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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A271714 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that (10*w+5*x)^2 + (12*y+36*z)^2 is a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 2, 1, 3, 1, 2, 1, 2, 3, 1, 4, 4, 2, 2, 1, 3, 3, 5, 2, 2, 5, 2, 1, 2, 3, 3, 3, 2, 3, 2, 3, 4, 4, 2, 3, 9, 2, 3, 1, 1, 6, 2, 3, 4, 6, 4, 1, 2, 5, 3, 3, 4, 3, 5, 1, 4, 5, 1, 3, 6, 6, 1, 3, 4, 5, 12, 2, 4, 6, 2, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 12 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 7, 9, 19, 49, 133, 589, 2^k, 2^k*3, 4^k*q (k = 0,1,2,... and q = 14, 67, 71, 199).
(ii) If P(y,z) is one of 2y-3z, 2y-8z and 4y-6z, then any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w-x)^2 + P(y,z)^2 is a square.
(iii) For each triple (a,b,c) = (1,4,4), (1,12,12), (2,4,8), (2,6,6), (2,12,12), (3,4,4), (3,4,8), (3,8,8), (3,12,12), (3,12,36), (5,4,4), (5,4,8), (5,8,16), (5,36,36), (6,4,4), (7,12,12), (7,20,20), (7,24,24), (9,4,4), (9,12,12),(9,36,36), (11,12,12), (13,4,4), (15,12,12), (16,12,12), (21,20,20), (21,24,24), (23,12,12), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w+a*x)^2 + (b*y-c*z)^2 is a square.
See also A271510, A271513, A271518, A271644, A271665, A271721 and A271724 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(2) = 1 since 2 = 1^2 + 1^2 + 0^2 + 0^2 with (10*1+5*1)^2 + (12*0+36*0)^2 = 15^2 + 0^2 = 15^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with (10*1+5*1)^2 + (12*0+36*1)^2 = 15^2 + 36^2 = 39^2.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with (10*2+5*0)^2 + (12*0+36*0)^2 = 20^2 + 0^2 = 20^2.
a(6) = 1 since 6 = 2^2 + 0^2 + 1^2 + 1^2 with (10*2+5*0)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with (10*1+5*2)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(9) = 1 since 9 = 3^2 + 0^2 + 0^2 + 0^2 with (10*3+5*0)^2 + (12*0+36*0)^2 = 30^2 + 0^2 = 30^2.
a(19) = 1 since 19 = 3^2 + 0^2 + 3^2 + 1^2 with (10*3+5*0)^2 + (12*3+36*1)^2 = 30^2 + 72^2 = 78^2.
a(49) = 1 since 49 = 7^2 + 0^2 + 0^2 + 0^2 with (10*7+5*0)^2 + (12*0+36*0)^2 = 70^2 + 0^2 = 70^2.
a(133) = 1 since 133 = 9^2 + 0^2 + 6^2 + 4^2 with (10*9+5*0)^2 + (12*6+36*4)^2 = 90^2 + 216^2 = 234^2.
a(589) = 1 since 589 = 17^2 + 10^2 + 2^2 + 14^2 with (10*17+5*10)^2 + (12*2+36*14)^2 = 220^2 + 528^2 = 572^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(10*Sqrt[n-x^2-y^2-z^2]+5x)^2+(12y+36z)^2],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A271724 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w*(x+2*y+3*z) a square, where w,x,y,z are nonnegative integers with x > 0.

Original entry on oeis.org

1, 3, 2, 1, 4, 4, 1, 3, 4, 6, 4, 2, 4, 7, 1, 1, 10, 8, 5, 6, 8, 5, 1, 4, 7, 10, 7, 2, 11, 13, 2, 3, 8, 9, 8, 6, 7, 13, 3, 6, 15, 8, 4, 4, 13, 8, 1, 2, 8, 15, 11, 4, 14, 18, 5, 7, 6, 6, 12, 5, 12, 17, 5, 1, 16, 21, 3, 11, 16, 12, 1, 8, 8, 18, 16, 5, 16, 12, 4, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 13 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 7, 15, 47, 151, 4^k*q (k = 0,1,2,... and q = 1, 23, 71).
(ii) For positive integers a,b,c with gcd(a,b,c) squarefree, any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and w*(a*x+b*y+c*z) a square, if and only if {a,b,c} is among {1,2,3}, {1,3,6}, {1,6,9}, {5,6,9}, {18,30,114}.
(iii) For each quadruple (a,b,c,d) = (1,1,2,12), (1,2,7,60), (1,3,9,48), (1,4,11,48), (1,5,8,24), (1,8,11,24), (2,6,8,15), (3,5,6,24), (3,6,15,40), (3,6,18,40), (3,12,15,20), (4,4,8,15), (4,8,12,21), (4,8,12,45), (4,8,20,15), (4,8,36,45), (5,10,15,24), (6,9,15,20), (7,14,28,60), (7,21,28,60), (7,21,42,60), (12,36,48,55), (14,21,28,60), (3,9,18,112), (3,21,33,80), (4,5,9,120), (4,12,16,105), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (a*x+b*y+c*z)^2 + (d*w)^2 is a square.
See also A271510, A271513, A271518, A271644, A271665, A271714 and A271721 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(1) = 1 since 1 = 0^2 + 1^2 + 0^2 + 0^2 with 1 > 0 and 0*(1+2*0+3*0) = 0^2.
a(3) = 2 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1*(1+2*0+3*1) = 2^2, and 3 = 0^2 + 1^2 + 1^2 + 1^2 with 0*(1+2*1+3*1) = 0^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1*(1+2*1+3*2) = 3^2.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^2 + 1^2 with 2*(3+2*1+3*1) = 4^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1*(3+2*2+3*3) = 4^2.
a(31) = 2 since 31 = 2^2 + 1^2 + 1^2 + 5^2 with 2*(1+2*1+3*5) = 6^2, and also 31 = 2^2 + 3^2 + 3^2 + 3^2 with 2*(3+2*3+3*3) = 6^2.
a(47) = 1 since 47 = 1^2 + 1^2 + 3^2 + 6^2 with 1*(1+2*3+3*6) = 5^2.
a(71) = 1 since 71 = 1^2 + 6^2 + 5^2 + 3^2 with 1*(6+2*5+3*3) = 5^2.
a(151) = 1 since 151 = 9^2 + 6^2 + 5^2 + 3^2 with 9*(6+2*5+3*3) = 15^2.
		

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](x+2y+3z)],r=r+1],{x,1,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Label[aa];Continue,{n,1,80}]

A271775 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 (x >= y >= z <= w) with x - y a square, where w,x,y,z are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 13 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 3, 11, 47, 2^{4k+3}*m (k = 0,1,2,... and m = 1, 3, 7, 15, 79).
(ii) Let a and b be positive integers with a <= b and gcd(a,b) squarefree. Then any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x-b*y a square, if and only if (a,b) is among the ordered pairs (1,1), (2,1), (2,2), (4,3), (6,2). Verified for all nonnegative integers up to 10^11. - Mauro Fiorentini, Jun 14 2024
(iii) Let a and b be positive integers with gcd(a,b) squarefree. Then any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x+b*y a square, if and only if {a,b} is among {1,2}, {1,3} and {1,24}. Verified for all nonnegative integers up to 10^11. - Mauro Fiorentini, Jun 14 2024
(iv) Let a,b,c be positive integers with a <= b and gcd(a,b,c) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x+b*y-c*z a square, if and only if (a,b,c) is among the triples (1,1,1), (1,1,2), (1,2,1), (1,2,2), (1,2,3), (1,3,1), (1,3,3), (1,4,4), (1,5,1), (1,6,6), (1,8,6), (1,12,4), (1,16,1), (1,17,1), (1,18,1), (2,2,2), (2,2,4), (2,3,2), (2,3,3), (2,4,1), (2,4,2), (2,6,1), (2,6,2), (2,6,6), (2,7,4), (2,7,7), (2,8,2), (2,9,2), (2,32,2), (3,3,3), (3,4,2), (3,4,3), (3,8,3), (4,5,4), (4,8,3), (4,9,4), (4,14,14), (5,8,5), (6,8,6), (6,10,8), (7,9,7), (7,18,7), (7,18,12), (8,9,8), (8,14,14), (8,18,8), (14,32,14), (16,18,16), (30,32,30), (31,32,31), (48,49,48), (48,121,48). Verified for all nonnegative integers up to 10^11. - Mauro Fiorentini, Jun 14 2024
(v) Let a,b,c be positive integers with b <= c and gcd(a,b,c) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x-b*y-c*z a square, if and only if (a,b,c) is among the triples (1,1,1), (2,1,1), (2,1,2), (3,1,2) and (4,1,2).
(vi) Let a,b,c,d be positive integers with a <= b, c <= d and gcd(a,b,c,d) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x+b*y-(c*z+d*w) a square, if and only if (a,b,c,d) is among the quadruples (1,2,1,1), (1,2,1,2), (1,3,1,2), (1,4,1,3), (2,4,1,2), (2,4,2,4), (8,16,7,8), (9,11,2,9) and (9,16,2,7).
(vii) Let a,b,c,d be positive integers with a <= b <= c and gcd(a,b,c,d) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x+b*y+c*z-d*w a square, if and only if (a,b,c,d) is among the quadruples (1,1,2,1), (1,2,3,1), (1,2,3,3), (1,2,4,2), (1,2,4,4), (1,2,5,5), (1,2,6,2), (1,2,8,1), (2,2,4,4), (2,4,6,4), (2,4,6,6), and (2,4,8,2).
It is known that any natural number not of the form 4^k*(16*m+14) (k,m = 0,1,2,...) can be written as x^2 + y^2 + 2*z^2 = x^2 + y^2 + z^2 + z^2 with x,y,z nonnegative integers.
See also A271510, A271513, A271518, A271644, A271665, A271714, A271721 and A271724 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1 = 1 > 0 < 1 and 1 - 1 = 0^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 = 1 = 1 < 2 and 1 - 1 = 0^2.
a(8) = 1 since 8 = 2^2 + 2^2 + 0^2 + 0^2 with 2 = 2 > 0 = 0 and 2 - 2 = 0^2.
a(11) = 1 since 11 = 1^2 + 1^2 + 0^2 + 3^2 with 1 = 1 > 0 < 3 and 1 - 1 = 0^2.
a(24) = 1 since 24 = 2^2 + 2^2 + 0^2 + 4^2 with 2 = 2 > 0 < 4 and 2 - 2 = 0^2.
a(47) = 1 since 47 = 3^2 + 3^2 + 2^2 + 5^2 with 3 = 3 > 2 < 5 and 3 - 3 = 0^2.
a(53) = 2 since 53 = 3^2 + 2^2 + 2^2 + 6^2 with 3 > 2 = 2 < 6 and 3 - 2 = 1^2, and also 53 = 6^2 + 2^2 + 2^2 + 3^2 with 6 > 2 = 2 < 3 and 6 - 2 = 2^2.
a(56) = 1 since 56 = 6^2 + 2^2 + 0^2 + 4^2 with 6 > 2 > 0 < 4 and 6 - 2 = 2^2.
a(120) = 1 since 120 = 8^2 + 4^2 + 2^2 + 6^2 with 8 > 4 > 2 < 6 and 8 - 4 = 2^2.
a(632) = 1 since 632 = 16^2 + 12^2 + 6^2 + 14^2 with 16 > 12 > 6 < 14 and 16 - 12 = 2^2.
		

References

  • L. E. Dickson, Modern Elementary Theory of Numbers, University of Chicago Press, Chicago, 1939, pp. 112-113.

Crossrefs

Programs

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

A271778 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and x^2 + 3*y^2 + 5*z^2 - 8*w^2 a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 14 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 15, 29, 33, 47, 53, 65, 89, 129, 689, 1553, 2^(2k+1)*m (k = 0,1,2,... and m = 1, 29).
(ii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x^2 + b*y^2 + c*z^2 - d*w^2 a square, if (a,b,c,d) is among the following quadruples: (1,3,6,3), (1,3,9,3), (1,3,30,3), (1,4,12,4), (1,4,20,4), (1,5,20,5), (1,5,35,20), (3,4,9,3), (3,9,40,3), (4,5,16,4), (4,11,33,11), (4,12,16,7), (5,16,20,20), (5,25,36,5), (6,10,25,10), (9,12,28,12), (9,21,28,21), (15,21,25,15), (15,24,25,15), (1,5,60,5), (1,20,60,20), (9, 28,63,63), (9,28,84,84), (12,33,64,12), (16,21,105,21), (16,33,64,16), (21,25,45,45), (24,25,75,75), (24,25,96,96), (25,40,96,40), (25,48,96,48), (25,60,84,60), (25,60,96,60), (25,75,126,75), (32,64,105,32).
(iii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x^2 + b*y^2 - c*z^2 -d*w^2 a square, whenever (a,b,c,d) is among the quadruples (3,9,3,20), (5,9,5,20), (5,25,4,5), (9,81,9,20),(12,16,3,12), (16,64,15,16), (20,25,4,20), (27,81,20,27), (30,64,15,30), (32,64,15,32), (48,64,15,48), (60,64,15,60), (60,81,20,60), (64,80,15,80).
(iv) For each triple (a,b,c) = (21,5,15), (36,3,8), (48,8,39), (64,7,8), (40,15,144), (45,20,144), (69,20,60), any natural number can be written as x^2 + y^2 + z^2 + w^2 with w,x,y,z nonnegative integers and a*x^2 - b*y^2 - c*z^2 a square.
See also A271510, A271513, A271518, A271665, A271714, A271721, A271724 and A271775 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(2) = 1 since 2 = 1^2 + 1^2 + 0^2 + 0^2 with 1^2 + 3*1^2 + 5*0^2 - 8*0^2 = 2^2.
a(15) = 1 since 15 = 1^2 + 3^2 + 1^2 + 2^2 with 1^2 + 3*3^2 + 5*1^2 - 8*2^2 = 1^2.
a(29) = 1 since 29 = 3^2 + 4^2 + 0^2 + 2^2 with 3^2 + 3*4^2 + 5*0^2 - 8*2^2 = 5^2.
a(33) = 1 since 33 = 2^2 + 4^2 + 2^2 + 3^2 with 2^2 + 3*4^2 + 5*2^2 - 8*3^2 = 0.
a(47) = 1 since 47 = 5^2 + 3^2 + 2^2 + 3^2 with 5^2 + 3*3^2 + 5*2^2 - 8*3^2 = 0^2.
a(53) = 1 since 53 = 3^2 + 2^2 + 6^2 + 2^2 with 3^2 + 3*2^2 + 5*6^2 - 8*2^2 = 13^2.
a(58) = 1 since 58 = 4^2 + 1^2 + 5^2 + 4^2 with 4^2 + 3*1^2 + 5*5^2 - 8*4^2 = 4^2.
a(65) = 1 since 65 = 3^2 + 6^2 + 2^2 + 4^2 with 3^2 + 3*6^2 + 5*2^2 - 8*4^2 = 3^2.
a(89) = 1 since 89 = 6^2 + 4^2 + 6^2 + 1^2 with 6^2 + 3*4^2 + 5*6^2 - 8*1^2 = 16^2.
a(129) = 1 since 129 = 9^2 + 4^2 + 4^2 + 4^2 with 9^2 + 3*4^2 + 5*4^2 - 8*4^2 = 9^2.
a(689) = 1 since 689 = 11^2 + 18^2 + 10^2 + 12^2 with 11^2 + 3*18^2 + 5*10^2 - 8*12^2 = 21^2.
a(1553) = 1 since 1553 = 21^2 + 6^2 + 26^2 + 20^2 with 21^2 + 3*6^2 + 5*26^2 - 8*20^2 = 27^2.
		

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^2+3*y^2+5*z^2-8*(n-x^2-y^2-z^2)],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,0,70}]

A271824 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (x+2*y)^2 + 8*z^2 + 40*w^2 a square, where x is a positive integer and y,z,w are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 14 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 9, 11, 15, 23, 33, 71, 129, 167, 187, 473, 4^k*m (k = 0,1,2,... and m = 1, 22, 38, 278). Also, any positive integer can be written as x^2 + y^2 + z^2 + w^2 with 9*(x+2*y)^2 + 16*z^2 + 24*w^2 a square, where x is a positive integer and y,z,w are nonnegative integers.
(ii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and (a*x-b*y)^2 + c*z^2 + d*w^2 a square, provided that (a,b,c,d) is among the quadruples (4,8,1,8), (12,24,1,24), (2,4,5,40), (3,6,7,9), (3,9,7,9), (3,6,7,63), (1,2,8,16), (1,2,8,40), (3,6,8,40), (2,6,9,12), (3,5,9,15), (4,8,9,16), (12,24,9,16), (3,6,15,25), (3,6,16,24), (3,12,16,24), (6,9,16,24), (9,12,16,24), (4,8,16,41), (8,12,16,41), (3,6,16,48), (6,9,16,48), (2,3,16,56), (3,6,28,63), (2,4,36,45), (6,12,40,45), (7,14,56,64) and (2,6,57,60).
(iii) Let a and be positive integers with a <= b and gcd(a,b) squarefree. Then any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and (a*x+b*y)*z a square, if and only if (a,b) is among the ordered pairs (1,1), (1,2), (1,3), (2,5), (3,3), (3,6), (3,15), (5,6), (5,11), (5,13), (6,15), (8,46) and (9,23).
(iv) Let a and b be positive integers with a <= b and gcd(a,b) squarefree. Then, any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and (a*x^2+b*y^2)*z a square, if and only if (a,b) is among the ordered pairs (3,13), (5,11), (15,57), (15,165) and (138,150).
There are many ordered pairs (a,b) of integers with gcd(a,b) squarefree such that any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers and a*x^2 + b*y^2 a square. For example, we have shown that (1,-1), (2,-2), (3,-3) and (1,2) are indeed such ordered pairs.
See also A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775 and A271778 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(9) = 1 since 9 = 3^2 + 0^2 + 0^2 + 0^2 with (3+2*0)^2 + 8*0^2 + 40*02 = 3^2.
a(11) = 1 since 11 = 1^2 + 1^2 + 3^2 + 0^2 with (1+2*1)^2 + 8*3^2 + 40*0^2 = 9^2.
a(15) = 1 since 15 = 1^2 + 3^2 + 2^2 + 1^2 with (1+2*3)^2 + 8*2^2 + 40*1^2 = 11^2.
a(22) = 1 since 22 = 3^2 + 2^2 + 3^2 + 0^2 with (3+2*2)^2 + 8*3^2 + 40*0^2 = 11^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with (1+2*3)^2 + 8*2^2 + 40*3^2 = 21^2.
a(33) = 1 since 33 = 4^2 + 1^2 + 0^2 + 4^2 with (4+2*1)^2 + 8*0^2 + 40*4^2 = 26^2.
a(38) = 1 since 38 = 5^2 + 2^2 + 0^2 + 3^2 with (5+2*2)^2 + 8*0^2 + 40*3^2 = 21^2.
a(71) = 1 since 71 = 1^2 + 6^2 + 5^2 + 3^2 since (1+2*6)^2 + 8*5^2 + 40*3^2 = 27^2.
a(129) = 1 since 129 = 5^2 + 6^2 + 8^2 + 2^2 with (5+2*6)^2 + 8*8^2 + 40*2^2 = 31^2.
a(167) = 1 since 167 = 11^2 + 1^2 + 3^2 + 6^2 with (11+2*1)^2 + 8*3^2 + 40*6^2 = 41^2.
a(187) = 1 since 187 = 3^2 + 5^2 + 12^2 + 3^2 with (3+2*5)^2 + 8*12^2 + 40*3^2 = 41^2.
a(278) = 1 since 278 = 3^2 + 0^2 + 10^2 + 13^2 with (3+2*0)^2 + 8*10^2 + 40*13^2 = 87^2.
a(473) = 1 since 473 = 7^2 + 10^2 + 0^2 + 18^2 with (7+2*10)^2 + 8*0^2 + 40*18^2 = 117^2.
		

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+2y)^2+8z^2+40(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]}];Print[n," ",r];Continue,{n,1,80}]

A262357 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w^2*x^2 + 5*x^2*y^2 + 80*y^2*z^2 + 20*z^2*w^2 a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 17 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m (k = 0,1,2,... and m = 1, 3, 11, 43, 547, 763, 1739, 6783).
(ii) For each quadruples (a,b,c,d) = (1,3,78,27), (1,3,222,75), (4,12,81,108), (6,27,25,75), (7,21,112,32), any positive integer can be written as w^2 + x^2 + y^2 + z^2 with a*w^2*x^2 + b*x^2*y^2 + c*y^2*z^2 + d*z^2*w^2 a square, where w is a positive integer and x,y,z are integers.
(iii) Each n = 0,1,2,.... can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z integers such that w^2*x^2 + 4*x^2*y^2 + 44*y^2*z^2 + 16*z^2*w^2 = 5*t^2 for some integer t.
See also A268507, A269400, A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775, A271778 and A271824 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 > 0 and 1^2*0^2 + 5*0^2*0^2 + 80*0^2*0^2 + 20*0^2*1^2 = 0^2.
a(2) = 2 since 2 = 1^2 + 0^2 + 1^2 + 0^2 with 1 > 0 and 1^2*0^2 + 5*0^2*1^2 + 80*1^2*0^2 + 20*0^2*1^2 = 0^2, and also 2 = 1^2 + 1^2 + 0^2 + 0^2 with 1 > 0 and 1^2*1^2 + 5*1^2*0^2 + 80*0^2*0^2 + 20*0^2*1^2 = 1^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1 > 0 and 1^2*0^2 + 5*0^2*1^2 + 80*1^2*1^2 + 20*1^2*1^2 = 10^2.
a(11) = 1 since 11 = 1^2 + 0^2 + 1^2 + 3^2 with 1 > 0 and
1^2*0^2 + 5*0^2*1^2 + 80*1^2*3^2 + 20*3^2*1^2 = 30^2.
a(43) = 1 since 43 = 3^2 + 0^2 + 3^2 + 5^2 with 3 > 0 and 3*0^2 + 5*0^2*3^2 + 80*3^2*5^2 + 20*5^2*3^2 = 150^2.
a(547) = 1 since 547 = 3^2 + 0^2 + 3^2 + 23^2 with 3 > 0 and 3^2*0^2 + 5*0^2*3^2 + 80*3^2*23^2 + 20*23^2*3^2 = 690^2.
a(763) = 1 since 763 = 13^2 + 20^2 + 13^2 + 5^2 with 13 > 0 and 13^2*20^2 + 5*20^2*13^2 + 80*13^2*5^2 + 20*5^2*13^2 = 910^2.
a(1739) = 1 since 1739 = 15^2 + 16^2 + 27^2 + 23^2 with 15 > 0 and 15^2*16^2 + 5*16^2*27^2 + 80*27^2*23^2 + 20*23^2*15^2 = 5850^2.
a(6783) = 1 since 6783 = 17^2 + 73^2 + 18^2 + 29^2 with 17 > 0 and 17^2*73^2 + 5*73^2*18^2 + 80*18^2*29^2 + 20*29^2*17^2 = 6069^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(n-x^2-y^2-z^2)*x^2+5*x^2*y^2+80*y^2*z^2+20*z^2*(n-x^2-y^2-z^2)],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,70}]

A268507 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w > 0, w >= x <= y <= z such that x^2*y^2 + y^2*z^2 + z^2*x^2 is a square, where w,x,y,z are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 16 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k, 4^k*m (k = 0,1,2,... and m = 3, 7, 23, 31, 39, 47, 55, 71, 79, 151, 191, 551).
(ii) For each triple (a,b,c) = (1,4,4), (1,4,16), (1,4,26), (1,4,31), (1,4,34), (1,9,9), (1,9,11), (1,9,17), (1,9,21), (1,9,27), (1,9,33), (1,9,41), (1,18,24), (1,36,44), (3,4,8), (4,6,9), (4,8,19), (4,8,27), (4,9,36), (4,16,41), (4,19,29), (5,9,25), (7,9,33), (7,25,49), (9,10,45), (9,12,28), (9,16,36), (9,21,49), (9,24,37), (9,25,27), (9,25,45), (9,30,40), (9,32,64), (9,34,36), (9,44,61), (14,25,40), (16,17,36), (16,20,25), (24,36,39), (25,40,64), (25,45,51), (27,36,37), (28,44,49), (32,49,64), (36,43,45), (36,54,58), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z integers such that a*x^2*y^2 + b*y^2*z^2 + c*z^2*x^2 is a square.
See also A269400, A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775, A271778 and A271824 for other conjectures refining Lagrange's four-square theorem.
The author has proved in arXiv:1604.06723 that a(n) > 0 for any positive integer n. - Zhi-Wei Sun, May 09 2016

Examples

			a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 > 0 = 0 < 1 and 0^2*0^2 + 0^2*1^2 + 1^2*0^2 = 0^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1 > 0 < 1 = 1 and 0^2*1^2 + 1^2*1^2 + 1^2*0^2 = 1^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 = 1 = 1 < 2 and 1^2*1^2 + 1^2*2^2 + 2^2*1^2 = 3^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 3 > 1 < 2 < 3 and 1^2*2^2 + 2^2*3^2 + 3^2*1^2 = 7^2.
a(31) = 1 since 31 = 5^2 + 1^2 + 1^2 + 2^2 with 5 > 1 = 1 < 2 and 1^2*1^2 + 1^2*2^2 + 2^2*1^2 = 3^2.
a(39) = 1 since 39 = 5^2 + 1^2 + 2^2 + 3^2 with 5 > 1 < 2 < 3 and 1^2*2^2 + 2^2*3^2 + 3^2*1^2 = 7^2.
a(47) = 1 since 47 = 3^2 + 2^2 + 3^2 + 5^2 with 3 > 2 < 3 < 5 and 2^2*3^2 + 3^2*5^2 + 5^2*2^2 = 19^2.
a(55) = 1 since 55 = 7^2 + 1^2 + 1^2 + 2^2 with 7 > 1 = 1 < 2 and 1^2*1^2 + 1^2*2^2 + 2^2*1^2 = 3^2.
a(71) = 1 since 71 = 3^2 + 1^2 + 5^2 + 6^2 with 3 > 1 < 5 < 6 and 1^2*5^2 + 5^2*6^2 + 6^2*1^2 = 31^2.
a(79) = 1 since 79 = 5^2 + 3^2 + 3^2 + 6^2 with 5 > 3 = 3 < 6 and 3^2*3^2 + 3^2*6^2 + 6^2*3^2 = 27^2.
a(151) = 1 since 151 = 5^2 + 3^2 + 6^2 + 9^2 with 5 > 3 < 6 < 9 and 3^2*6^2 + 6^2*9^2 + 9^2*3^2 = 63^2.
a(191) = 1 since 191 = 3^2 + 1^2 + 9^2 + 10^2 with 3 > 1 < 9 < 10 and 1^2*9^2 + 9^2*10^2 + 10^2*1^2 = 91^2.
a(551) = 1 since 551 = 15^2 + 3^2 + 11^2 + 14^2 with 15 > 3 < 11 < 14 and 3^2*11^2 + 11^2*14^2 + 14^2*3^2 = 163^2.
		

Crossrefs

Programs

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

A269400 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 6*w^2*x^2 + 12*x^2*y^2 + 52*y^2*z^2 + 27*z^2*w^2 a square, where w,x,y are nonnegative integers and z is a positive integer.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 16 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 15, 31, 39, 71, 79, 195, 311, 319, 403, 559, 591, 683, 719, 1031, 1439, 1643, 2519, 6879, 2^k, 2^(2k+1)*39 (k = 0,1,2,...). Also, any positive integer can be written as w^2 + x^2 + y^2 + z^2 with x a positive integer and w,y,z nonnegative integer such that 6*w^2*x^2 + 12*x^2*y^2 + 52*y^2*z^2 + 27*z^2*w^2 is a square.
(ii) For each triple (a,b,c) = (1,3,2), (1,11,9), (1,14,4),(1,20,25), (1,27,18), (1,36,9), (1,56,4), (4,32,25), (9,15,25), (9,35,25), (25,8,64), (25,15,54), (25,32,28), (25,35,49), (28,32,49), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z integers such that a*w^2*x^2 + b*x^2*y^2 + c*y^2*z^2 is a square.
See also A268507, A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775, A271778 and A271824 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 1 > 0 and 6*0^2*0^2 + 12*0^2*0^2 + 52*0^2*1^2 + 27*1^2*0^2 = 0^2.
a(2) = 1 since 2 = 0^2 + 1^2 + 0^2 + 1^2 with 1 > 0 and 6*0^2*1^2 + 12*1^2*0^2 + 52*0^2*1^2 + 27*1^2*0^2 = 0^2.
a(3) = 1 since 3 = 0^2 + 1^2 + 1^2 + 1^2 with 1 > 0 and 6*0^2*1^2 + 12*1^2*1^2 + 52*1^2*1^2 + 27*1^2*0^2 = 8^2.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^2 + 1^2 with 1 > 0 and 6*2^2*3^2 + 12*3^2*1^2 + 52*1^2*1^2 = 22^2.
a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 5 > 0 and 6*1^2*1^2 + 12*1^2*2^2 + 52*2^2*5^2 + 27*5^2*1^2 = 77^2.
a(39) = 1 since 39 = 2^2 + 1^2 + 5^2 + 3^2 with 3 > 0 and 6*2^2*1^2 + 12*1^2*5^2 + 52*5^2*3^2 + 27*3^2*2^2 = 114^2.
a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 5 > 0 and 6*3^2*1^2 + 12*1^2*6^2 + 52*6^2*5^2 + 27*5^2*3^2 = 231^2.
a(78) = 1 since 78 = 2^2 + 7^2 + 4^2 + 3^2 with 3 > 0 and 6*2^2*7^2 + 12*7^2*4^2 + 52*7^2*4^2 + 27*3^2*2^2 = 138^2.
a(79) = 1 since 79 = 2^2 + 5^2 + 7^2 + 1^2 with 1 > 0 and 6*2^2*5^2 + 12*5^2*7^2 + 52*7^2*1^2 + 27*1^2*2^2 = 134^2.
a(195) = 1 since 195 = 3^2 + 7^2 + 4^2 + 11^2 with 11 > 0 and 6*3^2*7^2 + 12*7^2*4^2 + 52*4^2*11^2 + 27*11^2*3^2 = 377^2.
a(311) = 1 since 311 = 14^2 + 9^2 + 3^2 + 5^2 with 5 > 0 and 6*14^2*9^2 + 12*9^2*3^2 + 52*3^2*5^2 + 27*5^2*14^2 = 498^2.
a(319) = 1 since 319 = 6^2 + 3^2 + 7^2 + 15^2 with 15 > 0 and 6*6^2*3^2 + 12*3^2*7^2 + 52*7^2*15^2 + 27*15^2*6^2 = 894^2.
a(403) = 1 since 403 = 3^2 + 13^2 + 12^2 + 9^2 with 9 > 0 and 6*3^2*13^2 + 12*13^2*12^2 + 52*12^2*9^2 + 27*9^2*3^2 = 963^2.
a(559) = 1 since 559 = 5^2 + 23^2 + 2^2 + 1^2 with 1 > 0 and 6*5^2*23^2 + 12*23^2*2^2 + 52*2^2*1^2 + 27*1^2*5^2 = 325^2.
a(591) = 1 since 591 = 21^2 + 11^2 + 2^2 + 5^2 with 5 > 0 and 6*21^2*11^2 + 12*11^2*2^2 + 52*2^2*5^2 + 27*5^2*21^2 = 793^2.
a(683) = 1 since 683 = 0^2 + 11^2 + 21^2 + 11^2 with 11 > 0 and 6*0^2*11^2 + 12*11^2*21^2 + 52*21^2*11^2 + 27*11^2*0^2 = 1848^2.
a(719) = 1 since 719 = 10^2 + 3^2 + 21^2 + 13^2 with 13 > 0 and 6*10^2*3^2 + 12*3^2*21^2 + 52*21^2*13^2 + 27*13^2*10^2 = 2094^2.
a(1031) = 1 since 1031 = 26^2 + 15^2 + 9^2 + 7^2 with 7 > 0 and 6*26^2*15^2 + 12*15^2*9^2 + 52*9^2*7^2 + 27*7^2*26^2 = 1494^2.
a(1439) = 1 since 1439 = 13^2 + 27^2 + 10^2 + 21^2 with 21 > 0 and 6*13^2*27^2 + 12*27^2*10^2 + 52*10^2*21^2 + 27*21^2*13^2 = 2433^2.
a(1643) = 1 since 1643 = 36^2 + 17^2 + 3^2 + 7^2 with 7 > 0 and 6*36^2*17^2 + 12*17^2*3^2 + 52*3^2*7^2 + 27*7^2*36^2 = 2004^2.
a(2519) = 1 since 2519 = 27^2 + 7^2 + 30^2 + 29^2 with 29 > 0 and 6*27^2*7^2 + 12*7^2*30^2 + 52*30^2*29^2 + 27*29^2*27^2 = 7527^2.
a(6879) = 1 since 6879 = 38^2 + 53^2 + 49^2 + 15^2 with 15 > 0 and 6*38^2*53^2 + 12*53^2*49^2 + 52*49^2*15^2 + 27*15^2*38^2 = 11922^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[6*(n-x^2-y^2-z^2)*x^2+12*x^2*y^2+52*y^2*z^2+27*z^2*(n-x^2-y^2-z^2)],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}]

A272084 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 4*x^2 + 5*y^2 + 20*z*w a square, where x,y,z,w are nonnegative integers with z < w.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 19 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k*q (k = 0,1,2,... and q = 1, 3, 7), 4^k*m (k = 0,1,2,... and m = 21, 30, 38, 62, 70, 77, 142, 217, 237, 302, 382, 406, 453, 670).
(ii) For each triple (a,b,c) = (1,8,8), (7,9,-12), (9,40,-24), (9,40,-60), any positive integer can be written as x^2 + y^2 + z^2 + w^2 with a*x^2 + b*y^2 + c*z*w a square, where w is a positive integer and x,y,z are nonnegative integers.
(iii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with (3*x+5*y)^2 -24*z*w a square, where x,y,z,w are nonnegative integers. Also, for each ordered pair (a,b) = (1,4), (1,8), (1,12), (1,24), (1,32), (1,48), (25,24), (1,-4), (9,-4), (121,-20), every natural number can be written as x^2 + y^2 + z^2 + w^2 with a*x^2 + b*y*z a square, where x,y,z,w are nonnegative integers.
(iv) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with (x^2-y^2)*(w^2-2*z^2) (or (x^2-y^2)*(2*w^2-z^2) or (x^2-y^2)*(w^2-5*z^2)) a square, where w,x,y,z are integers.
See also A262357, A268507, A269400, A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775, A271778 and A271824 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 < 1 and 4*0^2 + 5*0^2 + 20*0*1 = 0^2.
a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 0 < 1 and 4*1^2 + 5*0^2 + 20*0*1 = 2^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 0 < 1 and 4*1^2 + 5*1^2 + 20*0*1 = 3^2.
a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 0 < 2 and 4*1^2 + 5*1^2 + 20*0*2 = 3^2.
a(14) = 1 since 14 = 1^2 + 3^2 + 0^2 + 2^2 with 0 < 2 and 4*1^2 + 5*3^2 + 20*0*2 = 7^2.
a(21) = 1 since 21 = 0^2 + 2^2 + 1^2 + 4^2 with 1 < 4 and 4*0^2 + 5*2^2 + 20*1*4 = 10^2.
a(30) = 1 since 30 = 4^2 + 2^2 + 1^2 + 3^2 with 1 < 3 and 4*4^2 + 5*2^2 + 20*1*3 = 12^2.
a(38) = 1 since 38 = 1^2 + 1^2 + 0^2 + 6^2 with 0 < 6 and 4*1^2 + 5*1^2 + 20*0*6 = 3^2.
a(62) = 1 since 62 = 1^2 + 3^2 + 4^2 + 6^2 with 4 < 6 and 4*1^2 + 5*3^2 + 20*4*6 = 23^2.
a(70) = 1 since 70 = 7^2 + 1^2 + 2^2 + 4^2 with 2 < 4 and 4*7^2 + 5*1^2 + 20*2*4 = 19^2.
a(77) = 1 since 77 = 4^2 + 6^2 + 3^2 + 4^2 with 3 < 4 and 4*4^2 + 5*6^2 + 20*3*4 = 22^2.
a(142) = 1 since 142 = 4^2 + 6^2 + 3^2 + 9^2 with 3 < 9 and 4*4^2 + 5*6^2 + 20*3*9 = 28^2.
a(217) = 1 since 217 = 6^2 + 6^2 + 8^2 + 9^2 with 8 < 9 and 4*6^2 + 5*6^2 + 20*8*9 = 42^2.
a(237) = 1 since 237 = 5^2 + 8^2 + 2^2 + 12^2 with 2 < 12 and 4*5^2 + 5*8^2 + 20*2*12 = 30^2.
a(302) = 1 since 302 = 11^2 + 9^2 + 6^2 + 8^2 with 6 < 8 and 4*11^2 + 5*9^2 + 20*6*8 = 43^2.
a(382) = 1 since 382 = 11^2 + 7^2 + 4^2 + 14^2 with 4 < 14 and 4*11^2 + 5*4^2 + 20*4*14 = 43^2.
a(406) = 1 since 406 = 8^2 + 6^2 + 9^2 + 15^2 with 9 < 15 and 4*8^2 + 5*6^2 + 20*9*15 = 56^2.
a(453) = 1 since 453 = 8^2 + 14^2 + 7^2 + 12^2 with 7 < 12 and 4*8^2 + 5*14^2 + 20*7*12 = 54^2.
a(670) = 1 since 670 = 17^2 + 11^2 + 2^2 + 16^2 with 2 < 16 and 4*17^2 + 5*11^2 + 20*2*16 = 49^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[4x^2+5y^2+20*z*Sqrt[n-x^2-y^2-z^2]],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[(n-1-x^2-y^2)/2]}];Print[n," ",r];Continue,{n,1,100}]

A272332 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 36*x^2*y + 12*y^2*z + z^2*x a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 2, 2, 6, 4, 3, 3, 3, 8, 5, 2, 6, 6, 4, 1, 7, 10, 6, 8, 8, 5, 2, 2, 7, 16, 8, 3, 12, 6, 4, 3, 6, 13, 8, 8, 8, 6, 5, 7, 15, 14, 4, 2, 12, 7, 3, 2, 5, 18, 8, 12, 14, 8, 7, 4, 6, 8, 7, 5, 14, 8, 5, 2, 12, 18, 8, 12, 10, 6, 3, 5, 10, 19, 10, 3, 8, 3, 1, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 26 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m (k = 0,1,2,... and m = 1, 79, 591, 599, 1752, 1839, 10264).
We have verified that a(n) > 0 for all n = 1,...,400000.
For more refinements of Lagrange's four-square theorem, see arXiv:1604.06723.

Examples

			a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 > 0 and 36*0^2*0 + 12*0^2*0 + 0^2*0 = 0^2.
a(79) = 1 since 79 = 7^2 + 1^2 + 5^2 + 2^2 with 7 > 0 and 36*1^2*5 + 12*5^2*2 + 2^2*1 = 28^2.
a(591) = 1 since 591 = 23^2 + 1^2 + 6^2 + 5^2 with 23 > 0 and 36*1^2*6 + 12*6^2*5 + 5^2*1 = 49^2.
a(599) = 1 since 599 = 6^2 + 1^2 + 11^2 + 21^2 with 6 > 0 and 36*1^2*11 + 12*11^2*21 + 21^2*1 = 177^2.
a(1752) = 1 since 1752 = 10^2 + 4^2 + 40^2 + 6^2 with 10 > 0 and 36*4^2*40 + 12*40^2*6 + 6^2*10 = 372^2.
a(1839) = 1 since 1839 = 17^2 + 37^2 + 9^2 + 10^2 with 17 > 0 and 36*37^2*9 + 12*9^2*10 + 10^2*37 = 676^2.
a(10264) = 1 since 10264 = 96^2 + 30^2 + 2^2 + 12^2 with 96 > 0 and 36*30^2*2 + 12*2^2*12 + 12^2*30 = 264^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[36*x^2*y+12*y^2*z+z^2*x],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]
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