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.

Previous Showing 31-38 of 38 results.

A273458 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x-y+z+w a nonnegative cube, where x,y,z,w are integers with x >= y >= 0 and x >= |z| <= |w|.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 3, 3, 2, 2, 3, 2, 1, 5, 4, 3, 2, 1, 4, 3, 3, 6, 3, 2, 5, 3, 9, 3, 1, 1, 7, 5, 3, 7, 10, 4, 6, 2, 10, 2, 6, 2, 12, 7, 2, 5, 9, 3, 3, 6, 13, 3, 8, 3, 18, 3, 8, 5, 7, 3, 3, 5, 13, 8, 5, 3, 19, 4, 7, 7, 16, 1, 11, 5, 14, 7, 2, 3, 12, 5, 4
Offset: 0

Views

Author

Zhi-Wei Sun, May 22 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,....
In the latest version of arXiv:1605.03074, the authors showed that any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x + y + z + w is a cube (or a square).
For more conjectural refinements of Lagrange's four-square theorem, see the author's preprint arXiv:1604.06723.

Examples

			a(12) = 1 since 12 = 3^2 + 1^2 + (-1)^2 + (-1)^2 with 3 - 1 + (-1) + (-1) = 0^3.
a(17) = 1 since 17 = 2^2 + 0^2 + 2^2 + (-3)^2 with 2 - 0 + 2 + (-3) = 1^3.
a(28) = 1 since 28 = 3^2 + 1^2 + 3^2 + 3^2 with 3 - 1 + 3 + 3 = 2^3.
a(29) = 1 since 29 = 3^2 + 0^2 + 2^2 + (-4)^2 with 3 - 0 + 2 + (-4) = 1^3.
a(71) = 1 since 71 = 5^2 + 1^2 + 3^2 + (-6)^2 with 5 - 1 + 3 + (-6) = 1^3.
a(149) = 1 since 149 = 8^2 + 0^2 + 2^2 + (-9)^2 with 8 - 0 + 2 + (-9) = 1^3.
a(188) = 1 since 188 = 13^2 + 3^2 + 1^2 + (-3)^2 with 13 - 3 + 1 + (-3) = 2^3.
a(284) = 1 since 284 = 15^2 + 5^2 + 3^2 + (-5)^2 with 15 - 5 + 3 + (-5) = 2^3.
		

Crossrefs

Programs

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

A273568 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w + x + 2*y - 4*z twice a nonnegative cube, where w is an integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 2, 2, 2, 4, 3, 3, 4, 1, 2, 2, 1, 4, 6, 2, 4, 5, 3, 5, 5, 4, 1, 4, 5, 3, 3, 3, 1, 5, 4, 4, 4, 6, 8, 5, 1, 5, 4, 3, 13, 9, 2, 6, 2, 4, 7, 9, 8, 7, 8, 5, 6, 2, 4, 5, 7, 9, 11, 5, 2, 5, 10, 6, 12, 9, 4
Offset: 0

Views

Author

Zhi-Wei Sun, May 25 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,....
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 + 0^2 + 1^2 + 0^2 with 0 + 0 + 2*1 - 4*0 = 2*1^3.
a(3) = 1 since 3 = (-1)^2 + 1^2 + 1^2 + 0^2 with (-1) + 1 + 2*1 - 4*0 = 2*1^3.
a(13) = 1 since 13 = (-2)^2 + 2^2 + 2^2 + 1^2 with (-2) + 2 + 2*2 - 4*1 = 2*0^3.
a(16) = 1 since 16 = 2^2 + 2^2 + 2^2 + 2^2 with 2 + 2 + 2*2 - 4*2 = 2*0^3.
a(26) = 1 since 26 = 3^2 + 3^2 + 2^2 + 2^2 with 3 + 3 + 2*2 - 4*2 = 2*1^3.
a(32) = 1 since 32 = (-4)^2 + 4^2 + 0^2 + 0^2 with (-4) + 4 + 2*0 - 4*0 = 2*0^3.
a(40) = 1 since 40 = (-2)^2 + 4^2 + 4^2 + 2^2 with (-2) + 4 + 2*4 - 4*2 = 2*1^3.
a(218) = 1 since 218 = (-6)^2 + 6^2 + 11^2 + 5^2 with (-6) + 6 + 2*11 - 4*5 = 2*1^3.
a(416) = 1 since 416 = (-4)^2 + 20^2 + 0^2 + 0^2 with (-4) + 20 + 2*0 - 4*0 = 2*2^3.
a(544) = 1 since 544 = (-4)^2 + 20^2 + 8^2 + 8^2 with (-4) + 20 + 2*8 - 4*8 = 2*0^3.
a(800) = 1 since 800 = (-20)^2 + 20^2 + 0^2 + 0^2 with (-20) + 20 + 2*0 - 4*0 = 2*0^3.
a(1184) = 1 since 1184 = (-28)^2 + 12^2 + 16^2 + 0^2 with (-28) + 12 + 2*16 - 4*0 = 2*2^3.
a(2080) = 1 since 2080 = (-20)^2 + 20^2 + 32^2 + 16^2 with (-20) + 20 + 2*32 - 4*16 = 2*0^3.
a(6304) = 1 since 6304 = (-36)^2 + 36^2 + 56^2 + 24^2 with (-36) + 36 + 2*56 - 4*24 = 2*2^3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    CQ[n_]:=CQ[n]=n>=0&&IntegerQ[n^(1/3)]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&CQ[(x+2y-4z+(-1)^k*Sqrt[n-x^2-y^2-z^2])/2],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]},{k,0,Min[1,n-x^2-y^2-z^2]}];Print[n," ",r];Continue,{n,0,70}]

A271719 Number of ordered ways to write n as x + y + z with x >= y > 0, z > 0 and gcd(x,y,z) = 1 such that x^2 + (2*y+z)^2 is a square.

Original entry on oeis.org

0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 1, 3, 1, 2, 4, 3, 1, 3, 4, 3, 2, 2, 1, 4, 2, 5, 4, 5, 2, 6, 4, 7, 4, 6, 4, 8, 5, 7, 7, 10, 3, 9, 7, 10, 9, 10, 4, 8, 5, 6, 4, 9, 1, 8, 5, 7, 6, 12, 4, 17, 11, 15, 10, 15, 8, 21, 12, 15, 9
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 12 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 10, and a(n) = 1 only for n = 11, 12, 14, 18, 24, 54.
See also A271714 for a similar conjecture refining Lagrange's four-square theorem.

Examples

			a(6) = 2 since 6 = 3 + 1 + 2 with 3 > 1, gcd(3,1,2) = 1 and 3^2 + (2*1+2)^2 = 5^2, and also 6 = 4 + 1 + 1 with 4 > 1, gcd(4,1,1) = 1 and 4^2 + (2*1+1)^2 = 5^2.
a(11) = 1 since 11 = 6 + 3 + 2 with 6 > 3, gcd(6,3,2) = 1 and 6^2 + (2*3+2)^2 = 10^2.
a(12) = 1 since 12 = 5 + 5 + 2 with 5 = 5, gcd(5,5,2) = 1 and 5^2 + (2*5+2)^2 = 13^2.
a(14) = 1 since 14 = 5 + 3 + 6 with 5 > 3, gcd(5,3,6) = 1 and 5^2 + (2*3+6)^2 = 13^2.
a(18) = 1 since 18 = 8 + 5 + 5 with 8 > 5, gcd(8,5,5) = 1 and 8^2 + (2*5+5)^2 = 17^2.
a(24) = 1 since 24 = 7 + 7 + 10 with 7 = 7, gcd(7,7,10) = 1 and 7^2 + (2*7+10)^2 = 25^2.
a(54) = 1 since 54 = 28 + 19 + 7 with 28 > 19, gcd(28,19,7) = 1 and 28^2 + (2*19+7)^2 = 53^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[GCD[x,y,n-x-y]==1&&SQ[x^2+(2y+(n-x-y))^2],r=r+1],{x,1,n-2},{y,1,Min[x,n-1-x]}];Print[n," ",r];Label[aa];Continue,{n,1,70}]

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

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}]

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}]

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

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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}]
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