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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A282545 Number of ways to write n as x^4 + y^2 + z^2 + w^2 with y^2 + 64*z^2 + 1024*y*z a square, where x,y,z,w are nonnegative integers with y > 0.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 18 2017

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any nonnegative integer can be written as x^4 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*y^2 + b*y*z + c*z^2 is a square, whenever (a,b,c) is among the ordered triples (1,484,44), (1,666,9), (16,1336,169), (25,900,36).
(iii) For each c = 1, 49, any nonnegative integer can be written as x^4 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 120*(x^2+y)*z + c*z^2 is a square.
By the linked JNT paper, each n = 0,1,2,... can be expressed as the sum of a fourth power and three squares.
See also A282463, A282494 and A282495 for similar conjectures.

Examples

			a(1) = 1 since 1 = 0^4 + 1^2 + 0^2 + 0^2 with 1^2 + 64*0^2 + 1024*1*0 = 1^2.
a(31) = 1 since 31 = 1^4 + 2^2 + 1^2 + 5^2 with 2^2 + 64*1^2 + 1024*2*1 = 46^2.
a(47) = 1 since 47 = 1^4 + 6^2 + 3^2 + 1^2 with 6^2 + 64*3^2 + 1024*6*3 = 138^2.
a(79) = 1 since 79 = 1^4 + 7^2 + 2^2 + 5^2 with 7^2 + 64*2^2 + 1024*7*2 = 121^2.
a(156) = 1 since 156 = 3^4 + 5^2 + 5^2 + 5^2 with 5^2 + 64*5^2 + 1024*5*5 = 165^2.
a(184) = 1 since 184 = 0^4 + 12^2 + 6^2 + 2^2 with 12^2 + 64*6^2 + 1024*12*6 = 276^2.
a(316) = 1 since 316 = 2^4 + 10^2 + 10^2 + 10^2 with 10^2 + 64*10^2 + 1024*10*10 = 330^2.
a(380) = 1 since 380 = 1^4 + 3^2 + 3^2 + 19^2 with 3^2 + 64*3^2 + 1024*3*3 = 99^2.
a(2383) = 1 since 2383 = 3^4 + 22^2 + 33^2 + 27^2 with 22^2 + 64*33^2 + 1024*22*33 = 902^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[y^2+1024y*z+64z^2],r=r+1],{x,0,(n-1)^(1/4)},{y,1,Sqrt[n-x^4]},{z,0,Sqrt[n-x^4-y^2]}];Print[n," ",r];Continue,{n,1,80}]

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}]

A282933 Number of ways to write n as x^4 + y^2 + z^2 + w^2 with x,y,z nonnegative integers and w a positive integer such that 8*y^2 - 8*y*z + 9*z^2 is a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 25 2017

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, 8, 15, 23, 39, 47, 71, 93, 239, 287, 311, 319, 383, 391, 591, 632, 1663, 2639, 5591, 6236).
(ii) Each n = 0,1,2,... can be written as x^4 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*y^2 - b*y*z + c*z^2 is a square, whenever (a,b,c) is among the ordered triples (6,21,19), (15,33,22), (16,54,39),(18,51,34), (19,53,34), (21,42,22), (22,69,51).
By the linked JNT paper, each n = 0,1,2,... is the sum of a fourth power and three squares, and we can also write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y*(y-z) = 0. Whether y = 0 or y = z, the number 8*y^2 - 8*y*z + 9*z^2 is definitely a square.
First occurrence of k: 1, 2, 5, 6, 10, 18, 26, 25, 41, 85, 81, 101, 105, 90, 201, 146, 321, 341, 261, 325, 297, 370, 585, 306, 906, ..., . Robert G. Wilson v, Feb 25 2017

Examples

			a(8) = 1 since 8 = 0^4 + 0^2 + 2^2 + 2^2 with 8*0^2 - 8*0*2 + 9*2^2 = 6^2.
a(15) = 1 since 15 = 1^4 + 2^2 + 1^2 + 3^2 with 8*2^2 - 8*2*1 + 9*1^2 = 5^2.
a(23) = 1 since 23 = 1^4 + 3^2 + 3^2 + 2^2 with 8*3^2 - 8*3*3 + 9*3^2 = 9^2.
a(591) = 1 since 591 = 3^4 + 5^2 + 1^2 + 22^2 with 8*5^2 - 8*5*1 + 9*1^2 = 13^2.
a(632) = 1 since 632 = 4^4 + 12^2 + 6^2 + 14^2 with 8*12^2 - 8*12*6 + 9*6^2 = 30^2.
a(1663) = 1 since 1663 = 3^4 + 27^2 + 23^2 + 18^2 with 8*27^2 - 8*27*23 + 9*23^2 = 75^2.
a(2639) = 1 since 2639 = 7^4 + 15^2 + 3^2 + 2^2 with 8*15^2 - 8*15*3 + 9*3^2 = 39^2.
a(5591) = 1 since 5591 = 5^4 + 6^2 + 21^2 + 67^2 with 8*6^2 - 8*6*21 + 9*21^2 = 57^2.
a(6236) = 1 since 6236 = 1^4 + 45^2 + 31^2 + 57^2 with 8*45^2 - 8*45*31 + 9*31^2 = 117^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[8y^2-8*y*z+9z^2],r=r+1],{x,0,(n-1)^(1/4)},{y,0,Sqrt[n-1-x^4]},{z,0,Sqrt[n-1-x^4-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A271381 Number of ordered ways to write n as u^2 + v^2 + x^3 + y^3, where u, v, x, y are nonnegative integers with 2 | u*v, u <= v and x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 06 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 except for n = 23, and a(n) = 1 only for n = 0, 3, 4, 7, 11, 12, 15, 19, 23, 30, 78, 203, 219.
(ii) Any natural number can be written as the sum of two squares and three fourth powers.

Examples

			a(3) = 1 since 3 = 0^2 + 1^2 + 1^3 + 1^3 with 0 even.
a(4) = 1 since 4 = 0^2 + 2^2 + 0^3 + 0^3 with 0 and 2 even.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^3 + 1^3 with 2 even.
a(11) = 1 since 11 = 0^2 + 3^2 + 1^3 + 1^3 with 0 even.
a(12) = 1 since 12 = 0^2 + 2^2 + 0^3 + 2^3 with 0 and 2 even.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^3 + 1^3 with 2 even.
a(19) = 1 since 19 = 1^2 + 4^2 + 1^3 + 1^3 with 4 even.
a(30) = 1 since 30 = 2^2 + 5^2 + 0^3 + 1^3 with 2 even.
a(78) = 1 since 78 = 2^2 + 3^2 + 1^3 + 4^3 with 2 even.
a(203) = 1 since 203 = 7^2 + 10^2 + 3^3 + 3^3 with 10 even.
a(219) = 1 since 219 = 8^2 + 8^2 + 3^3 + 4^3 with 8 even.
		

Crossrefs

Programs

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

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}]

A279522 Number of ways to write n as w^2 + x^2 + y^2 + z^2 with w + 2*x + 3*y + 5*z a square, where w,x,y,z are nonnegative integers.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 3, 1, 1, 2, 2, 2, 1, 3, 5, 2, 2, 5, 4, 1, 1, 2, 4, 1, 0, 7, 3, 1, 1, 1, 4, 6, 4, 5, 4, 4, 3, 3, 5, 2, 3, 5, 4, 2, 1, 4, 8, 5, 1, 5, 12, 1, 1, 2, 6, 3, 4, 3, 3, 9, 1, 6, 4, 2, 3, 8, 8, 2, 3, 7, 7, 7, 3, 8, 14, 3, 2
Offset: 0

Views

Author

Zhi-Wei Sun, Dec 14 2016

Keywords

Comments

Conjecture: (i) a(n) = 0 if and only if n = 16^k*28 for some k = 0,1,2,....
(ii) For any positive integers a,b,c,d, there are infinitely many positive integers which cannot be written as w^2 + x^2 + y^2 + z^2 with a*w + b*x + c*y + d*z a square, where w,x,y,z are nonnegative integers.

Examples

			a(27) = 1 since 27 = 1^2 + 5^2 + 0^2 + 1^2 with 1 + 2*5 + 3*0 + 5*1 = 4^2.
a(31) = 1 since 31 = 1^2 + 2^2 + 5^2 + 1^2 with 1 + 2*2 + 3*5 + 5*1 = 5^2.
a(33) = 1 since 33 = 0^2 + 4^2 + 4^2 + 1^2 with 0 + 2*4 + 3*4 + 5*1 = 5^2.
a(52) = 1 since 52 = 4^2 + 6^2 + 0^2 + 0^2 with 4 + 2*6 + 3*0 + 5*0 = 4^2.
a(55) = 1 since 55 = 1^2 + 5^2 + 5^2 + 2^2 with 1 + 2*5 + 3*5 + 5*2 = 6^2.
a(56) = 1 since 56 = 0^2 + 4^2 + 6^2 + 2^2 with 0 + 2*4 + 3*6 + 5*2 = 6^2.
a(88) = 1 since 88 = 4^2 + 8^2 + 2^2 + 2^2 with 4 + 2*8 + 3*2 + 5*2 = 6^2.
a(137) = 1 since 137 = 10^2 + 6^2 + 1^2 + 0^2 with 10 + 2*6 + 3*1 + 5*0 = 5^2.
a(164) = 1 since 164 = 12^2 + 2^2 + 0^2 + 4^2 with 12 + 2*2 + 3*0 + 5*4 = 6^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[Sqrt[n-x^2-y^2-z^2]+2x+3y+5z],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,0,80}]

A282161 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x and (12*x)^2 + (5*y-10*z)^2 both squares, where x,y,z are nonnegative integers and w is a positive integer.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 07 2017

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, 23, 28, 79, 119, 191, 223, 263, 463, 703, 860, 1052).
(ii) Any positive integer n can be written as x^2 + y^2 + z^2 + w^2 with x and (35*x)^2 + (12*y-24*z)^2 both squares, where x,y,z are nonnegative integers and w is a positive integer.
The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.
See also A281976, A281977, A282013 and A282014 for similar conjectures.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0^2 and (12*0)^2 + (5*0-10*0)^2 = 0^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1 = 1^2 and (12*1)^2 + (5*3-10*2)^2 = 13^2.
a(28) = 1 since 28 = 1^2 + 1^2 + 1^2 + 5^2 with 1 = 1^2 and (12*1)^2 + (5*1-10*1)^2 = 13^2.
a(79) = 1 since 79 = 1^2 + 5^2 + 2^2 + 7^2 with 1 = 1^2 and (12*1)^2 + (5*5-10*2)^2 = 13^2.
a(119) = 1 since 119 = 1^2 + 9^2 + 1^2 + 6^2 with 1 = 1^2 and (12*1)^2 + (5*9-10*1)^2 = 37^2.
a(191) = 1 since 191 = 9^2 + 5^2 + 7^2 + 6^2 with 9 = 3^2 and (12*9)^2 + (5*5-10*7)^2 = 117^2.
a(223) = 1 since 223 = 1^2 + 13^2 + 7^2 + 2^2 with 1 = 1^2 and (12*1)^2 + (5*13-10*7)^2 = 13^2.
a(263) = 1 since 263 = 9^2 + 13^2 + 2^2 + 3^2 with 9 = 3^2 and (12*9)^2 + (5*13-10*2)^2 = 117^2.
a(463) = 1 since 463 = 1^2 + 19^2 + 10^2 + 1^2 with 1 = 1^2 and (12*1)^2 + (5*19-10*10)^2 = 13^2.
a(703) = 1 since 703 = 1^2 + 13^2 + 7^2 + 22^2 with 1 = 1^2 and (12*1)^2 + (5*13-10*7)^2 = 13^2.
a(860) = 1 since 860 = 4^2 + 18^2 + 18^2 + 14^2 with 4 = 2^2 and (12*4)^2 + (5*18-10*18)^2 = 102^2.
a(1052) = 1 since 1052 = 4^2 + 30^2 + 6^2 + 10^2 with 4 = 2^2 and (12*4)^2 + (5*30-10*6)^2 = 102^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[144x^4+(5y-10z)^2],r=r+1],{x,0,(n-1)^(1/4)},{y,0,Sqrt[n-1-x^4]},{z,0,Sqrt[n-1-x^4-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A282226 Least nonnegative integer m which can be written in exactly n ways as x^2 + y^2 + z^2 + w^2 with both x and x + 24*y squares, where x,y,z,w are nonnegative integers with z <= w.

Original entry on oeis.org

0, 1, 2, 10, 18, 34, 41, 52, 66, 100, 90, 130, 261, 306, 226, 370, 426, 405, 612, 585, 661, 626, 770, 666, 756, 706, 810, 981, 882, 1026, 1266, 1170, 1330, 1530, 1476, 1426, 1881, 1701, 2650, 2410, 2506, 1666, 1386, 2226, 3861, 2626, 3366, 3006, 2106, 2610, 3346, 3186, 3226, 4410, 3786, 3850, 2826, 3762, 4026, 4500
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 09 2017

Keywords

Comments

Conjecture: a(n) exists for any n > 0.
See also A281976 for a related conjecture.

Examples

			a(1) = 0 since 0 = 0^2 + 0^2 + 0^2 + 0^2 with 0 = 0^2 and 0 + 24*0 = 0^2.
a(2) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0^2 and 0 + 24*0 = 0^2, and 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 = 1^2 and 1 + 24*0 = 1^2.
a(4) = 10 since 10 = 0^2 + 0^2 + 1^2 + 3^2 with 0 = 0^2 and 0 + 24*0 = 0^2, 10 = 1^2 + 0^2 + 0^2 + 3^2 with 1 = 1^2 and 1 + 24*0 = 1^2, 10 = 1^2 + 1^2 + 2^2 + 2^2 with 1 = 1^2 and 1 + 24*1 = 5^2, and 10 = 1^2 + 2^2 + 1^2 + 2^2 with 1 = 1^2 and 1 + 24*2 = 7^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Do[m=0;Label[aa];r=0;Do[If[SQ[m-x^4-y^2-z^2]&&SQ[x^2+24y],r=r+1;If[r>n,m=m+1;Goto[aa]]],{x,0,m^(1/4)},{y,0,Sqrt[m-x^4]},{z,0,Sqrt[(m-x^4-y^2)/2]}];If[r
    				

A282972 Number of ways to write n as 4*x^4 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 79*y^2 - 220*y*z + 205*z^2 is a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 25 2017

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,....
(ii) Any positive integer n can be written as 4*x^4 + y^2 + z^2 + w^2 with x,y,z nonnegative integers and w a positive integer such that 169*y^2 - 444*y*z + 396*z^2 (or 289*y^2 - 654*y*z + 401*z^2) is a square.
This is much stronger than Lagrange's four-square theorem, and we have verified parts (i) and (ii) of the conjecture for n up to 10^7 and 10^6 respectively.
By the linked JNT paper, any nonnegative integer n can be written as 4*x^4 + y^2 + z^2 + w^2 with x,y,z,w integers, and we can also write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and (y-z)*(y-2*z) = 0. Whether y = z or y = 2*z, the number 79*y^2 - 220*y*z + 205*z^2 is definitely a square.
See also A282933 for a similar conjecture.

Examples

			a(2) = 1 since 2 = 4*0^4 + 1^2 + 1^2 + 0^2 with 79*1^2 - 220*1*1 + 205*1^2 = 8^2.
a(35) = 1 since 35 = 4*0^4 + 3^2 + 1^2 + 5^2 with 79*3^2 - 220*3*1 + 205*1^2 = 16^2.
a(119) = 1 since 119 = 4*1^4 + 9^2 + 3^2 + 5^2 with 79*9^2 - 220*9*3 + 205*3^2 = 48^2.
a(124) = 1 since 124 = 4*1^4 + 4^2 + 2^2 + 10^2 with 79*4^2 - 220*4*2 + 205*2^2 = 18^2.
a(1564) = 1 since 1564 = 4*3^4 + 14^2 + 30^2 + 12^2 with 79*14^2 - 220*14*30 + 205*30^2 = 328^2.
a(4619) = 1 since 4619 = 4*2^4 + 51^2 + 27^2 + 35^2 with 79*51^2 - 220*51*27 + 205*27^2 = 228^2.
a(6127) = 1 since 6127 = 4*5^4 + 49^2 + 35^2 + 1^2 with 79*49^2 - 220*49*35 + 205*35^2 = 252^2.
a(7119) = 1 since 7119 = 4*1^4 + 51^2 + 17^2 + 65^2 with 79*51^2 - 220*51*17 + 205*17^2 = 272^2.
a(9087) = 1 since 9087 = 4*3^4 + 61^2 + 71^2 + 1^2 with 79*61^2 - 220*61*71 + 205*71^2 = 612^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-4x^4-y^2-z^2]&&SQ[79y^2-220*y*z+205z^2],r=r+1],{x,0,(n/4)^(1/4)},{y,0,Sqrt[n-4x^4]},{z,0,Sqrt[n-4x^4-y^2]}];Print[n," ",r];Continue,{n,0,80}]
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