A273429 -id:A273429 - cp's OEIS frontend

A273432 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 2*x + y - z a nonnegative cube, where x,y,z,w are nonnegative integers with y <= z.

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

Offset: 0

Zhi-Wei Sun, May 22 2016

nonn

Conjecture: (i) For each c = 1, 2, 4 and n = 0,1,2,..., we can write n as x^2 + y^2 + z^2 + w^2 with c*(2x+y-z) a nonnegative cube, where x,y,z,w are nonnegative integers with y <= z.
(ii) Each n = 0,1,2,.... can be written as x^2 + y^2 + z^2 + w^2 with x-y+z a nonnegative cube, where x,y,z,w are integers with x >= y >= 0 and x >= |z|.
The author proved in arXiv:1604.06723 that for each a = 1, 2 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 + a*z is a cube.
See also A273458 for a similar conjecture.
For more conjectural refinements of Lagrange's four-square theorem, see the author's preprint arXiv:1604.06723.

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0 and 2*0 + 0 - 0 = 0^3.
a(4) = 1 since 4 = 0^2 + 0^2 + 0^2 + 2^2 with 0 = 0 and 2*0 + 0 - 0 = 0^3.
a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 2 = 2 and 2*0 + 2 - 2 = 0^3.
a(10) = 1 since 10 = 1^2 + 1^2 + 2^2 + 2^2 with 1 < 2 and 2*1 + 1 - 2 = 1^3.
a(13) = 1 since 13 = 2^2 + 0^2 + 3^2 + 0^2 with 0 < 3 and 2*2 + 0 - 3 = 1^3.
a(23) = 1 since 23 = 1^2 + 2^2 + 3^2 + 3^2 with 2 < 3 and 2*1 + 2 - 3 = 1^3.
a(26) = 1 since 26 = 1^2 + 3^2 + 4^2 + 0^2 with 3 < 4 and 2*1 + 3 - 4 = 1^3.
a(28) = 1 since 28 = 4^2 + 2^2 + 2^2 + 2^2 with 2 = 2 and 2*4 + 2 - 2 = 2^3.
a(40) = 1 since 40 = 4^2 + 2^2 + 2^2 + 4^2 with 2 = 2 and 2*4 + 2 - 2 = 2^3.
a(104) = 1 since 104 = 4^2 + 6^2 + 6^2 + 4^2 with 6 = 6 and 2*4 + 6 - 6 = 2^3.
a(138) = 1 since 138 = 3^2 + 5^2 + 10^2 + 2^2 with 5 < 10 and 2*3 + 5 - 10 =1^3.
a(200) = 1 since 200 = 0^2 + 10^2 + 10^2 + 0^2 with 10 = 10 and 2*0 + 10 - 10 = 0^3.
a(296) = 1 since 296 = 8^2 + 6^2 + 14^2 + 0^2 with 6 < 14 and 2*8 + 6 - 14 = 2^3.
a(328) = 1 since 328 = 0^2 + 6^2 + 6^2 + 16^2 with 6 = 6 and 2*0 + 6 - 6 = 0^3.
a(520) = 1 since 520 = 4^2 + 2^2 + 10^2 + 20^2 with 2 < 10 and 2*4 + 2 - 10 = 0^3.
a(776) = 1 since 776 = 0^2 + 10^2 + 10^2 + 24^2 with 10 = 10 and 2*0 + 10 - 10 = 0^3.
a(1832) = 1 since 1832 = 4^2 + 30^2 + 30^2 + 4^2 with 30 = 30 and 2*4 + 30 - 30 = 2^3.
a(2976) = 1 since 2976 = 20^2 + 16^2 + 48^2 + 4^2 with 16 < 48 and 2*20 + 16 - 48 = 2^3.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A000578, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273458, A273568.

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

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

Zhi-Wei Sun, May 22 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Yu-Chen Sun and Zhi-Wei Sun, Two refinements of Lagrange's four-square theorem, arXiv:1605.03074 [math.NT], 2016.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A000578, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273432, A273568.

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 + 2y - 4z 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

Zhi-Wei Sun, May 25 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A000578, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273432, A273458.

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}]

A275297 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^3 with x + 2*y a square, where x,y,z,w are nonnegative integers with z >= w.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Jul 22 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 3, 7, 8, 11, 12, 15, 23, 24, 32, 39, 47, 71, 103, 120, 136, 159, 176, 183, 218, 359, 463.
Compare this conjecture with Conjecture 5.1 of the author's preprint arXiv:1604.06723. See also A275298 and A275299 for similar conjectures.
By Theorem 1.1 of arXiv:1604.06723, any natural number can be written as the sum of three squares and a sixth power.
Let c be 1 or 2. By the conjecture in A272979, any n = 0,1,2,... can be written as x^2 + 2*y^2 + z^3 + 2*c^2*w^4 with x,y,z,w nonnegative integers, and hence n = x^2 + (y+c*w^2)^2 + (y-c*w^2)^2 + z^3 with (y+c*w^2)-(y-c*w^2) = 2*c*w^2. If n > 0 is not among the 174 terms in the b-file of A275169, then the conjecture in A275169 implies that n can be written as x^2 + y^2 + z^2 + w^3 with x - y = 0^2, where x,y,z,w are nonnegative integers. If n is among the 174 terms in the b-file of A275169, then we may use a computer to verify that n can be written as x^2 + y^2 + z^2 + w^3 with c*(x-y) a square, where x,y,z,w are nonnegative integers.

			a(0) = 1 since 0 = 0^2 + 0^2 + 0^2 + 0^3 with 0 + 2*0 = 0^2 and 0 = 0.
a(1) = 2 since 1 = 0^2 + 0^2 + 1^2 + 0^3 with 0 + 2*0 = 0^2 and 1 > 0, and also 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 + 2*0 = 1^2 and 0 = 0.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^3 with 1 + 2*0 = 1^2 and 1 = 1.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^3 with 2 + 2*1 = 2^2 and 1 = 1.
a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^3 with 0 + 2*2 = 2^2 and 2 > 0.
a(11) = 1 since 11 = 1^2 + 0^2 + 3^2 + 1^3 with 1 + 2*0 = 1^2 and 3 > 1.
a(12) = 1 since 12 = 0^2 + 0^2 + 2^2 + 2^3 with 0 + 2*0 = 0^2 and 2 = 2.
a(15) = 1 since 15 = 2^2 + 1^2 + 3^2 + 1^3 with 2 + 2*1 = 2^2 and 3 > 1.
a(23) = 1 since 23 = 3^2 + 3^2 + 2^2 + 1^3 with 3 + 2*3 = 3^2 and 2 > 1.
a(24) = 1 since 24 = 0^2 + 0^2 + 4^2 + 2^3 with 0 + 2*0 = 0^2 and 4 > 2.
a(32) = 1 since 32 = 4^2 + 0^2 + 4^2 + 0^3 with 4 + 2*0 = 2^2 and 4 > 0.
a(39) = 1 since 39 = 5^2 + 2^2 + 3^2 + 1^3 with 5 + 2*2 = 3^2 and 3 > 1.
a(47) = 1 since 47 = 0^2 + 2^2 + 4^2 + 3^3 with 0 + 2*2 = 2^2 and 4 > 3.
a(71) = 1 since 71 = 6^2 + 5^2 + 3^2 + 1^3 with 6 + 2*5 = 4^2 and 3 > 1.
a(103) = 1 since 103 = 2^2 + 7^2 + 7^2 + 1^3 with 2 + 2*7 = 4^2 and 7 > 1.
a(120) = 1 since 120 = 5^2 + 2^2 + 8^2 + 3^3 with 5 + 2*2 = 3^2 and 8 > 3.
a(136) = 1 since 136 = 0^2 + 8^2 + 8^2 + 2^3 with 0 + 2*8 = 4^2 and 8 > 2.
a(159) = 1 since 159 = 10^2 + 3^2 + 7^2 + 1^3 with 10 + 2*3 = 4^2 and 7 > 1.
a(176) = 1 since 176 = 2^2 + 1^2 + 12^2 + 3^3 with 2 + 2*1 = 2^2 and 12 > 3.
a(183) = 1 since 183 = 6^2 + 5^2 + 11^2 + 1^3 with 6 + 2*5 = 4^2 and 11 > 1.
a(218) = 1 since 218 = 5^2 + 2^2 + 8^2 + 5^3 with 5 + 2*2 = 3^2 and 8 > 5.
a(359) = 1 since 359 = 11^2 + 7^2 + 8^2 + 5^3 with 11 + 2*7 = 5^2 and 8 > 5.
a(463) = 1 since 463 = 2^2 + 17^2 + 13^2 + 1^3 with 2 + 2*17 = 6^2 and 13 > 1.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000290, A000578, A271518, A272979, A273404, A273429, A275169, A275298, A275299.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0;Do[If[CQ[n-x^2-y^2-z^2]&&SQ[x+2y]&&(n-x^2-y^2-z^2)^(1/3)<=z,r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,Floor[(n-x^2-y^2)^(1/3)],Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,0,80}]

A273915 Number of ordered ways to write n as w^5 + x^2 + y^2 + z^2, where w,x,y,z are nonnegative integers with x <= y <= z.

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 2, 1, 1, 3, 3, 2, 2, 2, 2, 1, 1, 3, 4, 3, 2, 2, 2, 1, 1, 3, 4, 4, 2, 2, 3, 1, 2, 4, 5, 4, 4, 4, 4, 2, 2, 6, 5, 3, 3, 4, 4, 1, 2, 5, 7, 6, 4, 4, 6, 3, 2, 5, 5, 5, 2, 4, 5, 2, 2, 6, 8, 5, 5, 5, 5, 1, 3, 7, 6, 6, 4, 5, 4, 1, 2

Offset: 0

Zhi-Wei Sun, Jun 03 2016

nonn

Let c be 1 or 4. Then any nonnegative integer n can be written as c*w^5 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers. We now prove this by induction on n. For n < 2^(10) this can be verified directly via a computer. If 2^(10) divides n, then by the induction hypothesis we can write n/2^(10) as c*w^5 + x^2 + y^2 + z^2 with w,x,y,z, nonnegative integers, and hence n = c*(2^2*w)^5 + (2^5*x)^2 + (2^5*y)^2 + (2^5*z)^2. If n is not of the form 4^k*(8m+7) with k and m nonnegative integers, then n is the sum of three squares and hence n = c*0^5 + x^2 + y^2 + z^2 for some integers x,y,z. When n = 4^k*(8m+7) > 2^(10) with k < 5, it is easy to see that n - c*1^5 or n - c*2^5 is the sum of three squares.
For any positive integer k and for each c = 2, 6, any natural number n can be written as c*w^k + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers. In fact, for every n = 0,1,2,... either n - c*0^k or n - c*1^k can be written as the sum of three squares.
See also  A270969 and  A273429 for similar results.
For some conjectural refinements of Lagrange's four-square theorem, one may consult the author's preprint arXiv:1604.06723

			a(0) = 1 since 0 = 0^5 + 0^2 + 0^2 + 0^2.
a(7) = 1 since 7 = 1^5 + 1^2 + 1^2 + 2^2.
a(8) = 1 since 8 = 0^5 + 0^2 + 2^2 + 2^2.
a(15) = 1 since 15 = 1^5 + 1^2 + 2^2 + 3^2.
a(16) = 1 since 16 = 0^5 + 0^2 + 0^2 + 4^2.
a(23) = 1 since 23 = 1^5 + 2^2 + 3^2 + 3^2.
a(24) = 1 since 24 = 0^2 + 2^2 + 2^2 + 4^2.
a(31) = 1 since 31 = 1^5 + 1^2 + 2^2 + 5^2.
a(47) = 1 since 47 = 1^5 + 1^2 + 3^2 + 6^2.
a(71) = 1 since 71 = 1^5 + 3^2 + 5^2 + 6^2.
a(79) = 1 since 79 = 1^5 + 2^2 + 5^2 + 7^2.
a(92) = 1 since 92 = 1^5 + 1^2 + 3^2 + 9^2.
a(112) = 1 since 112 = 2^5 + 0^2 + 4^2 + 8^2.
a(143) = 1 since 143 = 1^5 + 5^2 + 6^2 + 9^2.
a(191) = 1 since 191 = 1^5 + 3^2 + 9^2 + 10^2.
a(240) = 1 since 240 = 2^5 + 0^2 + 8^2 + 12^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.NT], 2016-2017.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.

Cf. A000118, A000290, A000584, A270969, A273429, A273917.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-w^5-x^2-y^2],r=r+1],{w,0,n^(1/5)},{x,0,Sqrt[(n-w^5)/3]},{y,x,Sqrt[(n-w^5-x^2)/2]}];Print[n," ",r];Label[aa];Continue,{n,0,80}]

A272979 Number of ways to write n as x^2 + 2y^2 + 3z^3 + 4*w^4 with x,y,z,w nonnegative integers.

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 3, 3, 3, 4, 2, 3, 4, 3, 1, 3, 4, 1, 3, 3, 2, 3, 4, 2, 3, 5, 3, 4, 4, 3, 4, 4, 4, 4, 4, 2, 7, 5, 2, 4, 6, 4, 3, 4, 3, 3, 4, 3, 4, 2, 3, 6, 3, 3, 5, 5, 2, 7, 5, 1, 5, 6, 3, 1, 6, 2, 5, 5, 5, 4, 5

Offset: 0

Zhi-Wei Sun, Jul 13 2016

nonn

Conjecture: For positive integers a,b,c,d, any natural number can be written as a*x^2 + b*y^2 + c*z^3 + d*w^4 with x,y,z,w nonnegative integers, if and only if (a,b,c,d) is among the following 49 quadruples: (1,2,1,1), (1,3,1,1), (1,6,1,1), (2,3,1,1), (2,4,1,1), (1,1,2,1), (1,4,2,1), (1,2,3,1), (1,2,4,1), (1,2,12,1), (1,1,1,2), (1,2,1,2), (1,3,1,2), (1,4,1,2), (1,5,1,2), (1,11,1,2), (1,12,1,2), (2,4,1,2), (3,5,1,2), (1,1,4,2), (1,1,1,3), (1,2,1,3), (1,3,1,3), (1,2,4,3), (1,2,1,4), (1,3,1,4), (2,3,1,4), (1,1,2,4), (1,2,2,4), (1,8,2,4), (1,2,3,4), (1,1,1,5), (1,2,1,5), (2,3,1,5), (2,4,1,5), (1,3,2,5), (1,1,1,6), (1,3,1,6), (1,1,2,6), (1,2,1,8), (1,2,4,8), (1,2,1,10), (1,1,2,10), (1,2,1,11), (2,4,1,11), (1,2,1,12), (1,1,2,13), (1,2,1,14),(1,2,1,15).

See also A262824, A262827, A262857 and A273917 for similar conjectures.

			a(0) = 1 since 0 = 0^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(1) = 1 since 1 = 1^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(2) = 1 since 2 = 0^2 + 2*1^2 + 3*0^3 + 4*0^4.
a(14) = 1 since 14 = 3^2 + 2*1^2 + 3*1^3 + 4*0^4.
a(17) = 1 since 17 = 3^2 + 2*2^2 + 3*0^3 + 4*0^4.
a(59) = 1 since 59 = 3^2 + 2*5^2 + 3*0^3 + 4*0^4.
a(63) = 1 since 63 = 3^2 + 2*5^2 + 3*0^2 + 4*1^4.
a(287) = 1 since 287 = 11^2 + 2*9^2 + 3*0^2 + 4*1^4.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000290, A000578, A000583, A262824, A262827, A262857, A270969, A273429, A273915, A273917.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-4w^4-3z^3-2y^2],r=r+1],{w,0,(n/4)^(1/4)},{z,0,((n-4w^4)/3)^(1/3)},{y,0,((n-4w^4-3z^3)/2)^(1/2)}];Print[n," ",r];Continue,{n,0,100}]

A273616 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (3x^2+13y^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

Zhi-Wei Sun, May 26 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273432, A273458, A273568.

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}]

A273917 Number of ordered ways to write n as w^2 + 3*x^2 + y^4 + z^5, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 2, 1, 2, 4, 2, 1, 2, 2, 2, 1, 1, 3, 3, 1, 2, 5, 3, 1, 4, 4, 2, 2, 1, 2, 3, 1, 4, 8, 4, 1, 4, 4, 1, 1, 5, 8, 5, 3, 3, 3, 2, 1, 6, 6, 1, 1, 4, 7, 5, 3, 8, 10, 5, 2, 1, 3, 3, 2, 5, 5, 2, 3, 8, 8, 4, 2, 7, 8, 1, 1, 1, 3, 3, 2, 7, 7, 4, 3, 6

Offset: 1

Zhi-Wei Sun, Jun 04 2016

nonn

Conjectures:
(i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 7, 11, 12, 15, 19, 24, 27, 31, 34, 35, 43, 46, 47, 56, 70, 71, 72, 87, 88, 115, 136, 137, 147, 167, 168, 178, 207, 235, 236, 267, 286, 297, 423, 537, 747, 762, 1017.
(ii) Any positive integer n can be written as w^2 + x^4 + y^5 + pen(z), where w is a positive integer, x,y,z are nonnegative integers, and pen(z) denotes the pentagonal number z*(3*z-1)/2.
Conjectures a(n) > 0 and (ii) verified up to 10^11. - Mauro Fiorentini, Jul 19 2023
See also A262813, A262857, A270566, A271106 and A271325 for some other conjectures on representations.

			a(1) = 1 since 1 = 1^2 + 3*0^2 + 0^4 + 0^5.
a(3) = 1 since 3 = 1^2 + 3*0^2 + 1^4 + 1^5.
a(7) = 1 since 7 = 2^2 + 3*1^2 + 0^4 + 0^5.
a(11) = 1 since 11 = 3^2 + 3*0^2 + 1^4 + 1^5.
a(12) = 1 since 12 = 3^2 + 3*1^2 + 0^4 + 0^5.
a(15) = 1 since 15 = 1^2 + 3*2^2 + 1^4 + 1^5.
a(19) = 1 since 19 = 4^2 + 3*1^2 + 0^4 + 0^5.
a(24) = 1 since 24 = 2^2 + 3*1^2 + 2^4 + 1^5.
a(27) = 1 since 27 = 5^2 + 3*0^2 + 1^4 + 1^5.
a(31) = 1 since 31 = 2^2 + 3*3^2 + 0^4 + 0^5.
a(34) = 1 since 34 = 1^2 + 3*0^2 + 1^4 + 2^5.
a(35) = 1 since 35 = 4^2 + 3*1^2 + 2^4 + 0^5.
a(43) = 1 since 43 = 4^2 + 3*3^2 + 0^4 + 0^5.
a(46) = 1 since 46 = 1^2 + 3*2^2 + 1^4 + 2^5.
a(47) = 1 since 47 = 2^2 + 3*3^2 + 2^4 + 0^5.
a(56) = 1 since 56 = 6^2 + 3*1^2 + 2^4 + 1^5.
a(70) = 1 since 70 = 5^2 + 3*2^2 + 1^4 + 2^5.
a(71) = 1 since 71 = 6^2 + 3*1^2 + 0^4 + 2^5.
a(72) = 1 since 72 = 6^2 + 3*1^2 + 1^4 + 2^5.
a(87) = 1 since 87 = 6^2 + 2*1^2 + 2^4 + 2^5.
a(88) = 1 since 88 = 2^2 + 3*1^2 + 3^4 + 0^5.
a(115) = 1 since 115 = 8^2 + 3*1^2 + 2^4 + 2^5.
a(136) = 1 since 136 = 10^2 + 3*1^2 + 1^4 + 2^5.
a(137) = 1 since 137 = 11^2 + 3*0^2 + 2^4 + 0^5.
a(147) = 1 since 147 = 12^2 + 3*1^2 + 0^4 + 0^5.
a(167) = 1 since 167 = 2^2 + 3*7^2 + 2^4 + 0^5.
a(168) = 1 since 168 = 2^2 + 3*7^2 + 2^4 + 1^5.
a(178) = 1 since 178 = 7^2 + 3*4^2 + 3^4 + 0^5.
a(207) = 1 since 207 = 10^2 + 3*5^2 + 0^4 + 2^5.
a(235) = 1 since 235 = 12^2 + 3*5^2 + 2^4 + 0^5.
a(236) = 1 since 236 = 12^2 + 3*5^2 + 2^4 + 1^5.
a(267) = 1 since 267 = 12^2 + 3*5^2 + 2^4 + 2^5.
a(286) = 1 since 286 = 4^2 + 3*3^2 + 0^4 + 3^5.
a(297) = 1 since 297 = 3^2 + 3*0^2 + 4^4 + 2^5.
a(423) = 1 since 423 = 11^2 + 3*10^2 + 1^4 + 1^5.
a(537) = 1 since 537 = 21^2 + 3*4^2 + 2^4 + 2^5.
a(747) = 1 since 747 = 11^2 + 3*0^2 + 5^4 + 1^5.
a(762) = 1 since 762 = 27^2 + 3*0^2 + 1^4 + 2^5.
a(1017) = 1 since 1017 = 27^2 + 3*0^2 + 4^4 + 2^5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.NT], 2016-2017.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. (See Remark 1.1.)
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. (See Conjecture 3.2.)

Cf. A000118, A000290, A000326, A000583, A000584, A262813, A262827, A262857, A270566, A270969, A271076, A271106, A273429, A273915.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-3*x^2-y^4-z^5],r=r+1],{x,0,Sqrt[(n-1)/3]},{y,0,(n-1-3x^2)^(1/4)},{z,0,(n-1-3x^2-y^4)^(1/5)}];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 xy + yz + 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

Zhi-Wei Sun, May 31 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..7000
Yu-Chen Sun and Zhi-Wei Sun, Two refinements of Lagrange's four-square theorem, arXiv:1605.03074 [math.NT], 2016.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A000578, A000583, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273432, A273458, A273568, A273616.

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 xy + 2y*z + 4zx 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

Zhi-Wei Sun, Jun 02 2016

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Yu-Chen Sun and Zhi-Wei Sun, Two refinements of Lagrange's four-square theorem, arXiv:1605.03074 [math.NT], 2016.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A000578, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273432, A273458, A273568, A273616, A273826.

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}]

cp's OEIS Frontend

A273432 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 2*x + y - z a nonnegative cube, where x,y,z,w are nonnegative integers with y <= z.

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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|.

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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.

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A275297 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^3 with x + 2*y a square, where x,y,z,w are nonnegative integers with z >= w.

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A273915 Number of ordered ways to write n as w^5 + x^2 + y^2 + z^2, where w,x,y,z are nonnegative integers with x <= y <= z.

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A272979 Number of ways to write n as x^2 + 2*y^2 + 3*z^3 + 4*w^4 with x,y,z,w nonnegative integers.

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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.

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A273917 Number of ordered ways to write n as w^2 + 3*x^2 + y^4 + z^5, where w is a positive integer and x,y,z are nonnegative integers.

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

A272979 Number of ways to write n as x^2 + 2y^2 + 3z^3 + 4*w^4 with x,y,z,w nonnegative integers.

A273616 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (3x^2+13y^2)*z a square, where x,y,z,w are nonnegative integers.

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