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.

Showing 1-4 of 4 results.

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]

A301375 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z nonnegative integers and w a positive integer such that x*(y+3*z) is a cube or half a cube and y <= z <= w if x = 0.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 19 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
We have verified this for all n = 1..3*10^6.

Examples

			a(14) = 1 since 14 = 0^2 + 1^2 + 2^2 + 3^2 with 0*(1+3*2) = 0^3.
a(15) = 1 since 15 = 2^2 + 1^2 + 1^2 + 3^2 with 2*(1+3*1) = 2^3.
a(16) = 1 since 16 = 0^2 + 0^2 + 0^2 + 4^2 with 0*(0+3*0) = 0^3.
a(60) = 1 since 60 = 4^2 + 2^2 + 2^2 + 6^2 with 4*(2+3*2) = 4^3/2.
a(92) = 1 since 92 = 6^2 + 6^2 + 4^2 + 2^2 with 6*(6+3*4) = 6^3/2.
a(240) = 1 since 240 = 2^2 + 14^2 + 6^2 + 2^2 with 2*(14+3*6) = 4^3.
a(807) = 1 since 807 = 1^2 + 21^2 + 2^2 + 19^2 with 1*(21+3*2) = 3^3.
		

Crossrefs

Programs

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