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.

A272332 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 36*x^2*y + 12*y^2*z + z^2*x a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 2, 2, 6, 4, 3, 3, 3, 8, 5, 2, 6, 6, 4, 1, 7, 10, 6, 8, 8, 5, 2, 2, 7, 16, 8, 3, 12, 6, 4, 3, 6, 13, 8, 8, 8, 6, 5, 7, 15, 14, 4, 2, 12, 7, 3, 2, 5, 18, 8, 12, 14, 8, 7, 4, 6, 8, 7, 5, 14, 8, 5, 2, 12, 18, 8, 12, 10, 6, 3, 5, 10, 19, 10, 3, 8, 3, 1, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 26 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m (k = 0,1,2,... and m = 1, 79, 591, 599, 1752, 1839, 10264).
We have verified that a(n) > 0 for all n = 1,...,400000.
For more refinements of Lagrange's four-square theorem, see arXiv:1604.06723.

Examples

			a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 > 0 and 36*0^2*0 + 12*0^2*0 + 0^2*0 = 0^2.
a(79) = 1 since 79 = 7^2 + 1^2 + 5^2 + 2^2 with 7 > 0 and 36*1^2*5 + 12*5^2*2 + 2^2*1 = 28^2.
a(591) = 1 since 591 = 23^2 + 1^2 + 6^2 + 5^2 with 23 > 0 and 36*1^2*6 + 12*6^2*5 + 5^2*1 = 49^2.
a(599) = 1 since 599 = 6^2 + 1^2 + 11^2 + 21^2 with 6 > 0 and 36*1^2*11 + 12*11^2*21 + 21^2*1 = 177^2.
a(1752) = 1 since 1752 = 10^2 + 4^2 + 40^2 + 6^2 with 10 > 0 and 36*4^2*40 + 12*40^2*6 + 6^2*10 = 372^2.
a(1839) = 1 since 1839 = 17^2 + 37^2 + 9^2 + 10^2 with 17 > 0 and 36*37^2*9 + 12*9^2*10 + 10^2*37 = 676^2.
a(10264) = 1 since 10264 = 96^2 + 30^2 + 2^2 + 12^2 with 96 > 0 and 36*30^2*2 + 12*2^2*12 + 12^2*30 = 264^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[36*x^2*y+12*y^2*z+z^2*x],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A280831 Number of ways to write 8*n+7 as x^2 + y^2 + z^2 + w^2 with x^4 + 1680*y^3*z a square, where x,y,z,w are positive integers.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 2, 1, 2, 1, 1, 1, 4, 2, 1, 4, 5, 3, 3, 1, 3, 2, 3, 2, 6, 5, 3, 4, 4, 3, 12, 6, 2, 7, 5, 3, 10, 4, 5, 2, 7, 5, 4, 5, 3, 8, 2, 2, 3, 4, 6, 7, 8, 1, 5, 2, 6, 9, 6, 5, 9, 9, 4, 6, 1, 4, 14, 5, 4, 12, 3, 11, 12, 1, 4, 8, 6, 7, 4, 6, 7
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 08 2017

Keywords

Comments

Conjecture: Let a and b be nonzero integers with gcd(a,b) squarefree. Then any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x^4 + b*y^3*z a square, if and only if (a,b) is among the ordered pairs (1,1), (1,15), (1,20), (1,36), (1,60), (1,1680) and (9,260).
If a natural number n is not of the form 4^k*(8m+7) (k,m = 0,1,...), then by the Gauss-Legendre theorem, there are nonnegative integers w,x,y such that n = w^2 + x^2 + y^2 + 0^2 and hence x^4 + 1680*y^3*0 is a square. Thus, the conjecture for (a,b) = (1,1680) has the following equivalent form: a(n) > 0 for all n = 0,1,...
See also A272336 for a similar conjecture.
Concerning the author's 1680-conjecture which states that each n = 0,1,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x^4 + 1680*y^3*z is a square, Qing-Hu Hou at Tianjin Univ. has verified it for n up to 10^8. The author would like to offer 1680 RMB as the prize for the first complete solution of the 1680-conjecture. - Zhi-Wei Sun, Jun 22 2020

Examples

			a(0) = 1 since 8*0+7 = 1^2 + 1^2 + 1^2 + 2^2 with 1^4 + 1680*1^3*1 = 41^2.
a(11) = 1 since 8*11 + 7 = 95 = 6^2 + 3^2 + 1^2 + 7^2 with 6^4 + 1680*3^3*1 = 216^2.
a(244) = 1 since 8*244 + 7 = 1959 = 13^2 + 13^2 + 39^2 + 10^2 with 13^4 + 1680*13^3*39 = 11999^2.
a(289) = 1 since 8*289 + 7 = 2319 = 14^2 + 7^2 + 45^2 + 7^2 with 14^4 + 1680*7^3*45 = 5096^2.
a(664) = 1 since 8*664 + 7 = 5319 = 3^2 + 6^2 + 45^2 + 57^2 with 3^4 + 1680*6^3*45 = 4041^2.
a(749) = 1 since 8*749 + 7 = 5999 = 31^2 + 18^2 + 15^2 + 67^2
with 31^4 + 1680*18^3*15 = 12161^2.
a(983) = 1 since 8*983 + 7 = 7871 = 27^2 + 54^2 + 1^2 + 65^2 with 27^4 + 1680*54^3*1 = 16281^2.
a(1228) = 1 since 8*1228 + 7 = 9831 = 35^2 + 10^2 + 91^2 + 15^2 with 35^4 + 1680*10^3*91 = 12425^2.
a(1819) = 1 since 8*1819 + 7 = 14559 = 34^2 + 1^2 + 39^2 + 109^2 with 34^4 + 1680*1^3*39 = 1184^2.
a(2503) = 1 since 8*2503 + 7 = 20031 = 97^2 + 7^2 + 13^2 + 102^2 with 97^4 + 1680*7^3*13 = 9799^2.
a(2506) = 1 since 8*2506 + 7 = 20055 = 47^2 + 6^2 + 77^2 + 109^2 with 47^4 + 1680*6^3*77 = 5729^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Do[r=0;Do[If[SQ[8n+7-x^2-y^2-z^2]&&SQ[x^4+1680y^3*z],r=r+1],{x,1,Sqrt[8n+6]},{y,1,Sqrt[8n+6-x^2]},{z,1,Sqrt[8n+6-x^2-y^2]}];Print[n," ",r];Continue,{n,0,80}]

A279056 Number of ways to write n as w^2 + x^2 + y^2 + z^2 with w a positive integer and x,y,z nonnegative integers such that x^3 + 4*y*z*(y-z) is a square.

Original entry on oeis.org

1, 3, 3, 2, 5, 5, 2, 2, 4, 9, 5, 3, 7, 4, 3, 1, 7, 13, 6, 7, 9, 4, 2, 4, 10, 13, 10, 4, 9, 6, 3, 3, 9, 15, 7, 10, 8, 6, 5, 6, 14, 14, 7, 3, 14, 7, 2, 3, 5, 14, 12, 11, 12, 9, 5, 5, 9, 12, 6, 6, 10, 5, 4, 2, 11, 20, 10, 10, 12, 4, 2, 6, 13, 14, 10, 4, 7, 5, 1, 5
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 05 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*q (k = 0,1,2,... and q = 1, 79, 143, 184, 575).
(ii) Any positive integer n can be written as w^2 + x^2 + y^2 + z^2 with w a positive integer and x,y,z nonnegative integers such that x^3 + 8*y*z*(2y-z) is a square.
We have verified a(n) > 0 and part (ii) of the conjecture for n up to 3*10^5.
For more conjectural refinements of Lagrange's four-square theorem, see Section 4 of arXiv:1604.06723.

Examples

			a(1) = 1 since  1 = 1^2 + 0^2 + 0^2 + 0^2 with 0^3 + 4*0*0*(0-0) = 0^2.
a(79) = 1 since 79 = 7^2 + 1^2 + 5^2 + 2^2 with 1^3 + 4*5*2*(5-2) = 11^2.
a(143) = 1 since 143 = 9^2 + 1^2 + 6^2 + 5^2 with 1^3 + 4*6*5*(6-5) = 11^2.
a(184) = 1 since 184 = 10^2 + 8^2 + 4^2 + 2^2 with 8^3 + 4*4*2*(4-2) = 24^2.
a(575) = 1 since 575 = 1^2 + 22^2 + 3^2 + 9^2 with 22^3 + 4*3*9*(3-9) = 100^2.
		

Crossrefs

Programs

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