A072068 Number of integer solutions to the equation 2x^2+y^2+8z^2=m for an odd number m=2n-1.

2, 4, 0, 0, 10, 12, 0, 0, 16, 12, 0, 0, 10, 16, 0, 0, 16, 24, 0, 0, 32, 12, 0, 0, 18, 24, 0, 0, 16, 36, 0, 0, 32, 12, 0, 0, 16, 28, 0, 0, 34, 36, 0, 0, 48, 24, 0, 0, 16, 36, 0, 0, 32, 36, 0, 0, 32, 24, 0, 0, 26, 24, 0, 0

Offset: 1

Author

T. D. Noe, Jun 13 2002

Keywords

• nonn

Comments

Related to primitive congruent numbers A006991.

Assuming the Birch and Swinnerton-Dyer conjecture, the odd number 2n-1 is a congruent number if it is squarefree and a(n) = 2*A072069(n).

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Examples

			a(2) = 4 because (1,1,0), (-1,1,0), (1,-1,0) and (-1,-1,0) are solutions when m=3.
G.f. = 2*x + 4*x^2 + 10*x^5 + 12*x^6 + 16*x^9 + 12*x^10 + 10*x^13 + 16*x^14 + 16*x^17 + ...
G.f. = 2*q + 4*q^3 + 10*q^9 + 12*q^11 + 16*q^17 + 12*q^19 + 10*q^25 + 16*q^27 + ...
		

Links

• T. D. Noe, Table of n, a(n) for n=1..10000
• Clay Mathematics Institute, The Birch and Swinnerton-Dyer Conjecture
• Department of Pure Maths., Univ. Sheffield, Pythagorean triples and the congruent number problem
• Karl Rubin, Elliptic curves and right triangles
• J. B. Tunnell, A classical Diophantine problem and modular forms of weight 3/2, Invent. Math., 72 (1983), 323-334.

Crossrefs

Cf. A006991, A003273, A072069, A072070, A072071, A080917.

Programs

Mathematica

maxN=128; soln1=Table[0, {maxN/2}]; xMax=Ceiling[Sqrt[maxN/2]]; yMax=Ceiling[Sqrt[maxN]]; zMax=Ceiling[Sqrt[maxN/8]]; Do[n=2x^2+y^2+8z^2; If[OddQ[n]&&nA072068 = CoefficientList[s, x] // Rest (* Jean-François Alcover, Feb 16 2015, after Michael Somos *)

PARI

{a(n) = my(A); n--; if( n<0, 0, A = x * O(x^n); polcoeff( 2 * eta(x^2 + A)^5 * eta(x^8 + A)^7 / (eta(x + A)^2 * eta(x^4 + A)^5 * eta(x^16 + A)^2), n))}; /* Michael Somos, Dec 26 2019 */

Formula

Expansion of 2 * x * phi(x) * psi(x^4) * phi(x^4) in powers of x where phi(), psi() are Ramanujan theta functions. - Michael Somos, Jun 08 2012

Expansion of 2 * q^(1/2) * eta(q^2)^5 * eta(q^8)^7 / (eta(q)^2 * eta(q^4)^5 * eta(q^16)^2) in powers of q. - Michael Somos, Feb 19 2015