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 1-4 of 4 results.

A271714 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that (10*w+5*x)^2 + (12*y+36*z)^2 is a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 2, 1, 3, 1, 2, 1, 2, 3, 1, 4, 4, 2, 2, 1, 3, 3, 5, 2, 2, 5, 2, 1, 2, 3, 3, 3, 2, 3, 2, 3, 4, 4, 2, 3, 9, 2, 3, 1, 1, 6, 2, 3, 4, 6, 4, 1, 2, 5, 3, 3, 4, 3, 5, 1, 4, 5, 1, 3, 6, 6, 1, 3, 4, 5, 12, 2, 4, 6, 2, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 12 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 7, 9, 19, 49, 133, 589, 2^k, 2^k*3, 4^k*q (k = 0,1,2,... and q = 14, 67, 71, 199).
(ii) If P(y,z) is one of 2y-3z, 2y-8z and 4y-6z, then any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w-x)^2 + P(y,z)^2 is a square.
(iii) For each triple (a,b,c) = (1,4,4), (1,12,12), (2,4,8), (2,6,6), (2,12,12), (3,4,4), (3,4,8), (3,8,8), (3,12,12), (3,12,36), (5,4,4), (5,4,8), (5,8,16), (5,36,36), (6,4,4), (7,12,12), (7,20,20), (7,24,24), (9,4,4), (9,12,12),(9,36,36), (11,12,12), (13,4,4), (15,12,12), (16,12,12), (21,20,20), (21,24,24), (23,12,12), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w+a*x)^2 + (b*y-c*z)^2 is a square.
See also A271510, A271513, A271518, A271644, A271665, A271721 and A271724 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(2) = 1 since 2 = 1^2 + 1^2 + 0^2 + 0^2 with (10*1+5*1)^2 + (12*0+36*0)^2 = 15^2 + 0^2 = 15^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with (10*1+5*1)^2 + (12*0+36*1)^2 = 15^2 + 36^2 = 39^2.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with (10*2+5*0)^2 + (12*0+36*0)^2 = 20^2 + 0^2 = 20^2.
a(6) = 1 since 6 = 2^2 + 0^2 + 1^2 + 1^2 with (10*2+5*0)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with (10*1+5*2)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(9) = 1 since 9 = 3^2 + 0^2 + 0^2 + 0^2 with (10*3+5*0)^2 + (12*0+36*0)^2 = 30^2 + 0^2 = 30^2.
a(19) = 1 since 19 = 3^2 + 0^2 + 3^2 + 1^2 with (10*3+5*0)^2 + (12*3+36*1)^2 = 30^2 + 72^2 = 78^2.
a(49) = 1 since 49 = 7^2 + 0^2 + 0^2 + 0^2 with (10*7+5*0)^2 + (12*0+36*0)^2 = 70^2 + 0^2 = 70^2.
a(133) = 1 since 133 = 9^2 + 0^2 + 6^2 + 4^2 with (10*9+5*0)^2 + (12*6+36*4)^2 = 90^2 + 216^2 = 234^2.
a(589) = 1 since 589 = 17^2 + 10^2 + 2^2 + 14^2 with (10*17+5*10)^2 + (12*2+36*14)^2 = 220^2 + 528^2 = 572^2.

Links

Crossrefs

Cf. A000118, A000290, A271510, A271513, A271518, A271608, A271644, A271665, A271719, A271721, A271724.

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(10*Sqrt[n-x^2-y^2-z^2]+5x)^2+(12y+36z)^2],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}]

A271721 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= z >= 0, x > 0 and w >= z such that (x-y)*(w-z) is a square.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 12 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 5, 11, 15, 23, 35, 95, 4^k*190 (k = 0,1,2,...).
(ii) For each k = 4, 5, 6, 7, 8, 11, 12, 13, 15, 17, 18, 20, 22, 25, 27, 29, 33, 37, 38, 41, 50, 61, any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that (x-y)*(w-k*z) is a square.
(iii) For each triple (a,b,c) = (3,1,1), (1,2,1), (2,2,1), (3,2,1), (2,2,2), (6,2,1), (1,3,1), (3,3,1), (15,3,1), (1,4,1), (2,4,1), (1,5,1), (3,5,1), (5,5,1), (1,5,2), (1,6,1), (2,6,1), (3,6,1), (15,6,1), (1,7,1), (5,7,1), (1,8,1), (1,8,5), (3,9,1), (1,10,1), (1,12,1), (1,13,1), (3,13,1), (1,14,1), (1,15,1), (1,15,2), (6,16,1), (2,18,1), (3,18,1), (1,20,2), (1,21,1), (3,21,1), (1,23,1), (1,24,1), (1,27,1), (3,27,1), (1,34,1), (1,45,1), (3,45,1), (3,48,1), (1,55,1), (1,60,1), (5,60,1), any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*(x+b*y)*(w-c*z) is a square.
This is stronger than Lagrange's four-square theorem. Note that for k = 2 or 3, any natural number n can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers and (x-y)*(w-k*z) = 0, for, if n cannot be represented by x^2 + y^2 + 2*z^2 then it has the form 4^k*(16*m+14) (k,m = 0,1,2,...) and hence it can be represented by x^2 + y^2 + (k^2+1)*z^2. It is known that natural numbers not represented by x^2 + y^2 + 5*z^2 have the form 4^k*(8*m+3), and that positive even numbers not represented by x^2 + y^2 + 10*z^2 have the form 4^k*(16*m+6) (as conjectured by S. Ramanujan and proved by L. E. Dickson).
See also A271510, A271513, A271518, A271644, A271714 and A271724 for other conjectures refining Lagrange's theorem.

Examples

			a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 0^2 with 1 = 1 > 0 = 0 and (1-1)*(0-0) = 0^2.
a(5) = 1 since 5 = 2^2 + 1^2 + 0^2 + 0^2 with 2 > 1 > 0 = 0 and (2-1)*(0-0) = 0^2.
a(11) = 1 since 11 = 1^2 + 1^2 + 0^2 + 3^2 with 1 = 1 > 0 < 3 and (1-1)*(3-0) = 0^2.
a(14) = 2 since 14 = 3^2 + 1^2 + 0^2 + 2^2 with 3 > 1 > 0 < 2 and (3-1)*(2-0) = 2^2, and also 14 = 3^2 + 2^2 + 0^2 + 1^2 with 3 > 2 > 0 < 1 and (3-2)*(1-0) = 1^2.
a(15) = 1 since 15 = 3^2 + 2^2 + 1^2 + 1^2 with 3 > 2 > 1 = 1 and (3-2)*(1-1) = 0^2.
a(23) = 1 since 23 = 3^2 + 3^2 + 1^2 + 2^2 with 3 = 3 > 1 < 2 and (3-3)*(2-1) = 0^2.
a(35) = 1 since 35 = 3^2 + 3^2 + 1^2 + 4^2 with 3 = 3 > 1 < 4 and (3-3)*(4-1) = 0^2.
a(95) = 1 since 95 = 5^2 + 5^2 + 3^2 + 6^2 with 5 = 5 > 3 < 6 and (5-5)*(6-3) = 0^2.
a(190) = 1 since 190 = 13^2 + 4^2 + 1^2 + 2^2 with 13 > 4 > 1 < 2 and (13-4)*(2-1) = 3^2.

References

  • L. E. Dickson, Integers represented by positive ternary quadratic forms, Bull. Amer. Math. Soc. 33(1927), 63-70.
  • L. E. Dickson, Modern Elementary Theory of Numbers, University of Chicago Press, Chicago, 1939, pp. 112-113.

Links

Crossrefs

Cf. A000118, A000290, A259789, A271510, A271513, A271518, A271608, A271644, A271665, A271714, A271719, A271724.

Programs

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

A340274 Number of ways to write n as x + y + z with x, y, z positive integers such that 3*x^2*y^2 + 5*y^2*z^2 + 8*z^2*x^2 is a square.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 4, 2, 5, 2, 1, 2, 2, 2, 2, 4, 2, 4, 3, 3, 3, 4, 5, 2, 3, 5, 5, 4, 4, 2, 4, 4, 5, 3, 4, 3, 6, 3, 2, 5, 2, 2, 7, 7, 1, 3, 6, 4, 4, 3, 3, 6, 2, 5, 5, 3, 6, 5, 4, 6, 6, 6, 3, 6, 6, 4, 5, 6, 2, 6, 3, 5, 4, 5, 3, 5, 12, 4, 4, 5, 1, 6, 6, 7, 9, 3, 3, 6, 5, 6, 7, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 24 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 2.
We have verified a(n) > 0 for all n = 3..10000. The conjecture holds if a(p) > 0 for every odd prime p. For any n > 0 we have a(3*n) > 0, since 3*n = n + n + n and 3 + 5 + 8 = 4^2.
It seems that a(n) = 1 only for n = 3..8, 10, 11, 19, 53, 89, 127, 178, 257, 461.
See also A343862 for similar conjectures.
Conjecture 1 holds for all n < 2^15. Note a(1823) = 1. - Martin Ehrenstein, May 03 2021

Examples

			a(4) = 1 with 4 = 2 + 1 + 1 and 3*2^2*1^2 + 5*1^2*1^2 + 8*1^2*2^2 = 7^2.
a(19) = 1 with 19 = 9 + 9 + 1 and 3*9^2*9^2 + 5*9^2*1^2 + 8*1^2*9^2 = 144^2.
a(53) = 1 with 53 = 23 + 7 + 23 and 3*23^2*7^2 + 5*7^2*23^2 + 8*23^2*23^2 = 1564^2.
a(89) = 1 with 89 = 2 + 58 + 29 and 3*2^2*58^2 + 5*58^2*29^2 + 8*29^2*2^2 = 3770^2.
a(257) = 1 with 257 = 11 + 164 + 82 and 3*11^2*164^2 + 5*164^2*82^2 + 8*82^2*11^2 = 30340^2.
a(461) = 1 with 461 = 186 + 165 + 110 and 3*186^2*165^2 + 5*165^2*110^2 + 8*110^2*186^2 = 88440^2.

Links

Crossrefs

Cf. A000041, A000290, A230121, A230747, A231168, A262357, A271719, A273278, A343862.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[3x^2*y^2+(n-x-y)^2*(5*y^2+8*x^2)],r=r+1],{x,1,n-2},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]

A343862 Number of ways to write n as x + y + z with x, y, z positive integers such that x^2*y^2 + 5*y^2*z^2 + 10*z^2*x^2 is a square.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 2, 1, 1, 4, 2, 4, 2, 2, 5, 5, 3, 3, 2, 6, 4, 3, 3, 6, 6, 5, 2, 6, 4, 10, 3, 6, 4, 6, 6, 8, 5, 6, 4, 9, 7, 6, 3, 7, 9, 5, 5, 15, 5, 12, 11, 10, 5, 6, 7, 10, 8, 9, 7, 15, 7, 6, 7, 10, 10, 7, 9, 10, 10, 12, 4, 15, 9, 9, 11, 9, 7, 12, 11, 15, 8, 9, 7, 12, 10, 3, 9, 11, 11, 19, 12, 12, 9, 10, 6, 23, 11, 6, 10, 18
Offset: 1

Views

Author

Zhi-Wei Sun, May 02 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 2.
We have verified a(n) > 0 for all n = 3..10000. Conjecture 1 holds if a(p) > 0 for each odd prime p. For any n > 0 we have a(3*n) > 0 since 3*n = n + n + n and 1 + 5 + 10 = 4^2.
See also A340274 for a similar conjecture.
Conjecture 2: There are infinitely many triples (a,b,c) of positive integers such that each n = 3,4,... can be written as x + y + z with x,y,z positive integers and a*x^2*y^2 + b*y^2*z^2 + c*z^2*x^2 a square.
Such triple candidates include (21,19,9), (23,17,9), (24,16,9), (25,14,10), (29,19,16), (33,27,21), (35,9,5), (37,32,31) etc.
Conjecture 1 holds for all n < 2^15. - Martin Ehrenstein, May 02 2021

Examples

			a(4) = 1 with 4 = 2 + 1 + 1 and 2^2*1^2 + 5*1^2*1^2 + 10*1^2*2^2 = 7^2.
a(5) = 1 with 5 = 1 + 3 + 1 and 1^2*3^2 + 5*3^2*1^2 + 10*1^1*1^2 = 8^2.
a(8) = 1 with 8 = 4 + 2 + 2 and 4^2*2^2 + 5*2^2*2^2 + 10*2^2*4^2 = 28^2.
a(9) = 1 with 9 = 3 + 3 + 3 and 3^2*3^2 + 5*3^2*3^2 + 10*3^2*3^2 = 36^2.
a(19) = 2. We have 19 = 4 + 5 + 10 with 4^2*5^2 + 5*5^2*10^2 + 10*10^2*4^2 = 170^2, and 19 = 4 + 13 + 2 with 4^2*13^2 + 5*13^2*2^2 + 10*2^2*4^2 = 82^2.

Links

Crossrefs

Cf. A000041, A000290, A230121, A230747, A231168, A262357, A271719, A273278, A340274.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[x^2*y^2+(n-x-y)^2*(5*y^2+10*x^2)],r=r+1],{x,1,n-2},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.