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.

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]

A343897 Number of ways to write n as 2*x + y + z with x,y,z positive integers such that 16*x^2*y^2 + 19*y^2*z^2 + 29*z^2*x^2 is a square.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 2, 4, 2, 1, 3, 1, 3, 1, 5, 2, 2, 6, 1, 1, 5, 5, 6, 2, 2, 4, 2, 4, 5, 6, 3, 2, 3, 2, 5, 2, 7, 10, 4, 1, 3, 3, 10, 9, 2, 5, 5, 10, 6, 7, 6, 7, 8, 8, 4, 7, 4, 5, 8, 2, 4, 4, 13, 9, 5, 6, 10, 11, 6, 11, 6, 6, 5, 4, 4, 10, 9, 8, 8, 8, 8, 9, 16, 5, 5, 6, 4, 7, 3, 12, 7, 11, 13
Offset: 1

Views

Author

Zhi-Wei Sun, May 03 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
We have verified a(n) > 0 for all n = 4..10000. The conjecture holds if a(p) > 0 for each odd prime p.
It seems that a(n) = 1 only for n = 4..9, 13, 17, 19, 21, 26, 27, 47.

Examples

			a(4) = 1 with 4 = 2*1 + 1 + 1 and 16*1^2*1^2 + 19*1^2*1^2 + 29*1^2*1^2 = 8^2.
a(6) = 1 with 6 = 2*1 + 2 + 2 and 16*1^2*2^2 + 19*2^2*2^2 + 29*2^2*1^2 = 22^2.
a(9) = 1 with 9 = 2*2 + 4 + 1 and 16*2^2*4^2 + 19*4^2*1^2 + 29*1^2*2^2 = 38^2.
a(13) = 1 with 13 = 2*5 + 2 + 1 and 16*5^2*2^2 + 19*2^2*1^2 + 29*1^2*5^2 = 49^2.
a(19) = 1 with 19 = 2*2 + 14 + 1 and 16*2^2*14^2 + 19*14^2*1^2 + 29*1^2*2^2 = 128^2.
a(47) = 1 with 47 = 2*13 + 13 + 8 and 16*13^2*13^2 + 19*13^2*8^2 + 29*8^2*13^2 = 988^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[16(x*y)^2+(n-2x-y)^2*(19*y^2+29x^2)],r=r+1],{x,1,(n-2)/2},{y,1,n-1-2x}];tab=Append[tab,r],{n,1,100}];Print[tab]

A343917 Positive integers m with 2*m^2 - 2^4 = x^4 + y^4 for some nonnegative integers x and y with |x-y| > 2.

Original entry on oeis.org

284, 1388, 2139, 4772, 8556, 8971, 10836, 21163, 28847, 45707, 54507, 71292, 73348, 95127, 101503, 104228, 131388, 136148, 263172, 350076, 638164, 982292, 1532148, 1687828, 1705407, 1958924, 2082188, 2299364, 2360347, 2728379, 3202356, 4042799, 5046771, 5165332, 5235323, 5560627, 7191079, 7740547, 8041364
Offset: 1

Views

Author

Zhi-Wei Sun, May 04 2021

Keywords

Comments

Conjecture: The sequence has infinitely many terms.
It is easy to see that no term is divisible by 5. In the b-file we list all the 62 terms not exceeding 10^8.
Note that 2*(n^2+3)^2 - 2^4 = (n+1)^4 + (n-1)^4 with (n+1) - (n-1) = 2. This implies that any integer n > 4 can be written as x + y + 2^(z-1) with x,y,z positive integers such that x^4 + y^4 + (2^z)^4 is twice a square.
See also A343913 for a similar conjecture.

Examples

			a(1) = 284, and 2*284^2 - 2^4 = 20^4 + 6^4 with |20-6| > 2.
a(62) = 97077407, and 2*97077407^2 - 2^4 = 18848045899687282 = 11563^4 + 5583^4 with |11563-5583| > 2.
		

Crossrefs

Programs

  • Mathematica
    QQ[n_]:=IntegerQ[n^(1/4)];
    n=0;Do[Do[If[QQ[2*m^2-16-x^4]&&(2*m^2-16-x^4)^(1/4)-x>2,n=n+1;Print[n," ",m];Goto[aa]],{x,0,(m^2-8)^(1/4)}];Label[aa],{m,3,8041364}]

A344058 Number of ways to write n as x + y + z with x*y + 2*y*z + 3*z*x a square, where x,y,z are positive integers with x or y a power of two (including 2^0 = 1).

Original entry on oeis.org

0, 0, 0, 1, 2, 1, 1, 1, 3, 2, 2, 5, 3, 2, 5, 1, 5, 5, 2, 8, 5, 3, 9, 5, 3, 8, 4, 7, 7, 6, 11, 1, 8, 5, 4, 14, 6, 2, 5, 8, 9, 6, 8, 11, 8, 10, 5, 5, 13, 5, 7, 18, 17, 6, 9, 7, 5, 7, 6, 14, 11, 12, 7, 1, 12, 10, 14, 9, 13, 6, 10, 14, 14, 11, 10, 9, 7, 6, 10, 8, 8, 12, 7, 12, 12, 10, 11, 11, 8, 10, 10, 25, 15, 7, 18, 5, 11, 13, 13, 12
Offset: 1

Views

Author

Zhi-Wei Sun, May 08 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
We have verified a(n) > 0 for all n = 4..10^5. Clearly, a(2*n) > 0 if a(n) > 0.

Examples

			a(6) = 1 with 6 = 3 + 2^0 + 2 and 3*2^0 + 2*2^0*2 + 3*2*3 = 5^2.
a(7) = 1 with 7 = 3 + 2^0 + 3 and 3*2^0 + 2*2^0*3 + 3*3*3 = 6^2.
For each k > 1, we have a(2^k) = 1 with 2^k = 2^(k-2) + 2^(k-1) + 2^(k-2) and 2^(k-2)*2^(k-1) + 2*2^(k-1)*2^(k-2) + 3*2^(k-2)*2^(k-2) = (3*2^(k-2))^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]];
    Pow[x_]:=x>0&&IntegerQ[Log[2,x]];
    tab={};Do[r=0;Do[If[(Pow[x]||Pow[y])&&SQ[x*y+(2y+3x)*(n-x-y)],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.