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If n == 10, 26, 40, 42 or 58 (mod 64) , or n == 160 (mod 256), then a(n) = 0. - Robert Israel, Jun 18 2019

If n == 10, 26, 40, 42 or 58 (mod 64) then a(n) = 0. - Robert Israel, Jun 18 2019

Robert Israel, <a href="/A323381/b323381.txt">Table of n, a(n) for n = 0..10000</a>

N:= 100: # for a(0)..a(N)

V:= Array(0..N):

for z from 0 to floor(sqrt(4*N/19)) do

if z = 0 then mz:= 1 else mz:= 2 fi;

for y from 0 to floor(sqrt((N - 19/4*z^2)/2)) do

if y = 0 then my:= 1 else my:= 2 fi;

r:= sqrt(N - 2*y^2 - 19/4*z^2);

for x from ceil(-r -z/2) to floor(r-z/2) do

v:= x^2 + 2*y^2 + 5*z^2 + x*z;

V[v]:= V[v] + my*mz;

od

od

od:

convert(V, list); # Robert Israel, Jun 18 2019

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allocated for Michael SomosNumber of solutions to n = x^2 + 2*y^2 + 5*z^2 + x*z in integers.

1, 2, 2, 4, 2, 4, 4, 12, 2, 14, 0, 8, 4, 16, 0, 8, 2, 8, 6, 22, 4, 4, 12, 12, 4, 30, 0, 20, 12, 8, 8, 8, 2, 16, 4, 16, 14, 28, 12, 8, 0, 12, 0, 40, 8, 12, 16, 12, 4, 22, 2, 28, 16, 24, 8, 32, 0, 48, 0, 4, 8, 20, 8, 36, 2, 16, 8, 36, 8, 12, 16, 8, 6, 32, 0, 28

0,2

If n<0, then a(n)=0. If n>0, then a(n) is even since (-x, -y, -z) is a solution if (x, y, z) is.

Rouse [2014] conjectures that the ternary quadratic form x^2 + 2y^2 + 5z^2 + xz represents all positive odd integers.

J. Rouse, <a href="http://users.wfu.edu/rouseja/cv/451thm.pdf">Quadratic forms representing all odd positive integers</a> Amer. J. Math, 136 (2014), no. 6, 1693-1745.

K. S. Williams, <a href="http://dx.doi.org/10.1080/00029890.2018.1503003">Eveything You Wanted to Know about ax^2+by^2+cz^2+dt^2 But Were Afraid To Ask</a>, Amer. Math. Monthly, Vol. 125, No. 9, (2018), 797-810. See page 803.

G.f. = 1 + 2*x + 2*x^2 + 4*x^3 + 2*x^4 + 4*x^5 + 4*x^6 + 12*x^7 + ...

a[ n_] := Length @ FindInstance[ x^2 + 2 y^2 + 5 z^2 + x z == n, {x, y, z}, Integers, 10^8];

(PARI) {a(n) = if( n<1, n==0, 2*qfrep([4, 0, 0; 0, 2, 1; 0, 1, 10], 2*n)[2*n])};

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Michael Somos, Jan 12 2019

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