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.

Showing 1-10 of 10 results.

A300751 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x + 3*y + 5*z a positive square, where x,y,z,w are nonnegative integers such that 2*x or y or z is a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 11 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n = 1,2,3,..., and a(n) = 1 only for n = 16^k*m with k = 0,1,2,... and m = 0, 1, 4, 5, 6, 7, 8, 20, 24, 28, 31, 36, 43, 61, 71, 79, 100, 116, 157, 188, 200, 344, 351, 388, 632.
This is stronger than the author's 1-3-5 conjecture in A271518. See also A300752 for a similar conjecture stronger than the 1-3-5 conjecture.
a(n) > 0 for all n = 1..3*10^6. - Zhi-Wei Sun, Oct 06 2020

Examples

			a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 2*0 = 0^2 and 0 + 3*2 + 5*2 = 4^2.
a(61) = 1 since 61 = 0^2 + 0^2 + 5^2 + 6^2 with 0 = 0^2 and 0 + 3*0 + 5*5 = 5^2.
a(79) = 1 since 79 = 5^2 + 2^2 + 1^2 + 7^2 with 1 = 1^2 and 5 + 3*2 + 5*1 = 4^2.
a(188) = 1 since 188 = 7^2 + 9^2 + 3^2 + 7^2 with 9 = 3^2 and 7 + 3*9 + 5*3 = 7^2.
a(200) = 0 since 200 = 6^2 + 10^2 + 0^2 + 8^2 with 0 = 0^2 and 6 + 3*10 + 5*0 = 6^2.
a(632) = 1 since 632 = 6^2 + 16^2 + 18^2 + 4^2 with 16 = 4^2 and 6 + 3*16 + 5*18 = 12^2.
a(808) = 3 since 808 = 8^2 + 2^2 + 26^2 + 8^2 = 8^2 + 22^2 + 14^2 + 8^2 = 18^2 + 12^2 + 18^2 + 4^2  with 2*8 = 4^2, 2*18 = 6^2 and 8 + 3*2 + 5*26 = 8 + 3*22 + 5*14 = 18 + 3*12 + 5*18 = 12^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[2(m^2-3y-5z)]||SQ[y]||SQ[z])&&SQ[n-(m^2-3y-5z)^2-y^2-z^2],r=r+1],{m,1,(35n)^(1/4)},{y,0,Min[m^2/3,Sqrt[n]]},{z,0,Min[(m^2-3y)/5,Sqrt[n-y^2]]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300791 Number of ways to write n as x^2 + y^2 + z^2 + w^2, where w is a positive integer and x,y,z are nonnegative integers for which x or y or z is a square and (12*x)^2 + (15*y)^2 + (20*z)^2 is also a square.

Original entry on oeis.org

1, 3, 1, 1, 6, 1, 1, 3, 2, 8, 2, 2, 7, 2, 2, 1, 8, 6, 2, 8, 1, 3, 1, 1, 9, 8, 4, 3, 7, 3, 3, 3, 6, 9, 4, 4, 7, 5, 1, 8, 8, 4, 3, 3, 11, 2, 1, 1, 4, 11, 3, 8, 8, 4, 4, 2, 3, 8, 4, 2, 8, 3, 4, 1, 15, 9, 3, 9, 3, 5, 2, 6, 10, 11, 5, 3, 5, 6, 2, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 12 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 = 1, 3, 4, 6, 7, 21, 23, 24, 39, 47, 86, 95, 344, 651, 764.
By the author's 2017 JNT paper, each n = 0,1,2,... can be written as the sum of a fourth power and three squares.
See also A300792 for two similar conjectures.

Examples

			a(6) = 1 since 6 = 0^2 + 1^2 + 1^2 + 2^2 with 0 = 0^2 and (12*0)^2 + (15*1)^2 + (20*1)^2 = 25^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1 = 1^2 and (12*1)^2 + (15*2)^2 + (20*1)^2 = 38^2.
a(21) = 1 since 21 = 4^2 + 0^2 + 1^2 + 2^2 with 4 = 2^2 and (12*4)^2 + (15*0)^2 + (20*1)^2 = 52^2.
a(39) = 1 since 39 = 5^2 + 2^2 + 1^2 + 3^2 with 1 = 1^2 and (12*5)^2 + (15*2)^2 + (20*1)^2 = 70^2.
a(344) = 1 since 344 = 0^2 + 10^2 + 10^2 + 12^2 with 0 = 0^2 and (12*0)^2 + (15*10)^2 + (20*10)^2 = 250^2.
a(764) = 1 since 764 = 7^2 + 3^2 + 25^2 + 9^2 with 25 = 5^2 and (12*7)^2 + (15*3)^2 + (20*25)^2 = 509^2.
a(8312) = 2 since 8312 = 42^2 + 36^2 + 34^2 + 64^2 with 36 = 6^2 and (12*42)^2 + (15*36)^2 + (20*34)^2 = 1004^2, and 8312 = 66^2 + 16^2 + 44^2 + 42^2 with 16 = 4^2 and (12*66)^2 + (15*16)^2 + (20*44)^2 = 1208^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[x]||SQ[y]||SQ[z])&&SQ[(12x)^2+(15y)^2+(20z)^2]&&SQ[n-x^2-y^2-z^2],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300792 Number of ways to write n as x^2 + y^2 + z^2 + w^2, where w is a positive integer and x,y,z are nonnegative integers such that x or y or z is a square and 9*x^2 + 16*y^2 + 24*z^2 is also a square.

Original entry on oeis.org

1, 2, 1, 2, 4, 1, 2, 2, 2, 6, 2, 3, 5, 1, 4, 1, 5, 7, 4, 5, 1, 5, 2, 1, 9, 6, 5, 3, 4, 7, 2, 2, 6, 7, 3, 5, 7, 4, 4, 6, 6, 4, 5, 3, 9, 4, 2, 1, 4, 11, 5, 9, 5, 6, 4, 1, 9, 7, 3, 6, 5, 4, 4, 2, 14, 4, 6, 5, 2, 8, 2, 7, 9, 5, 5, 4, 3, 8, 1, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 12 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 = 1, 3, 6, 14, 21, 24, 56, 79, 119, 143, 248, 301, 383, 591, 728, 959, 1223, 1751, 2311, 6119.
Conjecture 2: Any positive integer n can be written as x^2 + y^2 + z^2 + w^2, where w is a positive integer and x,y,z are nonnegative integers such that x or y or z is a square and 36*x^2 + 40*y^2 + 45*z^2 is also a square.
See also A300791 for a similar conjecture.

Examples

			a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 1 = 1^2 and 9*1^2 + 16*1^2 + 24*0^2 = 5^2.
a(14) = 1 since 14 = 1^2 + 0^2 + 3^2 + 2^2 with 1 = 1^2 and 9*1^2 + 16*0^2 + 24*3^2 = 15^2.
a(728) = 1 since 728 = 10^2 + 0^2 + 12^2 + 22^2 with 0 = 0^2 and 9*10^2 + 16*0^2 + 24*12^2 = 66^2.
a(959) = 1 since 959 = 25^2 + 18^2 + 3^2 + 1^2 with 25 = 5^2 and 9*25^2 + 16*18^2 + 24*3^2 = 105^2.
a(1751) = 1 since 1751 = 19^2 + 25^2 + 18^2 + 21^2 with 25 = 5^2 and 9*19^2 + 16*25^2 + 24*18^2 = 145^2.
a(2311) = 1 since 2311 = 1^2 + 41^2 + 23^2 + 10^2 with 1 = 1^2 and 9*1^2 + 16*41^2 + 24*23^2 = 199^2.
a(6119) = 1 since 6119 = 1^2 + 5^2 + 3^2 + 78^2 with 1 = 1^2 and 9*1^2 + 16*5^2 + 24*3^2 = 25^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[x]||SQ[y]||SQ[z])&&SQ[9x^2+16y^2+24z^2]&&SQ[n-x^2-y^2-z^2],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300844 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x or 2*y or z is a square and (12*x)^2 + (21*y)^2 + (28*z)^2 is also a square.

Original entry on oeis.org

1, 4, 4, 2, 5, 6, 2, 2, 4, 5, 7, 1, 3, 7, 2, 3, 5, 7, 7, 2, 6, 1, 2, 2, 2, 11, 7, 3, 3, 8, 5, 1, 4, 5, 9, 4, 6, 8, 6, 4, 7, 9, 3, 3, 2, 9, 2, 1, 3, 6, 16, 5, 9, 7, 6, 5, 1, 5, 9, 4, 4, 7, 5, 5, 5, 17, 6, 4, 7, 3, 6, 3, 6, 11, 11, 4, 3, 1, 8, 2, 6
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 13 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n = 0,1,2,....
Conjecture 2: Any positive integer can be written as x^2 + y^2 + z^2 + w^2 with w a positive integer and x,y,z nonnegative integers such that x or y or z is a square and 144*x^2 + 505*y^2 + 720*z^2 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 x,y,z,w integers such that x (or 2*x) is a square.
In 2016, the author conjectured in A271510 that each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y >= z such that (3*x)^2 + (4*y)^2 + (12*z)^2 is a square.
See also A300791 and A300792 for similar conjectures.

Examples

			a(11) = 1 since 11 = 0^2 + 1^2 + 1^2 + 3^2 with 0 = 0^2 and (12*0)^2 + (21*1)^2 + (28*1)^2 = 35^2.
a(56) = 1 since 56 = 4^2 + 6^2 + 2^2 + 0^2 with 4 = 2^2 and (12*4)^2 + (21*6)^2 + (28*2)^2 = 146^2.
a(77) = 1 since 77 = 4^2 + 0^2 + 5^2 + 6^2 with 4 = 2^2 and (12*4)^2 + (21*0)^2 + (28*5)^2 = 148^2.
a(184) = 1 since 184 = 12^2 + 2^2 + 0^2 + 6^2 with 0 = 0^2 and (12*12)^2 + (21*2)^2 + (28*0)^2 = 150^2.
a(599) = 1 since 599 = 21^2 + 11^2 + 1^2 + 6^2 with 1 = 1^2 and (12*21)^2 + (21*11)^2 + (28*1)^2 = 343^2.
a(7836) = 1 since 7836 = 38^2 + 18^2 + 68^2 + 38^2 with 2*18 = 6^2 and (12*38)^2 + (21*18)^2 + (28*68)^2 = 1994^2.
a(15096) = 1 since 15096 = 16^2 + 6^2 + 52^2 + 110^2 with 16 = 4^2 and (12*16)^2 + (21*6)^2 + (28*52)^2 = 1474^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[x]||SQ[2y]||SQ[z])&&SQ[(12x)^2+(21y)^2+(28z)^2]&&SQ[n-x^2-y^2-z^2],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];tab=Append[tab,r],{n,0,80}]

A300908 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square , where w is a positive integer and x,y,z are nonnegative integers such that z or 2*z or 3*z is a square.

Original entry on oeis.org

1, 3, 1, 2, 6, 1, 2, 3, 2, 7, 2, 2, 7, 1, 5, 2, 7, 9, 3, 6, 2, 3, 4, 1, 9, 7, 3, 4, 5, 7, 2, 3, 5, 6, 3, 4, 7, 3, 8, 6, 7, 6, 4, 3, 7, 3, 2, 2, 4, 13, 5, 8, 6, 5, 3, 1, 8, 8, 3, 6, 5, 1, 11, 2, 16, 5, 4, 8, 1, 8, 2, 7, 11, 7, 5, 4, 2, 13, 2, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 15 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0. Moreover, any positive integer can be written as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square, where w is a positive integer and x,y,z are nonnegative integers for which one of z, z/2, z/3 is a square and z/2 (or z/3) is a power of 4 (including 1).
Conjecture 2: Each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square, where x,y,z,w are nonnegative integers for which x or z is a square or y = 2^k for some k = 0,1,2,....
Conjecture 3. Let a,b,c be positive integers with a <= b <= c and gcd(a,b,c) = 1. If each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that (a*x)^2 + (b*y)^2 + (c*z)^2 is a square, then (a,b,c) must be among the three triples (3,4,12), (12,15,20) and (12,21,28).
We have verified Conjectures 1 and 2 for n up to 5*10^5 and 10^6 respectively.
See also A300791 and A300844 for similar conjectures.

Examples

			a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 0 = 0^2 and (3*1)^2 + (4*1)^2 + (12*0)^2 = 5^2.
a(14) = 1 since 14 = 3^2 + 0^2 + 1^2 + 2^2 with 1 = 1^2 and (3*3)^2 + (4*0)^2 + (12*1)^2 = 15^2.
a(69) = 1 since 69 = 0^2 + 8^2 + 2^2 + 1^2 with 2*2 = 2^2 and (3*0)^2 + (4*8)^2 + (12*2)^2 = 40^2.
a(671) = 1 since 671 = 18^2 + 17^2 + 3^2 + 7^2 with 3*3 = 3^2 and (3*18)^2 + (4*17)^2 + (12*3)^2 = 94^2.
a(1175) = 1 since 1175 = 30^2 + 5^2 + 9^2 + 13^2 with 9 = 3^2 and (3*30)^2 + (4*5)^2 + (12*9)^2 = 142^2.
a(12151) = 1 since 12151 = 50^2 + 71^2 + 49^2 + 47^2 with 49 = 7^2 and (3*50)^2 + (4*71)^2 + (12*49)^2 = 670^2.
a(16204) = 1 since 16204 = 90^2 + 90^2 + 0^2 + 2^2 with 0 = 0^2 and (3*90)^2 + (4*90)^2 + (12*0)^2 = 450^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[z]||SQ[2z]||SQ[3z], Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(3x)^2+(4y)^2+(12z)^2],r=r+1],{x,0,Sqrt[n-1-z^2]},{y,0,Sqrt[n-1-x^2-z^2]}]],{z,0,Sqrt[n-1]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301303 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x^2 + 23*y^2 = 2^k*m^3 for some k = 0,1,2 and m = 1,2,3,....

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 17 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k (k = 0,1,2,...), 3, 7, 115, 151, 219, 267, 1151, 1367.
We have verified a(n) > 0 for all n = 1..10^8.
See also A301304 and A301314 for similar conjectures.

Examples

			a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 + 23*0 = 1^3.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 + 23*0 = 2^2*1^3.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 2^2 + 23*1^2 = 3^3.
a(48) = 2 since 48 = 4^2 + 0^2 + 4^2 + 4^2 = 6^2 + 2^2 + 2^2 + 2^2 with 4^2 + 23*0^2 = 2*2^3 and 6^2 + 23*2^2 = 2*4^3.
a(115) = 1 since 115 = 3^2 + 3^2 + 4^2 + 9^2 with 3^2 + 23*3^2 = 6^3.
a(267) = 1 since 267 = 3^2 + 1^2 + 1^2 + 16^2 with 3^2 + 23*1^2 = 2^2*2^3.
a(1151) = 1 since 1151 = 7^2 + 3^2 + 2^2 + 33^2 with 7^2 + 23*3^2 = 2^2*4^3.
a(1367) = 1 since 1367 = 17^2 + 5^2 + 18^2 + 27^2 with 17^2 + 23*5^2 = 2^2*6^3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
    QQ[n_]:=n>0&&(CQ[n]||CQ[n/2]||CQ[n/4]);
    tab={};Do[r=0;Do[If[QQ[x^2+23y^2],Do[If[SQ[n-x^2-y^2-z^2],r=r+1],{z,0,Sqrt[(n-x^2-y^2)/2]}]],{y,0,Sqrt[n/2]},{x,y,Sqrt[n-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301304 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x^2 + 7*y^2 = 2^k*m for some k = 0,1,2 and m = 1,2,3,....

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 1, 2, 3, 4, 3, 4, 2, 3, 2, 2, 4, 5, 3, 4, 4, 2, 1, 3, 2, 6, 5, 4, 3, 4, 2, 2, 6, 5, 4, 6, 3, 3, 3, 4, 6, 8, 2, 5, 5, 3, 2, 4, 4, 5, 6, 5, 5, 4, 3, 3, 6, 5, 2, 6, 3, 4, 3, 2, 6, 10, 3, 5, 8, 1, 2, 5, 5, 6, 5, 6, 5, 3, 1, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 17 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
We have verified this for n up to 10^8.
See also A301303 and A301314 for similar conjectures.

Examples

			a(79) = 1 since 79 = 5^2 + 1^2 + 2^2 + 7^2 with 5^2 + 7*1^2 = 2^2*2^3.
a(323) = 1 since 323 = 3^2 + 1^2 + 12^2 + 13^2 with 3^2 + 7*1^2 = 2*2^3.
a(646) = 1 since 646 = 22^2 + 11^2 + 4^2 + 5^2 with 22^2 + 7*11^2 = 11^3.
a(815) = 1 since 815 = 9^2 + 5^2 + 15^2 + 22^2 with 9^2 + 7*5^2 = 2^2*4^3.
a(1111) = 1 since 1111 = 1^2 + 1^2 + 22^2 + 25^2 with 1^2 + 7*1^2 = 2^3.
a(2822) = 1 since 2822 = 2^2 + 0^2 + 3^2 + 53^2 with 2^2 + 7*0^2 = 2^2*1^3.
		

Crossrefs

Programs

  • Mathematica
      SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
    QQ[n_]:=n>0&&(CQ[n]||CQ[n/2]||CQ[n/4]);
    tab={};Do[r=0;Do[If[QQ[x^2+7y^2],Do[If[SQ[n-x^2-y^2-z^2],r=r+1],{z,0,Sqrt[(n-x^2-y^2)/2]}]],{y,0,Sqrt[n/2]},{x,y,Sqrt[n-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301314 Number of ways to write n as x^2 + y^2 + z^2 + w^2, where w is a positive integer and x,y,z are nonnegative integers such that x + 3*y + 9*z = 2^k*m^3 for some k,m = 0,1,2,....

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 18 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: Any positive integer can be written as x^2 + y^2 + z^2 + w^2, where x is a positive integer and y,z,w are nonnegative integers such that 2*x + 7*y = 2^k*m^3 for some k = 0,1,2 and m = 1,2,3,....
We have verified a(n) > 0 for all n = 1..10^7.
See also A301303 and A301304 for similar conjectures.

Examples

			a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1 + 3*2 + 9*1 = 2*2^3.
a(19) = 1 since 19 = 4^2 + 1^2 + 1^2 + 1^2 with 4 + 3*1 + 9*1 = 2*2^3.
a(46) = 1 since 46 = 0^2 + 6^2 + 1^2 + 3^2 with 0 + 3*6 + 9*1 = 3^3.
a(79) = 1 since 79 = 2^2 + 7^2 + 1^2 + 5^2 with 2 + 3*7 + 9*1 = 2^2*2^3.
a(125) = 1 since 125 = 2^2 + 0^2 + 0^2 + 11^2 with 2 + 3*0 + 9*0 = 2*1^3.
a(736) = 1 since 736 = 0^2 + 24^2 + 4^2 + 12^2 with 0 + 3*24 + 9*4 = 2^2*3^3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
    QQ[n_]:=CQ[n]||CQ[n/2]||CQ[n/4];
    tab={};Do[r=0;Do[If[QQ[x+3y+9z]&&SQ[n-x^2-y^2-z^2],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300139 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with 4*x - 3*y a square, where x,y,z,w are nonnegative integers with z <= w such that 10*x or y is a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 12 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, 7, 8, 13, 22, 23, 24, 31, 44, 56, 71, 79, 88, 109, 120, 152, 184, 472, 1912, 6008, 9080.
Conjecture 2: Each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with 3*x - y twice a square, where x,y,z,w are nonnegative integers such that 5*x or y is 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, A300667, 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. - Zhi-Wei Sun, Oct 05 2020

Examples

			a(22) = 1 since 22 = 1^2 + 1^2 + 2^2 + 4^2 with 1 = 1^2 and 4*1 - 3*1 = 1^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 1 = 1^2 and 4*3 - 3*1 = 3^2.
a(109) = 1 since 109 = 0^2 + 0^2 + 3^2 + 10^2 with 0 = 0^2 and 4*0 - 3*0 = 0^2.
a(184) = 1 since 184 = 10^2 + 8^2 + 2^2 + 4^2 with 10*10 = 10^2 and 4*10 - 3*8 = 4^2.
a(6008) = 1 since 6008 = 12^2 + 16^2 + 42^2 + 62^2 with 16 = 4^2 and 4*12 - 3*16 = 0^2.
a(9080) = 1 since 9080 = 10^2 + 12^2 + 0^2 + 94^2 with 10*10 = 10^2 and 4*10 - 3*12 = 2^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[Mod[m^2+3y,4]==0&&(SQ[10(m^2+3y)/4]||SQ[y]), Do[If[SQ[n-((m^2+3y)/4)^2-y^2-z^2],r=r+1],{z,0,Sqrt[Max[0,(n-((m^2+3y)/4)^2-y^2)/2]]}]],{m,0,2n^(1/4)},{y,0,4/5*Sqrt[n-m^4/16]}];tab=Append[tab,r],{n,0,80}];Print[tab]

A301375 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z nonnegative integers and w a positive integer such that x*(y+3*z) is a cube or half a cube and y <= z <= w if x = 0.

Original entry on oeis.org

1, 2, 2, 2, 3, 2, 2, 2, 3, 5, 2, 3, 4, 1, 1, 1, 5, 7, 3, 3, 5, 3, 1, 4, 6, 7, 4, 3, 6, 1, 4, 2, 6, 7, 1, 5, 4, 4, 2, 5, 5, 4, 5, 2, 8, 4, 2, 1, 4, 7, 5, 7, 6, 4, 3, 3, 4, 7, 2, 1, 5, 3, 2, 2, 7, 10, 6, 4, 6, 3, 4, 5, 9, 8, 5, 4, 2, 6, 2, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 19 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
We have verified this for all n = 1..3*10^6.

Examples

			a(14) = 1 since 14 = 0^2 + 1^2 + 2^2 + 3^2 with 0*(1+3*2) = 0^3.
a(15) = 1 since 15 = 2^2 + 1^2 + 1^2 + 3^2 with 2*(1+3*1) = 2^3.
a(16) = 1 since 16 = 0^2 + 0^2 + 0^2 + 4^2 with 0*(0+3*0) = 0^3.
a(60) = 1 since 60 = 4^2 + 2^2 + 2^2 + 6^2 with 4*(2+3*2) = 4^3/2.
a(92) = 1 since 92 = 6^2 + 6^2 + 4^2 + 2^2 with 6*(6+3*4) = 6^3/2.
a(240) = 1 since 240 = 2^2 + 14^2 + 6^2 + 2^2 with 2*(14+3*6) = 4^3.
a(807) = 1 since 807 = 1^2 + 21^2 + 2^2 + 19^2 with 1*(21+3*2) = 3^3.
		

Crossrefs

Programs

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