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A072071 Number of integer solutions to the equation 4x^2+y^2+32z^2=n. 9
1, 2, 0, 0, 4, 4, 0, 0, 4, 2, 0, 0, 0, 4, 0, 0, 4, 4, 0, 0, 8, 0, 0, 0, 0, 6, 0, 0, 0, 4, 0, 0, 6, 4, 0, 0, 12, 12, 0, 0, 16, 8, 0, 0, 0, 12, 0, 0, 8, 10, 0, 0, 24, 4, 0, 0, 0, 12, 0, 0, 0, 12, 0, 0, 12, 8, 0, 0, 16, 8, 0, 0, 20, 12, 0, 0, 0, 8, 0, 0, 8, 6, 0, 0, 16, 16, 0, 0, 0, 4, 0, 0, 0, 8, 0, 0, 8 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Related to primitive congruent numbers A006991.
Assuming the Birch and Swinnerton-Dyer conjecture, the even number 2n is a congruent number if it is squarefree and 2 a(n) = A072070(n).
REFERENCES
J. B. Tunnell, A classical Diophantine problem and modular forms of weight 3/2, Invent. Math., 72 (1983), 323-334.
LINKS
Clay Mathematics Institute, The Birch and Swinnerton-Dyer Conjecture
Department of Pure Maths., Univ. Sheffield, Pythagorean triples and the congruent number problem
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions
FORMULA
Expansion of phi(x) * phi(x^4) * phi(x^32) in powers of x where phi() is a Ramanujan theta function.
a(4*n + 2) = a(4*n + 3) = 0. - Michael Somos, Jun 08 2012
EXAMPLE
a(4) = 4 because (1,0,0), (-1,0,0), (0,2,0) and (0,-2,0) are solutions.
1 + 2*x + 4*x^4 + 4*x^5 + 4*x^8 + 2*x^9 + 4*x^13 + 4*x^16 + 4*x^17 + 8*x^20 + ...
MATHEMATICA
J12[q_] := Sum[q^n^2, {n, -10, 10}]; CoefficientList[Series[J12[q]J12[q^4]J12[q^32], {q, 0, 100}], q]
PROG
(PARI) {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x + A)^-2 * eta(x^2 + A)^5 * eta(x^4 + A)^-4 * eta(x^8 + A)^5 * eta(x^16 + A)^-2 * eta(x^32 + A)^-2 * eta(x^64 + A)^5 * eta(x^128 + A)^-2, n))}
CROSSREFS
Sequence in context: A367054 A134014 A004531 * A329264 A045836 A182056
KEYWORD
nonn
AUTHOR
T. D. Noe, Jun 13 2002
EXTENSIONS
More terms from Vladeta Jovovic, Jun 16 2002
STATUS
approved

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Last modified May 1 12:36 EDT 2024. Contains 372170 sequences. (Running on oeis4.)