 54^3*17 = 4293^2.
		

Crossrefs

Programs

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

Extensions

Typo in example fixed by Zak Seidov, Apr 26 2016

A301303 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x^2 + 23*y^2 = 2^k*m^3 for some k = 0,1,2 and m = 1,2,3,....

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 17 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k (k = 0,1,2,...), 3, 7, 115, 151, 219, 267, 1151, 1367.
We have verified a(n) > 0 for all n = 1..10^8.
See also A301304 and A301314 for similar conjectures.

Examples

			a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 + 23*0 = 1^3.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 + 23*0 = 2^2*1^3.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 2^2 + 23*1^2 = 3^3.
a(48) = 2 since 48 = 4^2 + 0^2 + 4^2 + 4^2 = 6^2 + 2^2 + 2^2 + 2^2 with 4^2 + 23*0^2 = 2*2^3 and 6^2 + 23*2^2 = 2*4^3.
a(115) = 1 since 115 = 3^2 + 3^2 + 4^2 + 9^2 with 3^2 + 23*3^2 = 6^3.
a(267) = 1 since 267 = 3^2 + 1^2 + 1^2 + 16^2 with 3^2 + 23*1^2 = 2^2*2^3.
a(1151) = 1 since 1151 = 7^2 + 3^2 + 2^2 + 33^2 with 7^2 + 23*3^2 = 2^2*4^3.
a(1367) = 1 since 1367 = 17^2 + 5^2 + 18^2 + 27^2 with 17^2 + 23*5^2 = 2^2*6^3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
    QQ[n_]:=n>0&&(CQ[n]||CQ[n/2]||CQ[n/4]);
    tab={};Do[r=0;Do[If[QQ[x^2+23y^2],Do[If[SQ[n-x^2-y^2-z^2],r=r+1],{z,0,Sqrt[(n-x^2-y^2)/2]}]],{y,0,Sqrt[n/2]},{x,y,Sqrt[n-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301304 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x^2 + 7*y^2 = 2^k*m for some k = 0,1,2 and m = 1,2,3,....

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 1, 2, 3, 4, 3, 4, 2, 3, 2, 2, 4, 5, 3, 4, 4, 2, 1, 3, 2, 6, 5, 4, 3, 4, 2, 2, 6, 5, 4, 6, 3, 3, 3, 4, 6, 8, 2, 5, 5, 3, 2, 4, 4, 5, 6, 5, 5, 4, 3, 3, 6, 5, 2, 6, 3, 4, 3, 2, 6, 10, 3, 5, 8, 1, 2, 5, 5, 6, 5, 6, 5, 3, 1, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 17 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
We have verified this for n up to 10^8.
See also A301303 and A301314 for similar conjectures.

Examples

			a(79) = 1 since 79 = 5^2 + 1^2 + 2^2 + 7^2 with 5^2 + 7*1^2 = 2^2*2^3.
a(323) = 1 since 323 = 3^2 + 1^2 + 12^2 + 13^2 with 3^2 + 7*1^2 = 2*2^3.
a(646) = 1 since 646 = 22^2 + 11^2 + 4^2 + 5^2 with 22^2 + 7*11^2 = 11^3.
a(815) = 1 since 815 = 9^2 + 5^2 + 15^2 + 22^2 with 9^2 + 7*5^2 = 2^2*4^3.
a(1111) = 1 since 1111 = 1^2 + 1^2 + 22^2 + 25^2 with 1^2 + 7*1^2 = 2^3.
a(2822) = 1 since 2822 = 2^2 + 0^2 + 3^2 + 53^2 with 2^2 + 7*0^2 = 2^2*1^3.
		

Crossrefs

Programs

  • Mathematica
      SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
    QQ[n_]:=n>0&&(CQ[n]||CQ[n/2]||CQ[n/4]);
    tab={};Do[r=0;Do[If[QQ[x^2+7y^2],Do[If[SQ[n-x^2-y^2-z^2],r=r+1],{z,0,Sqrt[(n-x^2-y^2)/2]}]],{y,0,Sqrt[n/2]},{x,y,Sqrt[n-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301314 Number of ways to write n as x^2 + y^2 + z^2 + w^2, where w is a positive integer and x,y,z are nonnegative integers such that x + 3*y + 9*z = 2^k*m^3 for some k,m = 0,1,2,....

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 18 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: Any positive integer can be written as x^2 + y^2 + z^2 + w^2, where x is a positive integer and y,z,w are nonnegative integers such that 2*x + 7*y = 2^k*m^3 for some k = 0,1,2 and m = 1,2,3,....
We have verified a(n) > 0 for all n = 1..10^7.
See also A301303 and A301304 for similar conjectures.

Examples

			a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1 + 3*2 + 9*1 = 2*2^3.
a(19) = 1 since 19 = 4^2 + 1^2 + 1^2 + 1^2 with 4 + 3*1 + 9*1 = 2*2^3.
a(46) = 1 since 46 = 0^2 + 6^2 + 1^2 + 3^2 with 0 + 3*6 + 9*1 = 3^3.
a(79) = 1 since 79 = 2^2 + 7^2 + 1^2 + 5^2 with 2 + 3*7 + 9*1 = 2^2*2^3.
a(125) = 1 since 125 = 2^2 + 0^2 + 0^2 + 11^2 with 2 + 3*0 + 9*0 = 2*1^3.
a(736) = 1 since 736 = 0^2 + 24^2 + 4^2 + 12^2 with 0 + 3*24 + 9*4 = 2^2*3^3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
    QQ[n_]:=CQ[n]||CQ[n/2]||CQ[n/4];
    tab={};Do[r=0;Do[If[QQ[x+3y+9z]&&SQ[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]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A273616 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (3*x^2+13*y^2)*z a square, where x,y,z,w are nonnegative integers.

Original entry on oeis.org

1, 4, 4, 2, 5, 8, 4, 2, 4, 8, 11, 4, 2, 10, 8, 1, 4, 12, 10, 8, 9, 8, 9, 1, 4, 17, 16, 6, 3, 16, 8, 1, 4, 8, 18, 10, 8, 12, 13, 2, 10, 18, 9, 8, 5, 17, 11, 3, 2, 15, 22, 7, 13, 15, 17, 4, 6, 10, 11, 14, 2, 18, 17, 1, 5, 23, 13, 9, 13, 14, 14, 1, 8, 16, 26, 8, 4, 16, 7, 1, 8
Offset: 0

Views

Author

Zhi-Wei Sun, May 26 2016

Keywords

Comments

Conjecture: For each ordered pair (a,b) = (3,13), (5,11), (15,57), (15,165), (138,150), any natural number can be written as x^2 + y^2 + z^2 + w^2 with (a*x^2+b*y^2)*z a square, where x,y,z,w are nonnegative integers.
For more conjectural refinements of Lagrange's four-square theorem, see the author's preprint arXiv:1604.06723.

Examples

			a(15) = 1 since 15 = 2^2 + 1^2 + 1^2 + 3^2 with (3*2^2+13*1^2)*1 = 5^2.
a(23) = 1 since 23 = 3^2 + 3^2 + 1^2 + 2^2 with (3*3^2+13*3^2)*1 = 12^2.
a(31) = 1 since 31 = 2^2 + 1^2 + 1^2 + 5^2 with (3*2^2+13*1^2)*1 = 5^2.
a(63) = 1 since 63 = 6^2 + 1^2 + 1^2 + 5^2 with (3*6^2+13*1^2)*1 = 11^2.
a(71) = 1 since 71 = 6^2 + 3^2 + 1^2 + 5^2 with (3*6^2+13*3^2)*1 = 15^2.
a(79) = 1 since 79 = 5^2 + 3^2 + 3^2 + 6^2 with (3*5^2+13*3^2)*3 = 24^2.
a(223) = 1 since 223 = 2^2 + 13^2 + 1^2 + 7^2 with (3*2^2+13*13^2)*1 = 47^2.
a(303) = 1 since 303 = 2^2 + 13^2 + 9^2 + 7^2 with (3*2^2+13*13^2)*9 = 141^2.
a(2703) = 1 since 2703 = 15^2 + 25^2 + 22^2 + 37^2 with (3*15^2+13*25^2)*22 = 440^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(3x^2+13y^2)z],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Label[aa];Continue,{n,0,80}]

A343950 Number of ways to write n as x + y + z with x^2 + 4*y^2 + 5*z^2 a square, where x,y,z are positive integers with y or z a positive power of two.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 05 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 7.
We have verified a(n) > 0 for all n = 8..50000. Clearly, a(2*n) > 0 if a(n) > 0.
Conjecture 2: For any integer n > 7, we can write n as x + y + z with x,y,z positive integers such that x^2 + 2*y^2 + 3*z^2 is a square.
Conjecture 3: For any integer n > 4, we can write n as x + y + z with x,y,z positive integers such that 3*x^2 + 4*y^2 + 5*z^2 (or x^2 + 3*y^2 + 5*z^2) is a square.

Examples

			a(4) = 1, and 4 = 1 + 1 + 2 with 1^2 + 4*1^2 + 5*2^2 = 5^2.
a(5) = 1, and 5 = 2 + 2 + 1 with 2^2 + 4*2^2 + 5*1^2 = 5^2.
a(9) = 1, and 9 = 4 + 1 + 4 with 4^2 + 4*1^2 + 5*4^2 = 10^2.
a(14) = 1, and 14 = 7 + 5 + 2 with 7^2 + 4*5^2 + 5*2^2 = 13^2.
a(23) = 1, and 23 = 7 + 8 + 8 with 7^2 + 4*8^2 + 5*8^2 = 25^2.
a(46) = 1, and 46 = 14 + 16 + 16 with 14^2 + 4*16^2 + 5*16^2 = 50^2.
a(71) = 1, and 71 = 42 + 8 + 21 with 42^2 + 4*8^2 + 5*21^2 = 65^2.
a(92) = 1, and 92 = 28 + 32 + 32 with 28^2 + 4*32^2 + 5*32^2 = 100^2.
a(142) = 1, and 142 = 84 + 16 + 42 with 84^2 + 4*16^2 + 5*42^2 = 130^2.
		

Crossrefs

Programs

  • Mathematica
    PowQ[n_]:=PowQ[n]=n>1&&IntegerQ[Log[2,n]];
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(PowQ[y]||PowQ[n-x-y])&&SQ[x^2+4*y^2+5*(n-x-y)^2],r=r+1],{x,1,n-3},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]

A273826 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x*y + y*z + z*w a fourth power, where x is a positive integer, y is a nonnegative integer, and z and w are integers.

Original entry on oeis.org

1, 5, 5, 3, 8, 6, 5, 4, 2, 11, 5, 5, 10, 1, 3, 1, 9, 15, 4, 9, 2, 4, 6, 2, 13, 13, 10, 7, 8, 6, 3, 5, 9, 14, 6, 9, 13, 9, 9, 10, 13, 11, 5, 4, 14, 5, 8, 5, 6, 15, 10, 17, 14, 13, 6, 1, 18, 17, 2, 8, 8, 5, 17, 3, 23, 15, 9, 17, 10, 9
Offset: 1

Views

Author

Zhi-Wei Sun, May 31 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m (k = 0,1,2,... and m = 1, 14, 56, 91, 184, 329, 355, 1016).
(ii) Any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x*y + y*z + z*w a nonnegative cube, where x is a positive integer, y is a nonnegative integer, and z and w are integers.
(iii) For each triple (a,b,c) = (1,1,2), (1,1,3), (1,2,2), (1,2,3), (1,3,4), (1,5,3), (1,6,2), (2,2,6), (4,4,12), (4,4,16), (4,8,8), (4,12,16), (4,20,12), (8,8,16), (8,8,24), (8,8,32), (8,24,16), any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that a*x*y + b*y*z + c*z*w is a fourth power.
For more conjectural refinements of Lagrange's four-square theorem, see the author's preprint arXiv:1604.06723.

Examples

			a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 > 0, 0 = 0 and 1*0 + 0*0 + 0*0 = 0^4.
a(14) = 1 since 14 = 3^2 + 1^2 + (-2)^2 + 0^2 with 3 > 0, 1 > 0 and 3*1 + 1*(-2) + (-2)*0 = 1^4.
a(56) = 1 since 56 = 6^2 + 4^2 + (-2)^2 + 0^2 with 6 > 0, 4 > 0 and 6*4 + 4*(-2) + (-2)*0 = 2^4.
a(91) = 1 since 91 = 4^2 + 7^2 + (-1)^2 + 5^2 with 4 > 0, 7 > 0 and 4*7 + 7*(-1) + (-1)*5 = 2^4.
a(184) = 1 since 184 = 10^2 + 4^2 + (-2)^2 + 8^2 with 10 > 0, 4 > 0 and 10*4 + 4*(-2) + (-2)*8 = 2^4.
a(329) = 1 since 329 = 18^2 + 1^2 + (-2)^2 + 0^2 with 18 > 0, 1 > 0 and 18*1 + 1*(-2) + (-2)*0 = 2^4.
a(355) = 1 since 355 = 17^2 + 1^2 + (-8)^2 + 1^2 with 17 > 0, 1 > 0 and 17*1 + 1*(-8) + (-8)*1 = 1^4.
a(1016) = 1 since 1016 = 2^2 + 20^2 + 6^2 + (-24)^2 with 2 > 0, 20 > 0 and 2*20 + 20*6 + 6*(-24) = 2^4.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    QQ[n_]:=QQ[n]=IntegerQ[n^(1/4)]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&QQ[x*y+y*(-1)^j*z+(-1)^(j+k)*z*Sqrt[n-x^2-y^2-z^2]],r=r+1],{x,1,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]},{j,0,Min[1,z]},{k,0,Min[1,Sqrt[n-x^2-y^2-z^2]]}];Print[n," ",r];Continue,{n,1,70}]

A273875 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with x*y + 2*y*z + 4*z*x a nonnegative cube, where w,x,y,z are integers with w >= 0 and x > 0.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jun 02 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any positive integer can be written as w^2 + x^2 + y^2 + z^2 with x*y + 2*y*z + 4*z*x = 4*t^3 for some t = 0,1,2,..., where w,x,y,z are integers with x > 0. Also, any natural number can be written as w^2 + x^2 + y^2 + z^2 with x*y + 3*y*z + 4*z*x = 3*t^3 for some t = 0,1,2,..., where w,x,y,z are integers with x >= 0.
(iii) For each triple (a,b,c) = (1,1,2), (1,2,3), (3,2,1), (4,1,1), any natural number can be written as w^2 + x^2 + y^2 + z^2 with a*x*y + b*y*z - c*z*w a nonnegative cube, where w,x,y are nonnegative integers and z is an integer.
For more conjectural refinements of Lagrange's four-square theorem, see the author's preprint arXiv:1604.06723.

Examples

			a(1) = 1 since 1 = 0^2 + 1^2 + 0^2 + 0^2 with 1*0 + 2*0*0 + 4*0*1 = 0^3.
a(7) = 1 since 7 = 2^2 + 1^2 + (-1)^2 + 1^2 with 1*(-1) + 2*(-1)*1 + 4*1*1 = 1^3.
a(8) = 1 since 8 = 2^2 + 2^2 + 0^2 + 0^2 with 2*0 + 2*0*0 + 4*0*2 = 0^3.
a(11) = 1 since 11 = 3^2 + 1^2 + 1^2 + 0^2 with 1*1 + 2*1*0 + 4*0*1 = 1^3.
a(12) = 1 since 12 = 3^2 + 1^2 + (-1)^2 + 1^2 with 1*(-1) + 2*(-1)*1 + 4*1*1 = 1^3.
a(15) = 1 since 15 = 1^2 + 1^2 + (-3)^2 + (-2)^2 with 1*(-3) + 2*(-3)*(-2) + 4*(-2)*1 = 1^3.
a(16) = 1 since 16 = 0^2 + 4^2 + 0^2 + 0^2 with 4*0 + 2*0*0 + 4*0*4 = 0^3.
a(48) = 1 since 48 = 4^2 + 4^2 + 0^2 + 4^2 with 4*0 + 2*0*4 + 4*4*4 = 4^3.
a(112) = 1 since 112 = 4^2 + 8^2 + (-4)^2 + 4^2 with 8*(-4) + 2*(-4)*4 + 4*4*8 = 4^3.
a(131) = 1 since 131 = 9^2 + 3^2 + (-4)^2 + 5^2 with 3*(-4) + 2*(-4)*5 + 4*5*3 = 2^3.
a(176) = 1 since 176 = 12^2 + 4^2 + 0^2 + 4^2 with 4*0 + 2*0*4 + 4*4*4 = 4^3.
a(224) = 1 since 224 = 0^2 + 8^2 + 4^2 + 12^2 with 8*4 + 2*4*12 + 4*12*8 = 8^3.
a(304) = 1 since 304 = 4^2 + 4^2 + (-16)^2 + (-4)^2 with 4*(-16) + 2*(-16)*(-4) + 4*(-4)*4 = 0^3.
a(944) = 1 since 944 = 20^2 + 12^2 + (-16)^2 + 12^2 with 12*(-16) + 2*(-16)*12 + 4*12*12 = 0^3.
a(4784) = 1 since 4784 = 60^2 + 28^2 + (-16)^2 + 12^2 with 28*(-16) + 2*(-16)*12 + 4*12*28 = 8^3.
a(8752) = 1 since 8752 = 92^2 + 4^2 + (-16)^2 + (-4)^2 with 4*(-16) + 2*(-16)*(-4) + 4*(-4)*4 = 0^3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    CQ[n_]:=QQ[n]=n>=0&&IntegerQ[n^(1/3)]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&CQ[x*(-1)^j*y+2(-1)^(j+k)*y*z+4*(-1)^k*z*x],r=r+1],{x,1,Sqrt[n]},{y,0,Sqrt[n-x^2]},{j,0,Min[1,y]},{z,0,Sqrt[n-x^2-y^2]},{k,0,Min[1,z]}];Print[n," ",r];Continue,{n,1,80}]
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