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 1-2 of 2 results.

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}]
Showing 1-2 of 2 results.