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 1-3 of 3 results.

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]

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]
Showing 1-3 of 3 results.