A283269 -id:A283269 - cp's OEIS frontend

A283273 Number of ways to write n as x^2 + y^2 + z^2 with x,y integers and z a nonnegative integer such that x + 2y + 3z is a square or twice a square.

1, 2, 4, 3, 2, 6, 3, 0, 4, 6, 4, 5, 3, 5, 7, 0, 2, 3, 4, 6, 6, 7, 5, 0, 3, 4, 9, 5, 0, 15, 4, 0, 4, 3, 6, 9, 6, 4, 7, 0, 4, 7, 7, 4, 5, 9, 3, 0, 3, 2, 6, 9, 5, 11, 12, 0, 7, 5, 4, 13, 0, 9, 6, 0, 2, 9, 11, 2, 3, 6, 5, 0, 4, 5, 12, 6, 6, 11, 5, 0, 6

Offset: 0

Zhi-Wei Sun, Mar 04 2017

nonn

Conjecture: (i) For any nonnegative integer n not of the form 4^k*(8*m+7) (k,m = 0,1,2,...), we have a(n) > 0.
(ii) Any nonnegative integer not of the form 4^k*(8*m+7) (k,m = 0,1,2,...) can be written as x^2 + y^2 + z^2 with x,y,z integers such that a*x + b*y + c*z is a square or twice a square, whenever (a,b,c) is among the triples (1,2,5), (1,3,4), (1,4,7), (1,6,7), (2,3,4), (2,3,9), (3,4,7), (3,4,9).
The Gauss-Legendre theorem states that a nonnegative integer can be written as the sum of three squares if and only if it is not of the form 4^k*(8*m+7) with k and m nonnegative integers. Thus a(4^k*(8*m+7)) = 0 for all k,m = 0,1,2,..., and part (i) of our conjecture (verified for n up to 1.6*10^6) is stronger than the Gauss-Legendre theorem.
See also A283269 for a similar conjecture.

			a(82) = 1 since 82 = 0^2 + (-1)^2 + 9^2 with 0 + 2*(-1) + 3*9 = 5^2.
a(328) = 1 since 328 = 0^2 + (-2)^2 + 18^2 with 0 + 2*(-2) + 3*18 = 2*5^2.
a(330) = 1 since 330 = 5^2 + 4^2 + 17^2 with 5 + 2*4 + 3*17 = 8^2.
a(466) = 1 since 466 = 21^2 + 0^2 + 5^2 with 21 + 2*0 + 3*5 = 6^2.
a(1320) = 1 since 1320 = 10^2 + 8^2 + 34^2 with 10 + 2*8 + 3*34 = 2*8^2.
a(1387) = 1 since 1387 = 33^2  + (-17)^2 + 3^2 with 33 + 2*(-17) + 3*3 = 2*2^2.
a(1857) = 1 since 1857 = (-37)^2 + (-2)^2 + 22^2 with (-37) + 2*(-2) + 3*22 = 5^2.
a(1864) = 1 since 1864 = 42^2 + 0^2 + 10^2 with 42 + 2*0 + 3*10 = 2*6^2.
a(2386) = 1 since 2386 = (-44)^2 + 3^2 + 21^2 with (-44) + 2*3 + 3*21 = 5^2.
a(5548) = 1 since 5548 = 66^2 + (-34)^2 + 6^2 with 66 + 2*(-34) + 3*6 = 4^2.
a(7428) = 1 since 7428 = (-74)^2 + (-4)^2 + 44^2 with (-74) + 2*(-4) + 3*44 = 2*5^2.
a(9544) = 1 since 9544 = (-88)^2 + 6^2 + 42^2 with (-88) + 2*6 + 3*42 = 2*5^2.

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. A000290, A005875, A271518, A275344, A283170, A283196, A283204, A283205, A283239, A283269.

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

A283299 Number of ways to write 2n + 1 as x^2 + 2y^2 + 3*z^2 with x,y,z integers such that x + y + z is a square or twice a square.

Original entry on oeis.org

1, 4, 3, 1, 6, 3, 1, 7, 1, 2, 8, 4, 4, 4, 3, 5, 4, 1, 4, 9, 3, 3, 9, 1, 4, 10, 3, 3, 11, 7, 4, 8, 5, 6, 7, 6, 2, 10, 3, 3, 14, 1, 2, 5, 3, 6, 12, 2, 4, 11, 3, 2, 5, 5, 7, 14, 6, 4, 6, 7, 4, 5, 4, 3, 13, 3, 3, 12, 3, 2, 15, 2, 2, 12, 3, 7, 4, 5, 4, 11, 8

Offset: 0

Zhi-Wei Sun, Mar 05 2017

nonn

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 3, 6, 8, 17, 23, 41, 128, 197, 372, 764, 1143, 1893, 3761, 4307, 6408, 6918.
(ii) Any positive odd integer can be written as x^2 + y^2 + 2*z^2 with x,y,z integers such that x + 2*y + 3*z is a square or twice a square.
By Dickson's book in the reference, any positive odd integer can be written as x^2 + 2*y^2 + 3*z^2 (or x^2 + y^2 + 2*z^2) with x,y,z integers.
We have verified a(n) > 0 for all n = 0..10^6.
See also A283366 for a similar conjecture.

			a(0) = 1 since 2*0 + 1 = 1^2 + 2*0^2 + 3*0^2 with 1 + 0 + 0 = 1^2.
a(3) = 1 since 2*3 + 1 = 2^2 + 2*0^2 + 3*(-1)^2 with 2 + 0 + (-1) = 1^2.
a(8) = 1 since 2*8 + 1 = 3^2 + 2*(-2)^2 + 3*0^2 with 3 + (-2) + 0 = 1^2.
a(17) = 1 since 2*17 + 1 = 0^2 + 2*(-2)^2 + 3^2 with 0 + (-2) + 3 = 1^2.
a(41) = 1 since 2*41 + 1 = 9^2 + 2*(-1)^2 + 3*0^2 with 9 + (-1) + 0 = 2*2^2.
a(128) = 1 since 2*128 + 1 = 3^2 + 2*10^2 + 3*(-4)^2 with 3 + 10 + (-4) = 3^2.
a(197) = 1 since 2*197 + 1 = 12^2 + 2*(-2)^2 + 3*(-9)^2 with 12 + (-2) + (-9) = 1^2.
a(372) = 1 since 2*372 + 1 = 22^2 + 2*3^2 + 3*(-9)^2 with 22 + 3 + (-9) = 4^2.
a(764) = 1 since 2*764 + 1 = 18^2 + 2*(-23)^2 + 3*7^2 with 18 + (-23) + 7 = 2*1^2.
a(3761) = 1 since 2*3761 + 1 = (-57)^2 + 2*31^2 + 3*28^2 with (-57) + 31 + 28 = 2*1^2.
a(6408) = 1 since 2*6408 + 1 = (-22)^2 + 2*75^2 + 3*19^2 with (-22) + 75 + 19 = 2*6^2.
a(6918) = 1 since 2*6918 + 1 = 100^2 + 2*9^2 + 3*35^2 with 100 + 9 + 35 = 12^2.

L. E. Dickson, Modern Elementary Theory of Numbers, University of Chicago Press, Chicago, 1939. (See pages 112-113.)

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. A000290, A271518, A283239, A283269, A283273, A283366.

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

A283366 Number of ways to write 2n + 1 as x^2 + y^2 + 2z^2 with x,y,z integers such that 2*x + y + z is a square or a power of two.

Original entry on oeis.org

2, 4, 3, 5, 4, 2, 7, 5, 5, 6, 3, 5, 7, 8, 3, 9, 7, 3, 11, 1, 2, 8, 9, 7, 6, 2, 3, 11, 7, 7, 7, 7, 1, 12, 7, 4, 12, 6, 7, 4, 8, 4, 8, 7, 7, 9, 3, 1, 15, 8, 2, 12, 4, 4, 4, 8, 5, 12, 11, 5, 7, 6, 5, 11, 2, 3, 12, 12, 9, 9, 9, 4, 12, 8, 5, 5, 7, 3, 18, 8, 6

Offset: 0

Zhi-Wei Sun, Mar 06 2017

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 19, 32, 47, 115, 200, 974, 1271, 2240, 2549, 3185, 4865, 9254, 15881.
By the Gauss-Legendre theorem, for any nonnegative integer n, we can write 4*n + 2 as u^2 + v^2 + (2*z)^2 with u,v,z integers and u == v (mod 2), and hence 2*n + 1 = x^2 + y^2 + 2*z^2 with x = (u+v)/2 and y = (u-v)/2.
The conjecture implies that any positive integer with even 2-adic order can be written as x^2 + y^2 + 2*z^2 with x,y,z integers such that 2*x + y + z is a square or twice a square.
See also A283299 for a similar conjecture.

			a(0) = 2 since 2*0 + 1 = 1^2 + 0^2 + 2*0^2 with 2*1 + 0 + 0 = 2^1, and 2*0 + 1 = 0^2 + 1^2 + 2*0^2 with 2*0 + 1 + 0 = 1^2.
a(19) = 1 since 2*19 + 1 = 1^2 + 6^2 + 2*1^2 with 2*1 + 6 + 1 = 3^2.
a(32) = 1 since 2*32 + 1 = 4^2 + (-7)^2 + 2*0^2 with 2*4 + (-7) + 0 = 1^2.
a(47) = 1 since 2*47 + 1 = 6^2 + (-3)^2 + 2*(-5)^2 with 2*6 + (-3) + (-5) = 2^2.
a(115) = 1 since 2*115 + 1 = 10^2 + (-9)^2 + 2*5^2 with 2*10 + (-9) + 5 = 4^2.
a(200) = 1 since 2*200 + 1 = (-3)^2 + 0^2 + 2*14^2 with 2*(-3) + 0 + 14 = 2^3.
a(974) = 1 since 2*974 + 1 = 26^2 + (-25)^2 + 2*(-18)^2 with 2*26 + (-25) + (-18) = 3^2.
a(1271) = 1 since 2*1271 + 1 = 14^2 + 13^2 + 2*(-33)^2 with 2*14 + 13 + (-33) = 2^3.
a(2240) = 1 since 2*2240 + 1 = 28^2 + (-13)^2 + 2*(-42)^2 with 2*28 + (-13) + (-42) = 1^2.
a(2549) = 1 since 2*2549 + 1 = 59^2 + (-40)^2 + 2*3^2 with 2*59 + (-40) + 3 = 9^2.
a(3185) = 1 since 2*3185 + 1 = 33^2 + (-72)^2 + 2*7^2 with 2*33 + (-72) + 7 = 1^2.
a(4865) = 1 since 2*4865 + 1 = 72^2 + (-63)^2 + 2*(-17)^2 with 2*72 + (-63) + (-17) = 8^2.
a(9254) = 1 since 2*9254 + 1 = 61^2 + 26^2 + 2*(-84)^2 with 2*61 + 26 + (-84) = 8^2.
a(15881) = 1 since 2*15881 + 1 = (-48)^2 + 153^2 + 2*(-55)^2 with 2*(-48) + 153 + (-55) = 2^1.

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. A000079, A000290, A283239, A283269, A283273, A283299.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Pow[n_]:=Pow[n]=n>0&&IntegerQ[Log[2,n]];
TQ[n_]:=TQ[n]=SQ[n]||Pow[n];
Do[r=0;Do[If[SQ[2n+1-2x^2-y^2]&&TQ[(-1)^i*x+(-1)^j*y+(-1)^k*2*Sqrt[2n+1-2x^2-y^2]],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[2n+1-2x^2]},{i,0,Min[x,1]},{j,0,Min[y,1]},{k,0,Min[Sqrt[2n+1-2x^2-y^2],1]}];Print[n," ",r],{n,0,80}]

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A283273 Number of ways to write n as x^2 + y^2 + z^2 with x,y integers and z a nonnegative integer such that x + 2y + 3z is a square or twice a square.

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A283299 Number of ways to write 2n + 1 as x^2 + 2y^2 + 3*z^2 with x,y,z integers such that x + y + z is a square or twice a square.

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A283366 Number of ways to write 2n + 1 as x^2 + y^2 + 2z^2 with x,y,z integers such that 2*x + y + z is a square or a power of two.

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cp's OEIS Frontend

A283273 Number of ways to write n as x^2 + y^2 + z^2 with x,y integers and z a nonnegative integer such that x + 2*y + 3*z is a square or twice a square.

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A283299 Number of ways to write 2*n + 1 as x^2 + 2*y^2 + 3*z^2 with x,y,z integers such that x + y + z is a square or twice a square.

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A283366 Number of ways to write 2*n + 1 as x^2 + y^2 + 2*z^2 with x,y,z integers such that 2*x + y + z is a square or a power of two.

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A283273 Number of ways to write n as x^2 + y^2 + z^2 with x,y integers and z a nonnegative integer such that x + 2y + 3z is a square or twice a square.

A283299 Number of ways to write 2n + 1 as x^2 + 2y^2 + 3*z^2 with x,y,z integers such that x + y + z is a square or twice a square.

A283366 Number of ways to write 2n + 1 as x^2 + y^2 + 2z^2 with x,y,z integers such that 2*x + y + z is a square or a power of two.