A301640 -id:A301640 - cp's OEIS frontend

A301471 Number of ways to write n^2 as x^2 + 2y^2 + 32^z with x,y,z nonnegative integers.

0, 1, 2, 1, 3, 4, 3, 1, 5, 4, 4, 4, 5, 4, 10, 1, 4, 7, 4, 4, 10, 4, 3, 4, 6, 6, 11, 4, 7, 10, 6, 1, 9, 5, 7, 7, 7, 6, 12, 4, 6, 12, 7, 4, 14, 4, 8, 4, 3, 8, 10, 6, 8, 13, 6, 4, 16, 8, 7, 10, 7, 6, 14, 1, 7, 11, 6, 5, 16, 9, 5, 7, 7, 7, 18, 6, 7, 14, 6, 4

Offset: 1

Zhi-Wei Sun, Mar 21 2018

nonn

Square Conjecture: a(n) > 0 for all n > 1. Moreover, for any integer n > 3 we can write n^2 as x^2 + 2*y^2 + 3*2^z, where x,y,z are nonnegative integers with y even and z > 1.
It is known that a positive integer n has the form x^2 + 2*y^2 with x and y integers if and only if the p-adic order of n is even for any prime p == 5 or 7 (mod 8).
See also A301472 for the list of positive integers not of the form x^2 + 2*y^2 + 3*2^z with x,y,z nonnegative integers.
If n^2 = x^2 + 2*y^2 + 3*2^z with x,y,z nonnegative integers, then it is easy to see that x is not divisible by 3.
The Square Conjecture implies that for each n = 1,2,3,... we can write 3*n^2 as x^2 + 2*y^2 + 2^z with x,y,z nonnegative integers. In fact, if (3*n)^2 = u^2 + 2*v^2 + 3*2^z with u,v,z integers and z >= 0, then u^2 == v^2 (mod 3) and thus we may assume u == v (mod 3) without loss of generality, hence 3*n^2 = (u^2+2*v^2)/3 + 2^z = x^2 + 2*y^2 + 2^z with x = (u+2*v)/3 and y = (u-v)/3 integers.
On March 25, 2018 Qing-Hu Hou at Tianjin Univ. finished his verification of the Square Conjecture for n <= 4*10^8. Then I used Hou's program to verify the conjecture for n <= 5*10^9. - Zhi-Wei Sun, Apr 10 2018
I have found a counterexample to the Square Conjecture, namely a(5884015571) = 0. Note that 5884015571 is the product of the three primes 7, 17 and 49445509. - Zhi-Wei Sun, Apr 15 2018

			a(2) = 1 with 2^2 = 1^2 + 2*0^2 + 3*2^0.
a(3) = 2 with 3^2 = 2^2 + 2*1^2 + 3*2^0 = 1^2 + 2*1^2 + 3*2^1.
a(4) = 1 with 4^2 = 2^2 + 2*0^2 + 3*2^2.
a(1131599953) = 1 with 1131599953^2 = 316124933^2 + 2*768304458^2 + 3*2^6.
a(5884015571) = 0 since there are no nonnegative integers x,y,z such that  x^2 + 2*y^2 + 3*2^z = 5884015571^2.

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], 2017-2018.

Cf. A000079, A000290, A002479, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301472, A301479, A301579, A301640, A302641.

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[(Mod[Part[Part[f[n],i],1],8]==5||Mod[Part[Part[f[n],i],1],8]==7)&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[QQ[n^2-3*2^k],Do[If[SQ[n^2-3*2^k-2x^2],r=r+1],{x,0,Sqrt[(n^2-3*2^k)/2]}]],{k,0,Log[2,n^2/3]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A302641 Number of nonnegative integers k such that n^2 - 32^k can be written as x^2 + 2y^2 with x and y integers.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Apr 10 2018

nonn

The author's Square Conjecture in A301471 implies that a(n) > 0 for all n > 1.

We have a(2^n) = 1 for all n > 0. In fact, (2^n)^2 = (2^(n-1))^2 + 2*0^2 + 3*2^(2*n-2). If k > 2*n-2 then 3*2^k >= 6*2^(2*n-2) > (2^n)^2. If 0 <= k < 2*n-2, then 2*n-k is at least 3 and hence (2^n)^2 - 3*2^k = 2^k*(2^(2*n-k)-3) cannot be written as x^2 + 2*y^2 with x and y integers.

			a(2) = 1 with 2^2 = 1^2 + 2*0^2 + 3*2^0.
a(3) = 2 with 3^2 = 2^2 + 2*1^2 + 3*2^0 = 1^2 + 2*1^2 + 3*2^1.
a(2857932461) = 1 since 3 is the only nonnegative integer k such that 2857932461^2 - 3*2^k has the form x^2 + 2*y^2 with x and y integers.
a(4428524981) = 2 since 3 and 8 are the only nonnegative integers k such that 4428524981^2 - 3*2^k has the form x^2 + 2*y^2 with x and y integers.
a(4912451281) = 3 since 3, 6 and 7 are the only nonnegative integers k with 4428524981^2 - 3*2^k = x^2 + 2*y^2 for some integers x and y.

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], 2017-2018.
Zhi-Wei Sun, My square conjecture with prize, a message to Number Theory List, April 7, 2018.

Cf. A000079, A000290, A002479, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301472, A301479, A301579, A301640.

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[(Mod[Part[Part[f[n],i],1],8]==5||Mod[Part[Part[f[n],i],1],8]==7)&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
tab={};Do[r=0;Do[If[QQ[n^2-3*2^k],r=r+1],{k,0,Log[2,n^2/3]}];tab=Append[tab,r],{n,1,80}];Print[tab]

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.

Original entry on oeis.org

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]

cp's OEIS Frontend

A301471 Number of ways to write n^2 as x^2 + 2*y^2 + 3*2^z with x,y,z nonnegative integers.

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A302641 Number of nonnegative integers k such that n^2 - 3*2^k can be written as x^2 + 2*y^2 with x and y integers.

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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.

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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.

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A301471 Number of ways to write n^2 as x^2 + 2y^2 + 32^z with x,y,z nonnegative integers.

A302641 Number of nonnegative integers k such that n^2 - 32^k can be written as x^2 + 2y^2 with x and y integers.

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.