A072069 Number of integer solutions to the equation 2x^2+y^2+32z^2=m for an odd number m=2n-1.

2, 4, 0, 0, 6, 4, 0, 0, 4, 4, 0, 0, 2, 8, 0, 0, 12, 8, 0, 0, 16, 12, 0, 0, 10, 16, 0, 0, 12, 20, 0, 0, 16, 4, 0, 0, 12, 12, 0, 0, 14, 20, 0, 0, 20, 8, 0, 0, 4, 20, 0, 0, 8, 12, 0, 0, 24, 8, 0, 0, 14, 8, 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 2 a(n) = A072068(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 + 6*x^5 + 4*x^6 + 4*x^9 + 4*x^10 + 2*x^13 + 8*x^14 + ... - _Michael Somos_, Dec 26 2019
G.f. = 2*q + 4*q^3 + 6*q^9 + 4*q^11 + 4*q^17 + 4*q^19 + 2*q^25 + 8*q^27 + 12*q^33
+ ...
		

References

• J. B. Tunnell, A classical Diophantine problem and modular forms of weight 3/2, Invent. Math., 72 (1983), 323-334.

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

Crossrefs

Cf. A006991, A003273, A072068, A072070, A072071, A080918.

Programs

Mathematica

maxN=128; soln2=Table[0, {maxN/2}]; xMax=Ceiling[Sqrt[maxN/2]]; yMax=Ceiling[Sqrt[maxN]]; zMax=Ceiling[Sqrt[maxN/32]]; Do[n=2x^2+y^2+32z^2; If[OddQ[n]&&n
				

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)^2 * eta(x^32 + A)^5 / (eta(x + A)^2 * eta(x^4 + A)^3 * eta(x^16 + A)^2 * eta(x^64 + A)^2), n))}; /* Michael Somos, Dec 26 2019 */

Formula

Expansion of 2 * x * phi(x) * psi(x^4) * phi(x^16) 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)^2 * eta(q^32)^5 / (eta(q)^2 * eta(q^4)^3 * eta(q^16)^2 * eta(q^64)^2) in powers of q. - Michael Somos, Dec 26 2019