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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A303234 Numbers of the form x*(x+1)/2 + 2^y with x and y nonnegative integers.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 14, 16, 17, 18, 19, 22, 23, 25, 26, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 42, 44, 46, 47, 49, 52, 53, 56, 57, 59, 60, 61, 63, 64, 65, 67, 68, 70, 71, 74, 77, 79, 80
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 20 2018

Keywords

Comments

Conjecture: Any integer n > 1 can be written as the sum of two terms of the current sequence.
This is equivalent to the author's conjecture in A303233.

Examples

			a(1) = 1 with 1 = 0*(0+1)/2 + 2^0.
a(2) = 2 with 2 = 1*(1+1)/2 + 2^0 = 0*(0+1)/2 + 2^1.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[Do[If[SQ[8(n-2^k)+1],tab=Append[tab,n];Goto[aa]],{k,0,Log[2,n]}];Label[aa],{n,1,80}];Print[tab]

A303389 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 5^c + 5^d, where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integers n > 1 can be written as the sum of two triangular numbers and two powers of 5.
This has been verified for all n = 2..10^10.
See A303393 for the numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.
See also A303401, A303432 and A303540 for similar conjectures.

Examples

			a(4) = 1 with 4 = 1*(1+1)/2 + 1*(1+1)/2 + 5^0 + 5^0.
a(5) = 1 with 5 = 0*(0+1)/2 + 2*(2+1)/2 + 5^0 + 5^0.
a(7) = 1 with 7 = 0*(0+1)/2 + 1*(1+1)/2 + 5^0 + 5^1.
a(25) = 1 with 25 = 0*(0+1)/2 + 5*(5+1)/2 + 5^1 + 5^1.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&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[4(n-5^j-5^k)+1],Do[If[SQ[8(n-5^j-5^k-x(x+1)/2)+1],r=r+1],{x,0,(Sqrt[4(n-5^j-5^k)+1]-1)/2}]],{j,0,Log[5,n/2]},{k,j,Log[5,n-5^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300396 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that 2*x or y is a power of 4 (including 4^0 = 1) and x + 63*y = 2^(2k+1) for some k = 0,1,2,....

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 05 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 5, 13, 25, 29, 59, 61, 79, 91, 95, 101, 103, 1315, 2^k (k = 1,2,3,...), 2^(2k+1)*m (k = 0,1,2,... and m = 3, 5, 7, 11, 15, 19, 23, 887).
This is stronger than the conjecture that A300360(n) > 0 for all n > 1. Note that a(387) = 3 < A300360(387) = 4 and a(1774) = 1 < A300360(1774) = 2.
We have verified that a(n) > 0 for all n = 2..10^7.
See also A299537, A299794 and A300219 for similar conjectures.

Examples

			a(29) = 1 since 29^2 = 2^2 + 2^2 + 7^2 + 28^2 with 2*2 = 4^1 and 2 + 63*2 = 2^7.
a(86) = 2 since 65^2 + 1^2 + 19^2 + 53^2 = 65^2 + 1^2 + 31^2 + 47^2 with 1 = 4^0 and 65 + 63*1 = 2^7.
a(1774) = 1 since 1774^2 = 8^2 + 520^2 + 14^2 + 1696^2 with 2*8 = 4^2 and 8 + 63*520 = 2^15.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]];
    tab={};Do[r=0;Do[If[Pow[y]||Pow[(2*4^k-63y)/2],Do[If[SQ[n^2-y^2-(2*4^k-63y)^2-z^2],r=r+1],{z,0,Sqrt[Max[0,(n^2-y^2-(2*4^k-63y)^2)/2]]}]],{k,0,Log[4,Sqrt[63^2+1]*n/2]},{y,0,Min[n,2*4^k/63]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A303393 Numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.

Original entry on oeis.org

1, 2, 4, 5, 6, 7, 8, 11, 15, 16, 20, 22, 25, 26, 28, 29, 31, 33, 35, 37, 40, 41, 46, 50, 53, 56, 60, 61, 67, 70, 71, 79, 80, 83, 91, 92, 96, 103, 106, 110, 116, 121, 125, 126, 128, 130, 131, 135, 137, 140, 141, 145, 146, 153, 154, 158, 161, 170, 172, 176
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

The author's conjecture in A303389 has the following equivalent version: Each integer n > 1 can be expressed as the sum of two terms of the current sequence.
This has been verified for all n = 2..2*10^8.

Examples

			a(1) = 1 with 1 = 0*(0+1)/2 + 5^0.
a(2) = 2 with 2 = 1*(1+1)/2 + 5^0.
a(3) = 4 with 4 = 2*(2+1)/2 + 5^0.
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
    tab={};Do[Do[If[TQ[m-5^k],tab=Append[tab,m];Goto[aa]],{k,0,Log[5,m]}];Label[aa],{m,1,176}];Print[tab]

A303399 Number of ordered pairs (a, b) with 0 <= a <= b such that n - 5^a - 5^b can be written as the sum of two triangular numbers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This is equivalent to the author's conjecture in A303389. It has been verified that a(n) > 0 for all n = 2..6*10^9.
Note that a nonnegative integer m is the sum of two triangular numbers if and only if 4*m + 1 can be written as the sum of two squares.

Examples

			a(6) = 2 with 6 - 5^0 - 5^0 = 1*(1+1)/2 + 2*(2+1)/2 and 6 - 5^0 - 5^1 = 0*(0+1)/2 + 0*(0+1)/2.
a(7) = 1 with 7 - 5^0 - 5^1 = 0*(0+1)/2 + 1*(1+1)/2.
a(25) = 1 with 25 - 5^1 - 5^1 = 0*(0+1)/2 + 5*(5+1)/2.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&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[4(n-5^j-5^k)+1],r=r+1],{j,0,Log[5,n/2]},{k,j,Log[5,n-5^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300441 Number of the integers 4^k*(4*u(m)^2+1) (k,m = 0,1,2,...) such that n^2 - 4^k*(4*u(m)^2+1) can be written as the sum of two squares, where u(0) = 0, u(1) = 1, and u(j+1) = 4*u(j) - u(j-1) for j = 1,2,3,....

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 05 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k*m (k = 0,1,2,... and m = 1, 7).
This curious conjecture indicates that any positive square can be written as (2^k)^2 + (2^(k+1)*u(m))^2 + x^2 + y^2 with k,m,x,y nonnegative integers. In the 2017 JNT paper, the author proved that each n = 1,2,3,... can be written as 4^k*(1+4*x^2+y^2)+z^2 with k,x,y,z nonnegative integers.
We have verified that a(n) > 0 for all n = 1..10^7.

Examples

			a(1) = 1 since 1^2 - 4^0*(4*u(0)^2+1) = 1 is 1^2 + 0^2.
a(5) = 3 since 5^2 - 4^0*(4*u(1)^2+1) = 20 = 4^2 + 2^2, 5^2 - 4^1*(4*u(1)^2+1) = 5 = 2^2 + 1^2, and 5^2 - 4^2*(4*u(0)^2+1) = 9 = 3^2 + 0^2.
a(7) = 1 since 7^2 - 4^1*(4*u(0)^2+1) = 45 = 6^2 + 3^2.
		

Crossrefs

Programs

  • Mathematica
    u[0]=0;
    u[1]=1;
    u[n_]:=u[n]=4u[n-1]-u[n-2];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[If[Mod[Part[Part[f[n],i],1]-3,4]==0&&Mod[Part[Part[f[n],i],2],2]==1,1,0],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
    tab={};Do[r=0;Do[m=0;Label[cc];If[4u[m]^2+1>n^2/4^k,Goto[bb]];If[QQ[n^2-4^k*(4u[m]^2+1)],r=r+1,m=m+1;Goto[cc]];
    Label[bb],{k,0,Log[2,n]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300708 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that x or y is a square and x - y is also a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 11 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 16^k*m with k = 0,1,2,... and m = 0, 8, 12, 23, 24, 44, 47, 56, 71, 79, 92, 95, 140, 168, 184, 248, 344, 428, 568, 632, 1144, 1544.
By the author's 2017 JNT paper, each nonnegative integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x - y (or x) is a square.
See also A281976, A300666, A300667 and A300712 for similar conjectures.
a(n) > 0 for all n = 0..10^8. - Zhi-Wei Sun, Oct 04 2020

Examples

			a(71) = 1 since 71 = 5^2 + 1^2 + 3^2 + 6^2 with 1 = 1^2 and 5 - 1 = 2^2.
a(95) = 1 since 95 = 2^2 + 1^2 + 3^2 + 9^2 with 1 = 1^2 and 2 - 1 = 1^2.
a(344) = 1 since 344 = 4^2 + 0^2 + 2^2 + 18^2 with 4 = 2^2 and 4 - 0 = 2^2.
a(428) = 1 since 428 = 13^2 + 9^2 + 3^2 + 13^2 with 9 = 3^2 and 13 - 9 = 2^2.
a(632) = 1 since 632 = 16^2 + 12^2 + 6^2 + 14^2 with 16 = 4^2 and 16 - 12 = 2^2.
a(1144) = 1 since 1144 = 20^2 + 16^2 + 2^2 + 22^2 with 16 = 4^2 and 20 - 16 = 2^2.
a(1544) = 1 since 1544 = 0^2 + 0^2 + 10^2 + 38^2 with 0 = 0^2 and 0 - 0 = 0^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[m^2+y]||SQ[y])&&SQ[n-(m^2+y)^2-y^2-z^2],r=r+1],{m,0,n^(1/4)},{y,0,Sqrt[(n-m^4)/2]},{z,0,Sqrt[Max[0,(n-(m^2+y)^2-y^2)/2]]}];tab=Append[tab,r],{n,0,80}];Print[tab]

A300666 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with nonnegative integers x,y,z,w and z <= w such that x or 2*y is a square and x + 3*y is also a square.

Original entry on oeis.org

1, 2, 3, 2, 2, 2, 2, 1, 1, 3, 4, 3, 1, 3, 3, 2, 2, 5, 6, 3, 3, 3, 3, 1, 1, 4, 7, 5, 2, 4, 3, 3, 3, 4, 7, 3, 5, 3, 5, 3, 2, 6, 6, 3, 1, 5, 3, 2, 2, 4, 7, 5, 6, 4, 4, 2, 1, 3, 5, 3, 2, 4, 4, 4, 2, 7, 10, 6, 3, 5, 4, 2, 2, 6, 9, 5, 3, 4, 5, 1, 2
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 10 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n = 0,1,2,....
Conjecture 2: Any nonnegative integer n not equal to 3 can be written as x^2 + y^2 + z^2 + w^2 with nonnegative integers x,y,z,w such that x or 2*y is a square and 3*x - y is also a square.
By the author's 2017 JNT paper, each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with integers x,y,z,w such that x (or 2*x) is a square.
See also A281976, A300667, A300708 and A300712 for similar conjectures.
a(n) > 0 for all n = 0..10^8. - Zhi-Wei Sun, Oct 04 2020

Examples

			a(8) = 1 since 8 = 0^2 + 0^2 + 2^2 + 2^2 with 0 = 0^2 and 0 + 3*0 = 0^2.
a(23) = 1 since 23 = 3^2 + 2^2 + 1^2 + 3^2 with 2*2 = 2^2 and 3 + 3*2 = 3^2.
a(56) = 1 since 56 = 4^2 + 0^2 + 2^2 + 6^2 with 4 = 2^2 and 4 + 3*0 = 2^2.
a(140) = 1 since 140 = 10^2 + 2^2 + 0^2 + 6^2 with 2*2 = 2^2 and 10 + 3*2 = 4^2.
a(472) = 1 since 472 = 0^2 + 12^2 + 2^2 + 18^2 with 0 = 0^2 and 0 + 3*12 = 6^2.
a(959) = 1 since 959 = 9^2 + 9^2 + 11^2 + 26^2 with 9 = 3^2 and 9 + 3*9 = 6^2.
a(1839) = 1 since 1839 = 1^2 + 5^2 + 7^2 + 42^2 with 1 = 1^2 and 1 + 3*5 = 4^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[m^2-3y]||SQ[2y])&&SQ[n-(m^2-3y)^2-y^2-z^2],r=r+1],{m,0,(10n)^(1/4)},{y,0,Min[m^2/3,Sqrt[n]]},{z,0,Sqrt[Max[0,(n-(m^2-3y)^2-y^2)/2]]}];tab=Append[tab,r],{n,0,80}];Print[tab]

A300667 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that 3*x or y is a square and x + 2*y is also a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 10 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 16^k*m with k = 0,1,2,... and m = 0, 3, 7, 11, 12, 15, 28, 39, 47, 60, 71, 92, 119, 172, 232, 253, 263, 316, 347, 515.
Conjecture 2: Each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 3*x or y is a square and 2*x - y is also a square.
By the author's 2017 JNT paper, any nonnegative integer can be written as the sum of a fourth power and three squares.
See also A281976, A300666, A300708 and A300712 for similar conjectures.
a(n) > 0 for all n = 0..10^8. Also, Conjecture 2 holds for all n = 0..10^8. In a 2018 paper Y.-C. Sun and Z.-W. Sun proved that any nonnegative integer can be written as x^2 + y^2 + z^2 + w^2 with x + 2*y a square, where x,y,z,w are nonnegative integers. - Zhi-Wei Sun, Oct 04 2020

Examples

			a(12) = 1 since 12 = 0^2 + 2^2 + 2^2 + 2^2 with 3*0 = 0^2 and 0 + 2*2 = 2^2.
a(39) = 1 since 39 = 2^2 + 1^2 + 3^2 + 5^2 with 1 = 1^2 and 2 + 2*1 = 2^2.
a(172) = 1 since 172 = 7^2 + 1^2 + 1^2 + 11^2 with 1 = 1^2 and 7 + 2*1 = 3^2.
a(232) = 1 since 232 = 0^2 + 0^2 + 6^2 + 14^2 with 0 = 0^2 and 0 + 2*0 = 0^2.
a(253) = 1 since 253 = 8^2 + 4^2 + 2^2 + 13^2 with 4 = 2^2 and 8 + 2*4 = 4^2.
a(263) = 1 since 263 = 3^2 + 3^2 + 7^2 + 14^2 with 3*3 = 3^2 and 3 + 2*3 = 3^2.
a(515) = 1 since 515 = 1^2 + 0^2 + 15^2 + 17^2 with 0 = 0^2 and 1 + 2*0 = 1^2.
		

References

  • Yu-Chen Sun and Zhi-Wei Sun, Some variants of Lagrange's four squares theorem, Acta Arith. 183(2018), 339-356.

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[3(m^2-2y)]||SQ[y])&&SQ[n-(m^2-2y)^2-y^2-z^2],r=r+1],{m,0,(5n)^(1/4)},{y,0,Min[m^2/2,Sqrt[n]]},{z,0,Sqrt[Max[0,(n-(m^2-2y)^2-y^2)/2]]}];tab=Append[tab,r],{n,0,80}];Print[tab]
  • PARI
    A300667(n)=sum(x=0,sqrtint(n),sum(y=0,sqrtint(n-x^2),if(issquare(x+2*y)&&(issquare(y)||issquare(3*x)),if(n>x^2+y^2,A000161(n-x^2-y^2),1)))) \\ M. F. Hasler, Mar 11 2018

A300712 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 3*x or y is a square and x - y is twice a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 11 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 16^k*m with k = 0,1,2,... and m = 0, 1, 3, 7, 21, 24, 46, 79, 88, 94, 142, 151, 184, 190, 193, 280, 286, 1336.
By the author's 2017 JNT paper, any nonnegative integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2*(x-y) (or x) is a square.
See also A281976, A300666, A300667 and A300708 for similar conjectures.
a(n) > 0 for all n = 0..10^8. - Zhi-Wei Sun, Oct 04 2020

Examples

			a(21) = 1 since 21 = 2^2 + 0^2 + 1^2 + 4^2 with 0 = 0^2 and 2 - 0 = 2*1^2.
a(79) = 1 since 79 = 3^2 + 3^2 + 5^2 + 6^2 with 3*3 = 3^2 and 3 - 3 = 2*0^2.
a(142) = 1 since 142 = 6^2 + 4^2 + 3^2 + 9^2 with 4 = 2^2 and 6 - 4 = 2*1^2.
a(190) = 1 since 190 = 3^2 + 1^2 + 6^2 + 12^2 with 1 = 1^2 and 3 - 1 = 2*1^2.
a(193) = 1 since 193 = 0^2 + 0^2 + 7^2 + 12^2 with 0 = 0^2 and 0 - 0 = 2*0^2.
a(280) = 1 since 280 = 12^2 + 10^2 + 0^2 + 6^2 with 3*12 = 6^2 and 12 - 10 = 2*1^2.
a(1336) = 1 since 1336 = 2^2 + 0^2 + 6^2 + 36^2 with 0 = 0^2 and 2 - 0 = 2*1^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[3(2m^2+y)]||SQ[y])&&SQ[n-(2m^2+y)^2-y^2-z^2],r=r+1],{m,0,(n/4)^(1/4)},{y,0,Sqrt[(n-4m^4)/2]},{z,0,Sqrt[Max[0,(n-(2m^2+y)^2-y^2)/2]]}];tab=Append[tab,r],{n,0,80}];Print[tab]
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