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.

A275344 Number of ordered ways to write n as x^2 + y^2 + z^2 + 2*w^2 with x + 2*y + 3*z a square, where x,y,z,w are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jul 24 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 3, 5, 7, 14, 15, 16, 25, 30, 33, 84, 169, 225.
(ii) For each ordered pair (a,b) = (1,2), (1,3), (1,12), (1,23), (2,3), (2,4), (2,6), (2,7), (2,15), (2,16), any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that a*x + b*y is a square.
This is similar to the conjecture in A271518. It is known that any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers.

Examples

			a(0) = 1 since 0 = 0^2 + 0^2 + 0^2 + 2*0^2 with 0 + 2*0 + 3*0 = 0^2.
a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 2*0^2 with 1 + 2*0 + 3*0 = 1^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 0^2 + 2*1^2 with 1 + 2*0 + 3*0 = 1^2.
a(5) = 1 since 5 = 2^2 + 1^2 + 0^2 + 2*0^2 with 2 + 2*1 + 3*0 = 2^2.
a(7) = 1 since 7 = 2^2 + 1^2 + 0^2 + 2*1^2 with 2 + 2*1 + 3*0 = 2^2.
a(14) = 1 since 14 = 1^2 + 1^2 + 2^2 + 2*2^2 with 1 + 2*1 + 3*2 = 3^2.
a(15) = 1 since 15 = 3^2 + 0^2 + 2^2 + 2*1^2 with 3 + 2*0 + 3*2 = 3^2.
a(16) = 1 since 16 = 4^2 + 0^2 + 0^2 + 2*0^2 with 4 + 2*0 + 3*0 = 2^2.
a(25) = 1 since 25 = 1^2 + 4^2 + 0^2 + 2*2^2 with 1 + 2*4 + 3*0 = 3^2.
a(30) = 1 since 30 = 3^2 + 2^2 + 3^2 + 2*2^2 with 3 + 2*2 + 3*3 = 4^2.
a(33) = 1 since 33 = 1^2 + 0^2 + 0^2 + 2*4^2 with 1 + 2*0 + 3*0 = 1^2.
a(84) = 1 since 84 = 4^2 + 6^2 + 0^2 + 2*4^2 with 4 + 2*6 + 3*0 = 4^2.
a(169) = 1 since 169 = 10^2 + 6^2 + 1^2 + 2*4^2 with 10 + 2*6 + 3*1 = 5^2.
a(225) = 1 since 225 = 10^2 + 6^2 + 9^2 + 2*2^2 with 10 + 2*6 + 3*9 = 7^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-2*w^2-x^2-y^2]&&SQ[x+2*y+3*Sqrt[n-2w^2-x^2-y^2]],r=r+1],{w,0,Sqrt[n/2]},{x,0,Sqrt[n-2*w^2]},{y,0,Sqrt[n-2*w^2-x^2]}];Print[n," ",r];Continue,{n,0,80}]

A275409 Number of ordered ways to write n as 2*w^2 + x^2 + y^2 + z^2 with w + x + 2*y + 4*z a square, where w,x,y,z are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jul 26 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 except for n = 3, 10, and a(n) = 1 only for n = 0, 2, 7, 8, 9, 12, 14, 15, 22, 23, 24, 25, 36, 39, 44, 45, 60, 87, 98, 106, 110, 111, 183.
(ii) Any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that x + 2*y + 3*z - 3*w is a square.
(iii) For each triple (a,b,c) = (1,2,1), (1,2,3), (1,3,1), (2,4,1), (2,4,2), (2,4,3), (2,4,4), (2,4,8), (8,9,5), any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that a*x + b*y - c*z is a square.
(iv) Any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that x + 2*y - 2*z is twice a nonnegative cube. Also, each natural number can be written as x^2 + y^2 + z^2 + 2*w^3 with x,y,z,w nonnegative integers such that x + 3*y - z is a square.
See also A275344 and A275301 for related conjectures. We are able to show that each natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w integers such that x + y + z = t^2 for some t = 0, 1, 2.

Examples

			a(2) = 1 since 2 = 2*1^2 + 0^2 + 0^2 + 0^2 with 1 + 0 + 2*0 + 4*0 = 1^2.
a(7) = 1 since 7 = 2*1^2 + 0^2 + 2^2 + 1^2 with 1 + 0 + 2*2 + 4*1 = 3^2.
a(8) = 1 since 8 = 2*1^2 + 2^2 + 1^2 + 1^2 with 1 + 2 + 2*1 + 4*1 = 3^2.
a(9) = 1 since 9 = 2*2^2 + 0^2 + 1^2 + 0^2 with 2 + 0 + 2*1 + 4*0 = 2^2.
a(12) = 1 since 12 = 2*2^2 + 2^2 + 0^2 + 0^2 with 2 + 2 + 2*0 + 4*0 = 2^2.
a(14) = 1 since 14 = 2*0^2 + 2^2 + 1^2 + 3^2 with 0 + 2 + 2*1 + 4*3 = 4^2.
a(15) = 1 since 15 = 2*1^2 + 2^2 + 3^2 + 0^2 with 1 + 2 + 2*3 + 4*0 = 3^2.
a(22) = 1 since 22 = 2*1^2 + 4^2 + 2^2 + 0^2 with 1 + 4 + 2*2 + 4*0 = 3^2.
a(23) = 1 since 23 = 2*3^2 + 2^2 + 0^2 + 1^2 with 3 + 2 + 2*0 + 4*1 = 3^2.
a(24) = 1 since 24 = 2*0^2 + 4^2 + 2^2 + 2^2 with 0 + 4 + 2*2 + 4*2 = 4^2.
a(25) = 1 since 25 = 2*0^2 + 4^2 + 0^2 + 3^2 with 0 + 4 + 2*0 + 4*3 = 4^2.
a(36) = 1 since 36 = 2*3^2 + 1^2 + 4^2 + 1^2 with 3 + 1 + 2*4 + 4*1 = 4^2.
a(39) = 1 since 39 = 2*1^2 + 6^2 + 1^2 + 0^2 with 1 + 6 + 2*1 + 4*0 = 3^2.
a(44) = 1 since 44 = 2*3^2 + 0^2 + 1^2 + 5^2 with 3 + 0 + 2*1 + 4*5 = 5^2.
a(45) = 1 since 45 = 2*0^2 + 5^2 + 2^2 + 4^2 with 0 + 5 + 2*2 + 4*4 = 5^2.
a(60) = 1 since 60 = 2*2^2 + 6^2 + 4^2 + 0^2 with 2 + 6 + 2*4 + 4*0 = 4^2.
a(87) = 1 since 87 = 2*3^2 + 2^2 + 8^2 + 1^2 with 3 + 2 + 2*8 + 4*1 = 5^2.
a(98) = 1 since 98 = 2*4^2 + 1^2 + 8^2 + 1^2 with 4 + 1 + 2*8 + 4*1 = 5^2.
a(106) = 1 since 106 = 2*2^2 + 8^2 + 3^2 + 5^2 with 2 + 8 + 2*3 + 4*5 = 6^2.
a(110) = 1 since 110 = 2*6^2 + 5^2 + 3^2 + 2^2 with 6 + 5 + 2*3 + 4*2 = 5^2.
a(111) = 1 since 111 = 2*5^2 + 3^2 + 6^2 + 4^2 with 5 + 3 + 2*6 + 4*4 = 6^2.
a(183) = 1 since 183 = 2*3^2 + 10^2 + 4^2 + 7^2 with 3 + 10 + 2*4 + 4*7 = 7^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-2*w^2-x^2-y^2]&&SQ[w+x+2y+4*Sqrt[n-2*w^2-x^2-y^2]],r=r+1],{w,0,Sqrt[n/2]},{x,0,Sqrt[n-2*w^2]},{y,0,Sqrt[n-2*w^2-x^2]}];Print[n," ",r];Continue,{n,0,80}]

A338263 Number of ways to write 8*n+7 as 2*w^2 + x^2 + y^2 + z^2 with w, x, y, z nonnegative integers such that w*(3*x+7*y+10*z) is a square and also one of w, x, y, z is a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 27 2020

Keywords

Comments

Conjecture 1: Each nonnegative integer can be written as 2*w^2 + x^2 + y^2 + z^2 with w, x, y, z nonnegative integers such that w*(3*x+7*y+10*z) is a square and one of w, x, y, z is also a square.
As all the nonnegative integers not of the form 4^k*(8*m+7) (k>=0, m>=0) can be written as 2*0^2 + x^2 + y^2 + z^2 with x, y, z integers, Conjecture 1 has the following equivalent version: a(n) > 0 for all n = 0,1,...
We have verified that a(n) > 0 for all n = 0..10^5.
Conjecture 2: If (a,b) is among the ordered pairs (1,2), (1,3), (2,4), (2,5), (2,8), (2,24), (6,8), (6,32), (9,12) and (18,24), then each n = 0,1,... can be written as 2*w^2 + x^2 + y^2 + z^2 with w, x, y, z nonnegative integers such that w*(a*x+b*y) is a square.

Examples

			a(7) = 1, and 8*7+7 = 63 = 2*3^2 + 6^2 + 0^2 + 3^2 with 0 = 0^2 and 3*(3*6+7*0+10*3) = 12^2.
a(11) = 1, and 8*11+7 = 95 = 2*1^2 + 2^2 + 5^2 + 8^2 with 1 = 1^2 and 1*(3*2+7*5+10*8) = 11^2.
a(15) = 1, and 8*15+7 = 127 = 2*7^2 + 3^2 + 2^2 + 4^2 with 4 = 2^2 and 7*(3*3+7*2+10*4) = 21^2.
a(64) = 1, and 8*64+7 = 519 = 2*3^2 + 1^2 + 20^2 + 10^2 with 1 = 1^2 and 3*(3*1+7*20+10*10) = 27^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[8n+7-2w^2-x^2-y^2]&&(SQ[w]||SQ[x]||SQ[y]||SQ[8n+7-2w^2-x^2-y^2])&&SQ[w(3x+7y+10*Sqrt[8n+7-2w^2-x^2-y^2])],r=r+1],{w,0,Sqrt[4n+3]},{x,0,Sqrt[8n+7-2w^2]},{y,0,Sqrt[8n+7-2w^2-x^2]}];tab=Append[tab,r],{n,0,80}];tab
Showing 1-3 of 3 results.