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.

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

Author

Zhi-Wei Sun, May 03 2021

Keywords

• nonn

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.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..1500

Crossrefs

Cf. A000041, A000290, A230121, A230747, A340274, A343862.

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]