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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A281659 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with 9*x^2 + 16*y^2 + 24*z^2 + 48*w^2 a square, where x,y,z,w are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 26 2017

Keywords

Comments

Conjecture: (i) a(n) > 0 except for n = 71, 85.
(ii) Any nonnegative integer n not among 39, 71 and 649 can be written as x^2 + y^2 + z^2 + w^2 with x^2 + 16*y^2 + 24*z^2 + 32*w^2 a square, where x,y,z,w are integers.
(iii) Each nonnegative integer n not among 5, 7, 23, 47, 93, 103, 109, 151, 191 and 911 can be written as x^2 + y^2 + z^2 + w^2 with x^2 + 3*y^2 + 8*z^2 + 16*w^2 a square, where x,y,z,w are integers.
(iv) Any nonnegative integer n not among 15, 19, 71, 103, 191 and 559 can be written as x^2 + y^2 + z^2 + w^2 with 9*x^2 + 48*y^2 + 64*z^2 + 96*w^2 a square, where x,y,z,w are integers.
(v) Any nonnegative integer n not among 5, 93, 95, 161 and 309 can be written as x^2 + y^2 + z^2 + w^2 with 16*x^2 + 25*y^2 + 48*z^2 + 128*w^2 a square, where x,y,z,w are integers.
(vi) Each nonnegative integer not among 13, 19, 39, 41, 71, 109, 131, 193, 233, 377, 415 and 941 can be written as x^2 + y^2 + z^2 + w^2 with x^2 + 24*y^2 + 48*z^2 + 144*w^2 a square, where x,y,z,w are integers.
We have verified part (i) of the conjecture for n up to 1.2*10^5.

Examples

			a(11) = 1 since 11 = 3^2 + 1^2 + 1^2 + 0^2 with 9*3^2 + 16*1^2 + 24*1^2 + 48*0 = 11^2.
a(170) = 1 since 170 = 3^2 + 6^2 + 2^2 + 11^2 with 9*3^2 + 16*6^2 + 24*2^2 + 48*11^2 = 81^2.
a(305) = 1 since 305 = 0^2 + 15^2 + 4^2 + 8^2 with 9*0^2 + 16*15^2 + 24*4^2 + 48*8^2 = 84^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[9x^2+16y^2+24z^2+48*(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]}];Print[n," ",r];Label[aa];Continue,{n,0,80}]

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]
Previous Showing 51-52 of 52 results.