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 1-8 of 8 results.

A349943 Number of ways to write n as a^4 + (b^4 + c^2 + d^2)/9, where a,b,c,d are nonnegative integers with c <= d.

Original entry on oeis.org

1, 3, 5, 4, 3, 4, 3, 1, 1, 6, 9, 6, 2, 4, 7, 3, 3, 7, 9, 7, 7, 5, 4, 2, 3, 10, 11, 8, 2, 10, 10, 1, 5, 9, 15, 14, 6, 5, 5, 1, 4, 9, 12, 8, 2, 11, 7, 1, 4, 11, 21, 8, 6, 9, 8, 3, 3, 7, 9, 9, 4, 11, 9, 2, 3, 13, 14, 7, 7, 10, 10, 4, 3, 10, 18, 16, 3, 10, 7, 1, 4, 10, 15, 12, 11, 12, 11, 3, 3, 16, 29, 17, 5, 6, 14, 10, 3, 10, 18, 15, 14
Offset: 0

Views

Author

Zhi-Wei Sun, Dec 05 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 7, 8, 31, 39, 47, 79, 519).
This implies that each nonnegative rational number can be written as x^4 + 9*y^4 + z^2 + w^2 with x,y,z,w rational numbers.
Conjecture 2: Each n = 0,1,2,... can be written as a^4 + (4*b^4 + c^2 + d^2)/81 with a,b,c,d nonnegative integers.
This implies that each nonnegative rational number can be written as x^4 + 4*y^4 + z^2 + w^2 with x,y,z,w rational numbers.
We have verified Conjectures 1 and 2 for n <= 10^5.
It seems that each n = 0,1,2,... can be written as a^4 + (b^4 + c^2 + d^2)/m^2 with a,b,c,d nonnegative integers, provided that m is among the odd numbers 7, 11, 15, 17, 19, 21, ....
See also A349942 for a similar conjecture.

Examples

			a(7) = 1 with 7 = 1^4 + (1^4 + 2^2 + 7^2)/9.
a(8) = 1 with 8 = 0^4 + (0^4 + 6^2 + 6^2)/9.
a(31) = 1 with 31 = 1^4 + (1^4 + 10^2 + 13^2)/9.
a(39) = 1 with 39 = 1^4 + (3^4 + 6^2 + 15^2)/9.
a(47) = 1 with 47 = 1^4 + (3^4 + 3^2 + 18^2)/9.
a(79) = 1 with 79 = 1^4 + (1^4 + 5^2 + 26^2)/9.
a(519) = 1 with 519 = 1^4 + (3^4 + 15^2 + 66^2)/9.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[9(n-x^4)-y^4-z^2],r=r+1],{x,0,n^(1/4)},{y,0,(9(n-x^4))^(1/4)},{z,0,Sqrt[(9(n-x^4)-y^4)/2]}];tab=Append[tab,r],{n,0,100}];Print[tab]

A349957 Number of ways to write n as x^4 + y^2 + (z^2 + 11*16^w)/60, where x,y,z are nonnegative integers, and w is 0 or 1.

Original entry on oeis.org

1, 2, 3, 5, 6, 4, 3, 3, 3, 4, 3, 4, 5, 3, 2, 2, 4, 4, 5, 9, 9, 3, 3, 4, 6, 5, 5, 9, 7, 4, 4, 6, 5, 2, 4, 8, 7, 3, 5, 7, 7, 4, 4, 4, 4, 4, 6, 9, 4, 3, 3, 9, 9, 4, 4, 5, 7, 2, 4, 4, 4, 2, 7, 7, 4, 3, 5, 12, 7, 3, 1, 6, 6, 4, 5, 8, 3, 1, 4, 5, 6, 3, 8, 14, 13, 6, 5, 5, 6, 6, 9, 8, 6, 3, 4, 8, 6, 6, 5, 12
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 06 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0.
This has been verified for n <= 10^6. It seems that a(n) = 1 only for n = 1, 71, 78, 247, 542, 1258, 1907, 5225, 19798.
Conjecture 2: (i) If a is 1 or 3, then each n = 0,1,2,... can be written as a*x^8 + y^2 + (7*z^4 + w^2)/64 with x,y,z,w nonnegative integers.
(ii) If a is among 1,2,5, then each n = 0,1,2,... can be written as a*x^8 + y^2 + (11*z^4 + w^2)/60 with x,y,z,w nonnegative integers.
Conjecture 3: If (a,b, m) is among the triples (1,7,8), (1,11,12), (2,7,8), (3,11,12), (5,7,8), then each n = 0,1,2,... can be written as a*x^4 + y^2 + (b*z^6 + w^2)/m with x,y,z,w nonnegative integers.
Conjecture 4: (i) If F(x,y,z,w) is x^6 + y^2 + (5*z^4 + 3*w^2)/16 or 3x^6 + 2*y^2 + (11*z^4 + w^2)/60, then each n = 0,1,2,... can be written as F(x,y,z,w) with x,y,z,w nonnegative integers.
(ii) If (a,b,m) is among the triples (1,7,64), (1,11,12), (1,11,60), (2,1,25), (2,1,65), (2,11,4), (2,11,20), (2,11,60), (4,2,9), (4,7,64), (5,11,60), (6,1,10), then each n = 0,1,2,... can be written as a*x^6 + y^2 + (b*z^4 + w^2)/m with x,y,z,w nonnegative integers.

Examples

			a(1) = 1 with 1 = 0^4 + 0^2 + (7^2 + 11*16^0)/60.
a(16) = 2 with 16 = 0^4 + 0^2 + (28^2 + 11*16)/60 = 1^4 + 2^2 + (22^2 + 11*16)/60.
a(71) = 1 with 71 = 0^4 + 2^2 + (62^2 + 11*16)/60.
a(78) = 1 with 78 = 2^4 + 5^2 + (47^2 + 11*16^0)/60.
a(247) = 1 with 247 = 3^4 + 3^2 + (97^2 + 11*16^0)/60.
a(542) = 1 with 542 = 3^4 + 21^2 + (32^2 + 11*16)/60.
a(1258) = 1 with 1258 = 2^4 + 15^2 + (247^2 + 11*16^0)/60.
a(1907) = 1 with 1907 = 0^4 + 0^2 + (338^2 + 11*16)/60.
a(5225) = 1 with 5225 = 5^4 + 58^2 + (272^2 + 11*16)/60.
a(19798) = 1 with 19798 = 1^4 + 137^2 + (248^2 + 11*16)/60.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[60(n-x^4-y^2)-11*16^z],r=r+1],{x,0,(n-1)^(1/4)},{y,0,Sqrt[n-1-x^4]},{z,0,1}];tab=Append[tab,r],{n,1,100}];Print[tab]

A349956 Number of ways to write n as x^2 + 2*y^2 + (z^4 + 4*w^4)/5 with x,y,z,w nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 06 2021

Keywords

Comments

a(n) > 0 for all n <= 10^5.
Conjecture: If m is 5 or 65 or 85, then each n = 0,1,2,... can be written as x^2 + 2*y^2 + (z^4 + 4*w^4)/m with x,y,z,w nonnegative integers.
It seems that there are infinitely many positive squarefree numbers m (including 3, 5, 15, 23, 31, 33, 37, 55, 59, 67, 69, 71, 89, 93, 97, 111, 113, 115) such that every n = 0,1,2,... can be written as x^4 + 2*y^4 + (z^2 + 11*w^2)/m with x,y,z,w nonnegative integers.

Examples

			a(11) = 1 with 11 = 3^2 + 2*1^2 + (0^4 + 4*0^4)/5.
a(14) = 1 with 14 = 1^2 + 2*0^2 + (1^4 + 4*2^4)/5.
a(78) = 1 with 78 = 7^2 + 2*0^2 + (3^4 + 4*2^4)/5.
a(155) = 1 with 155 = 11^2 + 2*3^2 + (2^4 + 4*2^4)/5.
a(174) = 1 with 174 = 7^2 + 2*0^2 + (5^4 + 4*0^4)/5.
		

Crossrefs

Programs

  • Mathematica
    QQ[n_]:=QQ[n]=IntegerQ[n^(1/4)];
    tab={};Do[r=0;Do[If[QQ[5(n-2x^2-y^2)-4z^4],r=r+1],{x,0,Sqrt[n/2]},{y,0,Sqrt[n-2x^2]},{z,0,(5(n-2x^2-y^2)/4)^(1/4)}];tab=Append[tab,r],{n,0,100}];Print[tab]

A350860 Number of ways to write n as w^4 + (x^4 + y^2 + z^2)/81, where w,x,y,z are nonnegative integers with y <= z.

Original entry on oeis.org

1, 6, 8, 5, 4, 7, 8, 3, 2, 5, 10, 10, 3, 4, 6, 2, 6, 14, 12, 10, 8, 19, 18, 4, 4, 11, 23, 15, 2, 7, 8, 3, 8, 13, 19, 18, 15, 21, 13, 4, 4, 17, 24, 10, 3, 6, 13, 7, 5, 10, 21, 23, 14, 15, 13, 3, 5, 17, 15, 12, 4, 13, 21, 4, 4, 13, 36, 25, 14, 20, 14, 3, 6, 13, 19, 18, 5, 14, 11, 3, 7, 32, 45, 19, 17, 22, 21, 8, 4, 17, 31
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 19 2022

Keywords

Comments

Conjecture 1: a(n) > 0 for any nonnegative integer n. Also, every n = 0,1,2,... can be written as 4*w^4 + (x^4 + y^2 + z^2)/81 with w,x,y,z integers.
This implies that each nonnegative rational number can be written as w^4 + x^4 + y^2 + z^2 (or 4*w^4 + x^4 + y^2 + z^2) with w,x,y,z rational numbers.
Conjecture 2: For any positive integer c, there is a positive integer m such that every n = 0,1,2,... can be written as w^4 + (c^2*x^4 + y^2 + z^2)/m^4 with w,x,y,z integers.
This implies that for any positive integer c each nonnegative rational number can be written as w^4 + c^2*x^4 + y^2 + z^2 with w,x,y,z rational numbers.
Conjecture 3: For any positive odd number d, there is a positive integer m such that every n = 0,1,2,... can be written as 4*w^4 + (d^2*x^4 + y^2 + z^2)/m^4 with w,x,y,z integers.
This implies that for any positive odd number d each nonnegative rational number can be written as 4*w^4 + d^2*x^4 + y^2 + z^2 with w,x,y,z rational numbers.

Examples

			a(8) = 2 with 8 = 0^4 + (0^4 + 18^2 + 18^2)/81 = 0^4 + (4^4 + 14^2 + 14^2)/81.
a(15) = 2 with 15 = 1^4 + (3^4 + 18^2 + 27^2)/81 = 1^4 + (5^4 + 5^2 + 22^2)/81.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[81(n-w^4)-x^4-y^2],r=r+1],{w,0,n^(1/4)},{x,0,3(n-w^4)^(1/4)},{y,0,Sqrt[(81(n-w^4)-x^4)/2]}];tab=Append[tab,r],{n,0,90}];Print[tab]

A347824 Number of ways to write n as x^4 + y^4 + (z^2 + 23*w^2)/16, where x,y,z,w are nonnegative integers with x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 23 2022

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,....
This has been verified for n up to 2*10^6. See also A347827 for a further refinement.
It seems that a(n) = 1 only for n = 0, 7, 15, 21, 22, 67, 77, 137, 252, 291, 437, 471, 477, 597, 1161, 4692, 7107.
For m = 32, 48, we also conjecture that every n = 0,1,2,... can be written as x^4 + y^4 + (z^2 + 23*w^2)/m, where x,y,z,w are nonnegative integers.

Examples

			a(7) = 1 with 7 = 0^4 + 1^4 + (2^2 + 23*2^2)/16.
a(15) = 1 with 15 = 1^4 + 1^4 + (1^2 + 23*3^2)/16.
a(67) = 1 with 67 = 1^4 + 2^4 + (15^2 + 23*5^2)/16.
a(477) = 1 with 477 = 0^4 + 2^4 + (27^2 + 23*17^2)/16.
a(597) = 1 with 597 = 2^4 + 4^4 + (5^2 + 23*15^2)/16.
a(1161) = 1 with 1161 = 2^4 + 2^4 + (89^2 + 23*21^2)/16.
a(4692) = 1 with 4692 = 2^4 + 5^4 + (248^2 + 23*12^2)/16.
a(7107) = 1 with 7107 = 1^4 + 5^4 + (239^2 + 23*45^2)/16.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[16(n-x^4-y^4)-23z^2],r=r+1],{x,0,(n/2)^(1/4)},{y,x,(n-x^4)^(1/4)},{z,0,Sqrt[16(n-x^4-y^4)/23]}];tab=Append[tab,r],{n,0,100}];Print[tab]

A349945 Number of ways to write n as a^4 + b^2 + (c^4 + d^2)/5 with a,b,c,d nonnegative integers.

Original entry on oeis.org

1, 3, 4, 3, 3, 6, 7, 3, 1, 4, 5, 4, 2, 3, 8, 5, 3, 9, 10, 6, 7, 11, 10, 3, 2, 6, 8, 9, 3, 9, 16, 5, 4, 11, 9, 7, 9, 9, 12, 7, 2, 8, 11, 7, 2, 11, 14, 4, 3, 10, 10, 9, 8, 9, 21, 9, 3, 9, 5, 7, 4, 10, 17, 8, 3, 15, 15, 9, 9, 16, 20, 5, 3, 5, 7, 11, 3, 11, 18, 4, 6, 22, 18, 11, 14, 15, 19, 10, 2, 9, 16, 10, 3, 9, 16, 11, 7, 19, 16, 13, 12
Offset: 0

Views

Author

Zhi-Wei Sun, Dec 06 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 2^(4*k+3) (k = 0,1,2,...).
This has been verified for all n <= 10^5.
Conjecture 2: Each n = 0,1,2,... can be written as a*x^4 + b*y^2 + (c*z^4 + w^2)/5 with x,y,z,w nonnegative integers, provided that (a,b,c) is among the four triples (1,2,4), (2,1,1), (6,1,1), (6,1,6).
See also A349942 for a similar conjecture.
Via a computer search, we have found many tuples (a,b,c,d,m) of positive integers (such as (1,1,4,2,3), (4,1,1,2,3) and (1,1,19,1,4900)) for which we guess that each n = 0,1,2,... can be written as a*x^4 + b*y^2 + (c*z^4 + d*w^2)/m with x,y,z,w nonnegative integers.

Examples

			a(8) = 1 with 8 = 0^4 + 2^2 + (2^4 + 2^2)/5.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[5(n-x^4-y^2)-z^4],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,(5(n-x^4-y^2))^(1/4)}];tab=Append[tab,r],{n,0,100}];Print[tab]

A349992 Number of ways to write n as x^4 + y^2 + (z^2 + 2*4^w)/3, where x, y, z are nonnegative integers, and w is 0 or 1.

Original entry on oeis.org

1, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 2, 2, 2, 4, 8, 7, 7, 6, 5, 6, 6, 6, 8, 7, 8, 6, 1, 4, 2, 6, 8, 6, 7, 5, 7, 6, 6, 6, 7, 7, 8, 7, 3, 5, 3, 4, 6, 6, 6, 7, 5, 3, 5, 4, 9, 8, 9, 8, 2, 4, 1, 2, 9, 8, 10, 8, 4, 6, 4, 9, 6, 6, 6, 4, 2, 2, 1, 2, 10, 10, 13, 8, 9, 7, 9, 9, 7, 10, 6, 10, 4, 3, 4, 3, 11, 10, 9
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 08 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 30, 64, 80, 302, 350, 472, 480, 847, 3497, 13582, 25630, 38064.
This has been verified for n up to 10^6.
Conjecture 2: If (a,b,c,m) is one of the ordered tuples (1,1,11,12), (1,1,11,60), (1,1,14,15), (1,1,23,24), (1,1,23,32), (1,1,23,48), (1,2,23,96), (2,1,11,60), (2,1,23,24), (2,1,23,48), (4,1,23,48), then each n = 1 2,3,... can be written as a*x^4 + b*y^2 + (z^2 + c*4^w)/m, where x,y,z are nonnegative integers, and w is 0 or 1.
We have verified Conjecture 2 for n up to 2*10^5.

Examples

			a(30) = 1 with 30 = 1^4 + 5^2 + (2^2 + 2*4)/3.
a(480) = 1 with 480 = 1^4 + 14^2 + (29^2 + 2*4)/3.
a(847) = 1 with 847 = 0^4 + 29^2 + (4^2 + 2*4^0)/3.
a(3497) = 1 with 3497 = 4^4 + 48^2 + (53^2 + 2*4^0)/3.
a(13582) = 1 with 13582 = 9^4 + 28^2 + (53^2 + 2*4^0)/3.
a(25630) = 1 with 25630 = 5^4 + 158^2 + (11^2 + 2*4^0)/3.
a(38064) = 1 with 38064 = 3^4 + 157^2 + (200^2 + 2*4^0)/3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[3(n-x^4-y^2)-2*4^z],r=r+1],{x,0,(n-1)^(1/4)},{y,0,Sqrt[n-1-x^4]},{z,0,1}];tab=Append[tab,r],{n,1,100}];Print[tab]

A351206 Least positive integer m such that n = x^4 + (y^4 + z^4 + 7*w^2)/m^4 for some nonnegative integers x,y,z,w with y <= z.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 4, 3, 1, 1, 1, 6, 2, 2, 10, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 3, 4, 2, 4, 10, 1, 1, 2, 4, 2, 1, 1, 1, 3, 1, 3, 2, 2, 2, 2, 2, 1, 3, 4, 2, 2, 1, 1, 3, 1, 1, 1, 1, 2, 4, 2, 2, 3, 8, 10, 2, 2, 1, 10, 5, 1, 1, 1, 1, 1, 4, 2, 2, 8, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 4, 4
Offset: 0

Views

Author

Zhi-Wei Sun, Feb 04 2022

Keywords

Comments

Conjecture: a(n) exists for any nonnegative integer n.
This implies that each nonnegative rational number can be written as 7*w^2 + x^4 + y^4 + z^4 with w,x,y,z rational numbers.

Examples

			a(6) = 2 with 6 = 1^4 + (1^4 + 2^4 + 7*3^2)/2^4.
a(19) = 6 with 19 = 0^4 + (1^4 + 4^4 + 7*59^2)/6^4.
a(22) = 10 with 22 = 2^4 + (2^4 + 13^4 + 7*67^2)/10^4.
a(5797) = 20 with 5797 = 0^4 + (81^4 + 164^4 + 7*4797^2)/20^4.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[m=1; Label[bb]; k=m^4; Do[If[SQ[(k*(n-x^4)-y^4-z^4)/7], tab=Append[tab,m]; Goto[aa]],  {x, 0, n^(1/4)}, {y, 0, (k*(n-x^4)/2)^(1/4)},{z,y,(k*(n-x^4)-y^4)^(1/4)}]; m=m+1; Goto[bb]; Label[aa], {n,0,100}];Print[tab]
Showing 1-8 of 8 results.