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 1-2 of 2 results.

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.
		

Crossrefs

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.
		

Crossrefs

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 1-2 of 2 results.