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 1-6 of 6 results.

A275656 Number of ordered ways to write n as 4^k*(1+x^2+y^2)+z^2, where k,x,y,z are nonnegative integers with x <= y <= z.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 04 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
This is stronger than Lagrange's four-square theorem and the conjecture in A275648.

Examples

			a(22) = 1 since 22 = 4^0*(1+1^2+2^2) + 4^2 with 1 < 2 < 4.
a(31) = 1 since 31 = 4^0*(1+1^2+2^2) + 5^2 with 1 < 2 < 5.
a(59) = 1 since 59 = 4^0*(1+0^2+3^2) + 7^2 with 0 < 3 < 7.
a(79) = 1 since 79 = 4^0*(1+2^2+5^2) + 7^2 with 2 < 5 < 7.
a(94) = 1 since 94 = 4^0*(1+2^2+5^2) + 8^2 with 2 < 5 < 8.
a(128) = 1 since 128 = 4^3*(1+0^2+0^2) + 8^2 with 0 = 0 < 8.
a(134) = 1 since 134 = 4^0*(1+4^2+6^2) + 9^2 with 4 < 6 < 9.
a(221) = 1 since 221 = 4*(1+3^2+5^2) + 9^2 with 3 < 5 < 9.
a(254) = 1 since 254 = 4^0*(1+3^2+10^2) + 12^2 with 3 < 10 < 12.
a(349) = 1 since 349 = 4*(1+5^2+7^2) + 7^2  with 5 < 7 = 7.
a(608) = 1 since 608 = 4^2*(1+0^2+1^2) + 24^2 with 0 < 1 < 24.
a(797) = 1 since 797 = 4*(1+0^2+4^2) + 27^2 with 0 < 4 < 27.
a(1181) = 1 since 1181 = 4*(1+9^2+9^2) + 23^2 with 9 = 9 < 23.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648.

Programs

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

A275675 Number of ordered ways to write n as 4^k*(1+x^2+y^2)+5*z^2, where k,x,y,z are nonnegative integers with x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 04 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
This is stronger than the known fact that any natural number can be written as w^2 + x^2 + y^2 + 5*z^2 with w,x,y,z integers.
See also A275656, A275676 and A275678 for similar conjectures.

Examples

			a(43) = 1 since 43 = 4^0*(1+1^2+6^2) + 5*1^2.
a(45) = 1 since 45 = 4*(1+0^2+3^2) + 5*1^2.
a(237) = 1 since 237 = 4^3*(1+1^2+1^2) + 5*3^2.
a(561) = 1 since 561 = 4*(1+8^2+8^2) + 5*3^2.
a(9777) = 1 since 9777 = 4*(1+11^2+31^2) + 5*33^2.
a(39108) = 1 since 39108 = 4^2*(1+11^2+31^2) + 5*66^2.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275676, A275678.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[(n-4^k*(1+x^2+y^2))/5],r=r+1],{k,0,Log[4,n]},{x,0,Sqrt[(n/4^k-1)/2]},{y,x,Sqrt[n/4^k-1-x^2]}];Print[n," ",r];Continue,{n,1,80}]

A275676 Number of ordered ways to write n as 4^k*(1+5*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 04 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any positive integer can be written as 4^k*(1+5*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with y <= z.
See also A275656, A275675 and A275678 for similar conjectures.

Examples

			a(4) = 1 since 4 = 4*(1+5*0^2+0^2) + 0^2 with 0 = 0.
a(259) = 1 since 259 = 4^0*(1+5*4^2+13^2) + 3^2 with 4 < 13.
a(333) = 1 since 333 = 4*(1+5*3^2+5^2) + 7^2 with 3 < 5.
a(621) = 1 since 621 = 4*(1+5*0^2+8^2) + 19^2 with 0 < 8.
a(717) = 1 since 717 = 4*(1+5*3^2+11^2) + 7^2 with 3 < 11.
a(1581) = 1 since 1581 = 4*(1+5*1^2+3^2) + 39^2 with 1 < 3.
a(2541) = 1 since 2541 = 4*(1+5*3^2+13^2) + 41^2 with 3 < 13.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275675, A275678

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-4^k*(1+5x^2+y^2)],r=r+1],{k,0,Log[4,n]},{x,0,Sqrt[(n/4^k-1)/6]},{y,x,Sqrt[n/4^k-1-5x^2]}];Print[n," ",r];Continue,{n,1,80}]

A275678 Number of ordered ways to write n as 4^k*(1+4*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 05 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any positive integer can be written as 4^k*(1+4*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with x <= z.
This is stronger than Lagrange's four-square theorem. We have shown 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.
See also A275656, A275675 and A275676 for similar conjectures.

Examples

			a(12) = 1 since 12 = 4*(1+4*0^2+1^2) + 2^2 with 0 < 1.
a(19) = 1 since 19 = 4^0*(1+4*0^2+3^2) + 3^2 with 0 < 3.
a(61) = 1 since 61 = 4*(1+4*1^2+2^2) + 5^2 with 1 < 2.
a(125) = 1 since 125 = 4*(1+4*0^2+0^2) + 11^2 with 0 = 0.
a(359) = 1 since 359 = 4^0*(1+4*7^2+9^2) + 9^2 with 7 < 9.
a(196253) = 1 since 196253 = 4*(1+4*0^2+0^2) + 443^2 with 0 = 0.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275675, A275676.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-4^k*(1+4x^2+y^2)],r=r+1],{k,0,Log[4,n]},{x,0,Sqrt[(n/4^k-1)/5]},{y,x,Sqrt[n/4^k-1-4x^2]}];Print[n," ",r];Continue,{n,1,80}]

A275738 Number of ordered ways to write n as w^2 + x^2*(1+y^2+z^2), where w,x,y,z are nonnegative integers with x > 0, y <= z and y == z (mod 2).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 07 2016

Keywords

Comments

Conjecture: For any n > 0, we have a(n) > 0, i.e., n can be written as w^2 + x^2*(1+(z-y)^2+(y+z)^2) = w^2 + x^2*(1+2*y^2+2*z^2), where w,x,y,z are nonnegative integers with x > 0 and y <= z. Moreover, any positive integer n not equal to 449 can be written as 4^k*(1+x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with x == y (mod 2).
This is stronger than Lagrange's four-square theorem, and we have verified it for n up to 10^6.
See also A275648, A275656, A275675, A275676 and A275678 for similar conjectures.

Examples

			a(2) = 1 since 2 = 1^2 + 1^2*(1+0^2+0^2) with 0 + 0 even.
a(7) = 1 since 7 = 2^2 + 1^2*(1+1^2+1^2) with 1 + 1 even.
a(59) = 1 since 59 = 0^2 + 1^2*(1+3^2+7^2) with 3 + 7 even.
a(71) = 1 since 71 = 6^2 + 1^2*(1+3^2+5^2) with 3 + 5 even.
a(113) = 2 since 113 = 7^2 + 8^2*(1+0^2+0^2) = 8^2 + 7^2*(1+0^2+0^2) with 0 + 0 even.
a(143) = 1 since 143 = 6^2 + 1^2*(1+5^2+9^2) with 5 + 9 even.
a(191) = 1 since 191 = 10^2 + 1^2*(1+3^2+9^2) with 3 + 9 even.
a(449) = 3 since 449 = 18^2 + 5^2*(1+0^2+2^2) with 0 + 2 even, and 449 = 7^2 + 20^2*(1+0^2+0^2) = 20^2 + 7^2*(1+0^2+0^2) with 0 + 0 even.
a(497) = 1 since 497 = 15^2 + 4^2*(1+0^2+4^2) with 0 + 4 even.
a(2033) = 1 since 2033 = 33^2 + 4^2*(1+3^2+7^2) with 3 + 7 even.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275675, A275676, A275678.

Programs

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

A275471 Number of ordered ways to write n as 4^k*(1+x^2+y^2)+z^2, where k,x,y,z are nonnegative integers with x <= y and x == y (mod 2).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Aug 11 2016

Keywords

Comments

Conjecture: a(n) > 0 except for n = 449.
See also A275656, A275678 and A275738 for related conjectures.
As x^2 + y^2 = 2*((x+y)/2)^2 + 2*((x-y)/2)^2, we see that {x^2 + y^2: x and y are integers with x == y (mod 2)} = {2*x^2 + 2*y^2: x and y are integers}.

Examples

			a(8) = 1 since 8 = 4*(1+0^2+0^2) + 2^2 with 0+0 even.
a(31) = 1 since 31 = 4^0*(1+1^2+5^2) + 2^2 with 1+5 even.
a(47) = 1 since 47 = 4^0*(1+1^2+3^2) + 6^2 with 1+3 even.
a(79) = 1 since 79 = 4^0*(1+5^2+7^2)+2^2 with 5+7 even.
a(1009) = 1 since 1009 = 4^2*(1+1^2+1^2) + 31^2 with 1+1 even.
a(7793) = 1 since 7793 = 4^2*(1+12^2+18^2) + 17^2 with 12+18 even.

Links

Crossrefs

Cf. A000118, A000290, A271518, A275648, A275656, A275675, A275676, A275678, A275738.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-4^k*(1+2x^2+2y^2)],r=r+1],{k,0,Log[4,n]},{x,0,Sqrt[(n/4^k-1)/4]},{y,x,Sqrt[(n/4^k-1-2x^2)/2]}];Print[n," ",r];Continue,{n,1,80}]
Showing 1-6 of 6 results.

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