A300510 -id:A300510 - cp's OEIS frontend

A301805 Number of ways to write 3n^2 as x^2 + 10y^2 + 2^z, where x, y and z are nonnegative integers with z > 3.

0, 0, 1, 1, 2, 2, 3, 3, 3, 2, 4, 4, 4, 4, 4, 4, 6, 4, 4, 3, 5, 4, 4, 5, 7, 5, 4, 4, 6, 4, 7, 5, 5, 7, 7, 5, 5, 4, 8, 5, 7, 6, 11, 6, 6, 5, 8, 5, 6, 7, 5, 7, 6, 5, 5, 5, 7, 7, 4, 4, 8, 8, 8, 6, 6, 6, 9, 8, 8, 7, 8, 6, 10, 6, 10, 6, 8, 8, 8, 5

Offset: 1

Zhi-Wei Sun, Mar 27 2018

nonn

It might seem that a(n) > 0 for all n > 2. However, we find that a(323525083) = 0, moreover 3*323525083^2 cannot be written as x^2 + 10*y^2 + 2^z with x,y,z nonnegative integers. We note also that a(270035155) = 0 but 3*270035155^2 - 2^0 has the form x^2 + 10*y^2 with x and y integers.
My way to check whether 3*n^2 can be written as x^2 + 10*y^2 + 2^z is to find z such that 3*n^2 - 2^z can be written as x^2 + 10*y^2. I observe that a positive integer n has the form x^2 + 10*y^2 with x and y integers if and only if the p-adic order ord_p(n) of n is even for any prime p == 3, 17, 21, 27, 29, 31, 33, 39 (mod 40) and the sum of those ord_p(n) with p prime and p == 2, 5, 7, 13, 23, 37 (mod 40) is even.
From David A. Corneth, Mar 27 2018: (Start)
If a(n) > 0 then a(2*n) > 0; 3*n^2 = x^2 + 10*y^2 + 2^z <=> 3*(2*n)^2 = 4 * 3*n^2 = 4 * (x^2 + 10*y^2 + 2^z) = (2*x)^2 + 10 * (2*y)^2 + 2^(z + 2).
So we just need to check odd n and as z > 0, 2 | 2^z and furthermore 2 | 10 * y^2 so 3*x^2 must be odd, i.e., x must be odd for 3*n^2 to be odd. Also, y must be odd. For odd n, 3*n^2 == 3 (mod 4), for odd x, x^2 == (1 mod 4), for z >= 3, 2^z == 0 (mod 4) so 10 * y^2 must be == 2 (mod 4) which happens if and only if y is odd. (End)

			a(1) = a(2) = 0 since 3*1^2 < 3*2^2 < 2^4.
a(3) = 1 since 3*3^2 = 1^2 + 10*1^2 + 2^4.
a(4) = 1 since 3*4^2 = 4^2 + 10*0^2 + 2^5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..800
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], 2017-2018.

Cf. A000079, A000290, A020673, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301471, A301472, A301479, A301579, A301640.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[3*n^2-2^k-10x^2],r=r+1],{k,4,Log[2,3n^2]},{x,0,(3*n^2-2^k)/10}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301891 Number of ways to write 2n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3y + 5z + 15w is a power of 4.

Original entry on oeis.org

2, 1, 1, 2, 3, 1, 2, 1, 3, 1, 2, 1, 6, 3, 8, 2, 5, 4, 6, 3, 4, 3, 6, 1, 3, 3, 7, 2, 8, 9, 8, 1, 15, 5, 8, 3, 11, 1, 5, 1, 4, 4, 2, 2, 10, 7, 17, 1, 18, 11, 14, 6, 16, 6, 17, 3, 21, 10, 16, 8, 19, 8, 30, 2, 15, 9, 18, 5, 28, 5, 27, 4, 13, 11, 24, 6, 28, 17, 20, 3

Offset: 1

Zhi-Wei Sun, Mar 28 2018

nonn

Conjecture 1: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m with k = 0,1,2,... and m = 2, 3, 6, 10, 38.

Conjecture 2: For any positive integer n, we can write 2*n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3*y + 5*z + 15*w is twice a power of 4.

			a(1) = 2 since 2*1^2 = 1^2 + 1^2 + 0^2 + 0^2 with 1 + 3*1 + 5*0 + 15*0 = 4, and 2*1^2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 + 3*0 + 5*0 + 15*1 = 4^2.
a(2) = 1 since 2*2^2 = 0^2 + 2^2 + 2^2 + 0^2 with 0 + 3*2 + 5*2 + 15*0 = 4^2.
a(3) = 1 since 2*3^2 = 1^2 + 1^2 + 0^2 + 4^2 with 1 + 3*1 + 5*0 + 15*4 = 4^3.
a(6) = 1 since 2*6^2 = 0^2 + 8^2 + 2^2 + 2^2 with 0 + 3*8 + 5*2 + 15*2 = 4^3.
a(10) = 1 since 2*10^2 = 10^2 + 8^2 + 6^2 + 0^2 with 10 + 3*8 + 5*6 + 15*0 = 4^3.
a(38) = 1 since 2*38^2 = 34^2 + 34^2 + 24^2 + 0^2 with 34 + 3*34 + 5*24 + 15*0 = 4^4.

Zhi-Wei Sun, Table of n, a(n) for n = 1..225
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], 2017-2018.

Cf. A000290, A000302, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301471, A301579, A301640.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]]
tab={};Do[r=0;Do[If[SQ[2n^2-x^2-y^2-z^2]&&Pow[x+3y+5z+15*Sqrt[2n^2-x^2-y^2-z^2]],r=r+1],{x,0,Sqrt[2]n},{y,0,Sqrt[2n^2-x^2]},{z,0,Sqrt[2n^2-x^2-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

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A301805 Number of ways to write 3n^2 as x^2 + 10y^2 + 2^z, where x, y and z are nonnegative integers with z > 3.

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A301891 Number of ways to write 2n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3y + 5z + 15w is a power of 4.

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cp's OEIS Frontend

A301805 Number of ways to write 3*n^2 as x^2 + 10*y^2 + 2^z, where x, y and z are nonnegative integers with z > 3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A301891 Number of ways to write 2*n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3*y + 5*z + 15*w is a power of 4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A301805 Number of ways to write 3n^2 as x^2 + 10y^2 + 2^z, where x, y and z are nonnegative integers with z > 3.

A301891 Number of ways to write 2n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3y + 5z + 15w is a power of 4.