A283239 -id:A283239 - cp's OEIS frontend

A283269 Number of ways to write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3y + 5z is a square.

1, 1, 2, 2, 0, 3, 5, 0, 3, 4, 4, 1, 0, 3, 7, 0, 1, 5, 3, 3, 0, 5, 3, 0, 6, 2, 8, 2, 0, 8, 3, 0, 2, 3, 6, 7, 0, 2, 6, 0, 6, 8, 4, 1, 0, 4, 3, 0, 2, 3, 7, 5, 0, 4, 13, 0, 8, 5, 2, 3, 0, 6, 6, 0, 0, 7, 13, 2, 0, 7, 3, 0, 5, 4, 9, 1, 0, 5, 3, 0, 3

Offset: 0

Zhi-Wei Sun, Mar 04 2017

nonn

Conjecture: (i) a(2^k*m) > 0 for any positive odd integers k and m. Also, a(4*n+1) = 0 only for n = 63.
(ii) For any positive odd integers k and m, we can write 2^k*m as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 5*z is twice a square.
(iii) For any positive odd integer n not congruent to 7 modulo 8, we can write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 4*z is a square.
(iv) Let n be any nonnegative integer. Then we can write 8*n + 1 as x^2 + y^2 + z^2 with x + 3*y a square, where x and y are integers, and z is a positive integer. Also, except for n = 2255, 4100 we can write 4*n + 2 as x^2 + y^2 + z^2 with x + 3*y a square, where x,y,z are integers.
(v) For each n = 0,1,2,..., we can write 8*n + 6 as x^2 + y^2 + z^2 with x + 2*y a square, where x,y,z are integers with y nonnegative and z odd.
The Gauss-Legendre theorem asserts 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*(8m+7) with k and m nonnegative integers. Thus a(4^k*(8m+7)) = 0 for all k,m = 0,1,2,....
See also A283273 for a similar conjecture.

			a(11) = 1 since 11 = 3^2 + 1^2 + (-1)^2 with 3 + 3*1 + 5*(-1) = 1^2.
a(43) = 1 since 43 = (-5)^2 + (-3)^2 + 3^2 with (-5) + 3*(-3) + 5*3 = 1^2.
a(75) = 1 since 75 = (-1)^2 + 5^2 + 7^2 with (-1) + 3*5 + 5*7 = 7^2.
a(262) = 1 since 262 = 1^2 + 15^2 + (-6)^2 with 1 + 3*15 + 5*(-6) = 4^2.
a(277) = 1 since 277 = (-6)^2 + 4^2 + 15^2 with (-6) + 3*4 + 5*15 = 9^2.
a(617) = 1 since 617 = 17^2 + 18^2 + 2^2 with 17 + 3*18 + 5*2 = 9^2.
a(1430) = 1 since 1430 = (-13)^2 + (-6)^2 + 35^2 with (-13) + 3*(-6) + 5*35 = 12^2.
a(5272) = 1 since 5272 = (-66)^2 + 30^2 + (-4)^2 with (-66) + 3*30 + 5*(-4) = 2^2.
a(7630) = 1 since 7630 = (-78)^2 + 39^2 + 5^2 with (-78) + 3*39 + 5*5 = 8^2.
a(7933) = 1 since 7933 = (-56)^2 + 69^2 + (-6)^2 with (-56) + 3*69 + 5*(-6) = 11^2.
a(14193) = 1 since 14193 = (-7)^2 + 112^2 + 40^2 with (-7) + 3*112 + 5*40 = 23^2.

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, A283170, A283196, A283204, A283205, A283239, A283273.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^2-y^2]&&SQ[(-1)^i*x+(-1)^j*3y+(-1)^k*5*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]},{k,0,Min[Sqrt[n-x^2-y^2],1]}];Print[n," ",r];Continue,{n,0,80}]

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.

Original entry on oeis.org

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}]

A340778 Decimal expansion of (2 - e*log(2))/4.

Original entry on oeis.org

0, 2, 8, 9, 5, 7, 6, 5, 3, 6, 5, 9, 0, 6, 9, 9, 7, 2, 5, 2, 4, 4, 6, 0, 2, 2, 0, 7, 6, 8, 6, 4, 9, 6, 6, 6, 3, 2, 5, 6, 8, 3, 7, 3, 5, 8, 8, 6, 3, 1, 4, 6, 5, 3, 4, 7, 6, 8, 4, 8, 3, 8, 0, 0, 4, 9, 2, 8, 0, 4, 4, 8, 6, 2, 8, 8, 2, 8, 8, 4, 9, 4, 3, 8, 8, 0, 5, 8

Offset: 0

Stefano Spezia, Jan 21 2021

This constant appears in an asymptotic formula proved by Linnik in 1960 in an additive problem of Hardy-Littlewood (see Formula 1 in Dimitrov and Formula 0.3 in Linnik).

			0.02895765365906997252446022076864966632568373588631465...

Stoyan I. Dimitrov, The ternary Piatetski-Shapiro inequality with one prime of the form p = x^2 + y^2 + 1, arXiv:2011.03967 [math.NT], 2020.
Juriĭ Vladimirovič Linnik, An asymptotic formula in an additive problem of Hardy-Littlewood, Izv. Akad. Nauk SSSR Ser. Mat., 24 (1960), 629-706 (in Russian).

Cf. A001113, A002162, A079544, A276415, A283239.

Mathematica
```
First[RealDigits[N[(2-E*Log[2])/4,87]]]
```

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A283269 Number of ways to write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 5*z is a square.

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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 + 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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A340778 Decimal expansion of (2 - e*log(2))/4.

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A283269 Number of ways to write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3y + 5z is a square.

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.