A282014 -id:A282014 - cp's OEIS frontend

A281976 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and x + 24*y are squares.

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

Offset: 0

Zhi-Wei Sun, Feb 04 2017

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 8, 12, 23, 24, 47, 71, 168, 344, 632, 1724).
By the linked JNT paper, any nonnegative integer can be written as the sum of a fourth power and three squares.
We have verified a(n) > 0 for all n = 0..10^7.
See also A281977, A282013 and A282014 for similar conjectures.
a(n) <= A273404(n). Starts to differ from A273404 at n=145. - R. J. Mathar, Feb 12 2017
Qing-Hu Hou at Tianjin Univ. has verified a(n) > 0 for all n = 0..10^10.
I would like to offer 2400 US dollars for the first proof of my conjecture that a(n) > 0 for any nonnegative integer n. - Zhi-Wei Sun, Feb 14 2017

			a(8) = 1 since 8 = 0^2 + 0^2 + 2^2 + 2^2 with 0 = 0^2 and 0 + 24*0 = 0^2.
a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1 = 1^2 and 1 + 24*1 = 5^2.
a(23) = 1 since 23 = 1^2 + 2^2 + 3^2 + 3^2 with 1 = 1^2 and 1 + 24*2 = 7^2.
a(24) = 1 since 24 = 4^2 + 0^2 + 2^2 + 2^2 with 4 = 2^2 and 4 + 24*0 = 2^2.
a(47) = 1 since 47 = 1^2 + 1^2 + 3^2 + 6^2 with 1 = 1^2 and 1 + 24*1 = 5^2.
a(71) = 1 since 71 = 1^2 + 5^2 + 3^2 + 6^2 with 1 = 1^2 and 1 + 24*5 = 11^2.
a(168) = 1 since 168 = 4^2 + 4^2 + 6^2 + 10^2 with 4 = 2^2 and 4 + 24*4 = 10^2.
a(344) = 1 since 344 = 4^2 + 0^2 + 2^2 + 18^2 with 4 = 2^2 and 4 + 24*0 = 2^2.
a(632) = 1 since 632 = 0^2 + 6^2 + 14^2 + 20^2 with 0 = 0^2 and 0 + 24*6 = 12^2.
a(1724) = 1 since 1724 = 25^2 + 1^2 + 3^2 + 33^2 with 25 = 5^2 and 25 + 24*1 = 7^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.
Zhi-Wei Sun, The 24-conjecture with $2400 prize, a message to Number Theory List, Feb. 14, 2017.

Cf. A000118, A000290, A000583, A270969, A273404, A281939, A281941, A281975, A281977, A281980, A282013, A282014.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[x^2+24y],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,Sqrt[(n-x^4-y^2)/2]}];Print[n," ",r];Continue,{n,0,80}]

A281977 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and -7x - 8y + 8z + 16w are squares.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Feb 04 2017

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,....
The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.
We have verified the conjecture for all n = 0..10^6.
See also A281976, A282013 and A282014 for similar conjectures.
Qing-Hu Hou at Tianjin University verified a(n) > 0 for n up to 10^8. - Zhi-Wei Sun, Jun 02 2019

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0^2 and -7*0 - 8*0 + 8*0 + 16*1 = 4^2.
a(12) = 1 since 12 = 1^2 + 1^2 + 3^2 + 1^2 with 1 = 1^2 and -7*1 - 8*1 + 8*3 + 16*1 = 5^2.
a(17) = 1 since 17 = 1^2 + 0^2 + 4^2 + 0^2 with 1 = 1^2 and -7*1 - 8*0 + 8*4 + 16*0 = 5^2.
a(28) = 1 since 28 = 4^2 + 2^2 + 2^2 + 2^2 with 4 = 2^2 and -7*4 - 8*2 + 8*2 + 16*2 = 2^2.
a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 1 = 1^2 and -7*1 - 8*1 + 8*2 + 16*5 = 9^2.
a(40) = 1 since 40 = 4^2 + 2^2 + 2^2 + 4^2 with 4 = 2^2 and -7*4 -8*2 + 8*2 + 16*4 = 6^2.
a(41) = 1 since 41 = 1^2 + 2^2 + 6^2 + 0^2 with 1 = 1^2 and -7*1 - 8*2 + 8*6 + 16*0 = 5^2.
a(49) = 1 since 49 = 0^2 + 6^2 + 2^2 + 3^2 with 0 = 0^2 and -7*0 - 8*6 + 8*2 + 16*3 = 4^2.
a(241) = 1 since 241 = 9^2 + 4^2 + 12^2 + 0^2 with 9 = 3^2 and -7*9 - 8*4 + 8*12 + 16*0 = 1^2.
a(433) = 1 since 433 = 16^2 + 8^2 + 8^2 + 7^2 with 16 = 4^2 and -7*16 - 8*8 + 8*8 + 16*7 = 0^2.
a(1113) = 1 since 1113 = 1^2 + 30^2 + 4^2 + 14^2 with 1 = 1^2 and -7*1 - 8*30 + 8*4 + 16*14 = 3^2.
a(1521) = 1 since 1521 = 0^2 + 22^2 + 14^2 + 29^2 with 0 = 0^2 and -7*0 - 8*22 + 8*14 + 16*29 = 20^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Cf. A000118, A000290, A270969, A281939, A281941, A281975, A281976, A282013, A282014.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[16*Sqrt[n-x^4-y^2-z^2]+8z-8y-7x^2],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,Sqrt[n-x^4-y^2]}];Print[n," ",r];Continue,{n,0,80}]

A282013 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 49x + 48(y-z) are squares.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Feb 04 2017

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 5, 8, 14, 31, 47, 71, 79, 143, 248, 463, 1039).
The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.
We have verified a(n) > 0 for all n = 0..10^7.
See also A281976, A281977 and A282014 for similar conjectures.
Qing-Hu Hou at Tianjin University verified a(n) > 0 for n up to 10^9. - Zhi-Wei Sun, Jun 02 2019

			a(5) = 1 since 5 = 1^2 + 0^2 + 0^2 + 2^2 with 1 = 1^2 and 49*1 + 48*(0-0) = 7^2.
a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 0 = 0^2 and 49*0 + 48*(2-2) = 0^2.
a(14) = 1 since 14 = 1^2 + 2^2 + 3^2 + 0^2 with 1 = 1^2 and 49*1 + 48*(2-3) = 1^2.
a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(1-2) = 1^2.
a(47) = 1 since 47 = 1^2 + 6^2 + 1^2 + 3^2 with 1 = 1^2 and 49*1 + 48*(6-1) = 17^2.
a(71) = 1 since 71 = 1^2 + 5^2 + 6^2 + 3^2 with 1 = 1^2 and 49*1 + 48*(5-6) = 1^2.
a(79) = 1 since 79 = 1^2 + 7^2 + 2^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(7-2) = 17^2.
a(143) = 1 since 143 = 1^2 + 5^2 + 6^2 + 9^2 with 1 = 1^2 and 49*1 + 48*(5-6) = 1^2.
a(248) = 1 since 248 = 4^2 + 6^2 + 0^2 + 14^2 with 4 = 2^2 and 49*4 + 48*(6-0) = 22^2.
a(463) = 1 since 463 = 9^2 + 6^2 + 15^2 + 11^2 with 9 = 3^2 and 49*9 + 48*(6-15) = 3^2.
a(1039) = 1 since 1039 = 1^2 + 22^2 + 23^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(22-23) = 1^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Cf. A000118, A000290, A270969, A271518, A281939, A281941, A281975, A281976, A281977, A282014.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[49x^2+48(y-z)],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,Sqrt[n-x^4-y^2]}];Print[n," ",r];Continue,{n,0,80}]

A282463 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers, x == y (mod 2) and z <= w such that both x and x^2 + 62xy + y^2 are squares.

Original entry on oeis.org

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

Offset: 0

Zhi-Wei Sun, Feb 16 2017

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 3, 43, 723, 1723, 3571, 3911 and 16^k*m (k = 0,1,2,... and m = 7, 23, 31, 71, 79, 143, 303, 1591).
(ii) Any nonnegative integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and a*x^2 + b*x*y + c*y^2 are squares, whenever (a,b,c) is among the ordered triples (84,84,1), (16,144,9), (153,36,100), (177,214,9), (249,114,121).
The author has proved that any nonnegative integer can be expressed as the sum of a fourth power and three squares.

			a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 == 1 (mod 2), 1 = 1^2 and 1^2 + 62*1*1 + 1^2 = 8^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1 == 1 (mod 3), 1 = 1^2 and 1^2 + 62*1*3 + 3^2 = 14^2.
a(30) = 2 since 30 = 0^2 + 2^2 + 1^2 + 5^2 with 0 == 2 (mod 3), 0 = 0^2 and 0^2 + 62*0*2 + 2^2 = 2^2, and 30 = 1^2 + 3^2 + 2^2 + 4^2 with 1 == 3 (mod 2), 1 = 1^2 and 1^2 + 62*1*3 + 3^2 = 14^2.
a(79) = 1 since 79 = 1^2 + 7^2 + 2^2 + 5^2 with 1 == 7 (mod 2), 1 = 1^2 and 1^2 + 62*1*7 + 7^2 = 22^2.
a(143) = 1 since 143 = 9^2 + 3^2 + 2^2 + 7^2 with 9 == 3 (mod 2), 9 = 3^2 and 9^2 + 62*9*3 + 3^2 = 42^2.
a(303) = 1 since 303 = 1^2 + 3^2 + 2^2 + 17^2 with 1 == 3 (mod 2), 1 = 1^2 and 1^2 + 62*1*3 + 3^2 = 14^2.
a(723) = 1 since 723 = 1^2 + 7^2 + 12^2 + 23^2 with 1 == 7 (mod 2), 1 = 1^2 and 1^2 + 62*1*7 + 7^2 = 22^2.
a(1591) = 1 since 1591 = 9^2 + 9^2 + 23^2 + 30^2 with 9 == 9 (mod 2), 9 = 3^2 and 9^2 + 62*9*9 + 9^2 = 72^2.
a(1723) = 1 since 1723 = 1^2 + 1^2 + 11^2 + 40^2 with 1 == 1 (mod 2), 1 = 1^2 and 1^2 + 62*1*1 + 1^2 = 8^2.
a(3571) = 1 since 3571 = 9^2 + 3^2 + 0^2 + 59^2 with 9 == 3 (mod 2), 9 = 3^2 and 9^2 + 62*9*3 + 3^2 = 42^2.
a(3911) = 1 since 9^2 + 3^2 + 10^2 + 61^2 with 9 == 3 (mod 2), 9 = 3^2 and 9^2 + 62*9*3 + 3^2 = 42^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Cf. A000118, A000290, A270969, A271518, A281976, A281977, A282013, A282014.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&Mod[x-y,2]==0&&SQ[x^4+62*x^2*y+y^2],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,Sqrt[(n-x^4-y^2)/2]}];Print[n," ",r];Continue,{n,0,80}]

A282494 Number of ways to write n as x^4 + y^2 + z^2 + w^2 with y(y+240z) a positive square, where x,y,z,w are nonnegative integers.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Feb 16 2017

nonn

Conjecture: a(n) > 0 for all n > 0.
By the linked JNT paper, any nonnegative integer can be expressed as the sum of a fourth power and three squares, and each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z*(y-2*z) = 0. Whether z = 0 or y = 2*z, the number y*(y+240*z) is definitely a square.
See also A282463 and A282495 for similar conjectures.

			a(3) = 1 since 3 = 1^4 + 1^2 + 0^2 + 1^2 with 1*(1+240*0) = 1^2.
a(4) = 1 since 4 = 0^4 + 2^2 + 0^2 + 0^2 with 2*(2+240*0) = 2^2.
a(39) = 1 since 39 = 1^4 + 2^2 + 3^2 + 5^2 with 2*(2+240*3) = 38^2.
a(188) = 1 since 188 = 3^4 + 5^2 + 1^2 + 9^2 with 5*(5+240*1) = 35^2.
a(399) = 1 since 399 = 3^4 + 10^2 + 7^2 + 13^2 with 10*(10+240*7) = 130^2.
a(428) = 1 since 428 = 0^4 + 10^2 + 2^2 + 18^2 with 10*(10+240*2) = 70^2.
a(439) = 1 since 439 = 1^4 + 10^2 + 7^2 + 17^2 with 10*(10+240*7) = 130^2.
a(508) = 1 since 508 = 1^4 + 5^2 + 11^2 + 19^2 with 5*(5+240*11) = 115^2.
a(748) = 1 since 748 = 3^4 + 1^2 + 21^2 + 15^2 with 1*(1+240*21) = 71^2.
a(1468) = 1 since 1468 = 2^4 + 10^2 + 26^2 + 26^2 with 10*(10+240*26) = 250^2.
a(2828) = 1 since 2828 = 3^4 + 5^2 + 11^2 + 51^2 with 5*(5+240*11) = 115^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Cf. A000118, A000290, A000583, A270969, A271518, A281976, A281977, A282013, A282014, A282463, A282495.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[y*(y+240*z)],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}]

A282495 Number of ways to write n as x^4 + y^2 + z^2 + w^2 with y^2 + 228yz + 60*z^2 a square, where x,y,z are nonnegative integers and w is a positive integer.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Feb 16 2017

nonn

Conjecture: a(n) > 0 for all n > 0.
By the linked JNT paper, any nonnegative integer can be expressed as the sum of a fourth power and three squares, and each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z*(y-z) = 0. Whether z = 0 or y = z, the number y^2 + 228*y*z + 60*z^2 is definitely a square.
See also A282463 and A282494 for similar conjectures.

			a(7) = 1 since 7 = 1^4 + 1^2 + 1^2 + 2^2 with 1^2 + 228*1*1 + 60*1^2 = 17^2.
a(8) = 1 since 8 = 0^4 + 2^2 + 0^2 + 2^2 with 2^2 + 228*2*0 + 60*0^2 = 2^2.
a(15) = 1 since 15 = 1^4 + 1^2 + 3^2 + 2^2 with 1^2 + 228*1*3 + 60*3^2 = 35^2.
a(23) = 1 since 23 = 1^4 + 3^2 + 3^2 + 2^2 with 3^2 + 228*3*3 + 60*3^2 = 51^2.
a(71) = 1 since 71 = 1^4 + 5^2 + 6^2 + 3^2 with 5^2 + 228*5*6 + 60*6^2 = 95^2.
a(159) = 1 since 159 = 3^4 + 7^2 + 2^2 + 5^2 with 7^2 + 228*7*2 + 60*2^2 = 59^2.
a(623) = 1 since 623 = 3^4 + 1^2 + 10^2 + 21^2 with 1^2 + 228*1*10 + 60*10^2 = 91^2.
a(879) = 1 since 879 = 5^4 + 5^2 + 15^2 + 2^2 with 5^2 + 228*5*15 + 60*15^2 = 175^2.
a(1423) = 1 since 1423 = 1^4 + 7^2 + 2^2 + 37^2 with 7^2 + 228*7*2 + 60*2^2 = 59^2.
a(3768) = 1 since 3768 = 0^4 + 2^2 + 20^2 + 58^2 with 2^2 + 228*2*20 + 60*20^2 = 182^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Cf. A000118, A000290, A000583, A271518, A281976, A281977, A282013, A282014, A282463, A282494.

Q[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[y^2+228*y*z+60*z^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}]

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 8y^2 - 8yz + 9z^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

Zhi-Wei Sun, Feb 25 2017

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, 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

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Cf. A000118, A000290, A000583, A270969, A271518, A281976, A281977, A282013, A282014, A282494, A282495.

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}]

A282161 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x and (12x)^2 + (5y-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

Zhi-Wei Sun, Feb 07 2017

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, 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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Cf. A000118, A000290, A270969, A271714, A273108, A281939, A281941, A281975, A281976, A281977, A282013, A282014.

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}]

cp's OEIS Frontend

A281976 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and x + 24*y are squares.

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A281977 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and -7*x - 8*y + 8*z + 16*w are squares.

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A282013 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 49*x + 48*(y-z) are squares.

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A282463 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers, x == y (mod 2) and z <= w such that both x and x^2 + 62*x*y + y^2 are squares.

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A282494 Number of ways to write n as x^4 + y^2 + z^2 + w^2 with y*(y+240*z) a positive square, where x,y,z,w are nonnegative integers.

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A282495 Number of ways to write n as x^4 + y^2 + z^2 + w^2 with y^2 + 228*y*z + 60*z^2 a square, where x,y,z are nonnegative integers and w is a positive integer.

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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.

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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.

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A281977 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and -7x - 8y + 8z + 16w are squares.

A282013 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 49x + 48(y-z) are squares.

A282463 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers, x == y (mod 2) and z <= w such that both x and x^2 + 62xy + y^2 are squares.

A282494 Number of ways to write n as x^4 + y^2 + z^2 + w^2 with y(y+240z) a positive square, where x,y,z,w are nonnegative integers.

A282495 Number of ways to write n as x^4 + y^2 + z^2 + w^2 with y^2 + 228yz + 60*z^2 a square, where x,y,z are nonnegative integers and w is a positive integer.

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 8y^2 - 8yz + 9z^2 is a square.

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