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A072069
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Number of integer solutions to the equation 2x^2+y^2+32z^2=m for an odd number m=2n-1.
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9
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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
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OFFSET
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1,1
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COMMENTS
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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).
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REFERENCES
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J. B. Tunnell, A classical Diophantine problem and modular forms of weight 3/2, Invent. Math., 72 (1983), 323-334.
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LINKS
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FORMULA
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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
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EXAMPLE
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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
+ ...
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MATHEMATICA
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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<maxN, s=8; If[x==0, s=s/2]; If[y==0, s=s/2]; If[z==0, s=s/2]; soln2[[(n+1)/2]]+=s], {x, 0, xMax}, {y, 0, yMax}, {z, 0, zMax}]
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PROG
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(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 */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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