login
A085409
Sum of three solutions of the Diophantine equation x^2 - y^2 = z^3.
6
0, 12, 84, 270, 624, 1200, 2052, 3234, 4800, 6804, 9300, 12342, 15984, 20280, 25284, 31050, 37632, 45084, 53460, 62814, 73200, 84672, 97284, 111090, 126144, 142500, 160212, 179334, 199920, 222024, 245700, 271002, 297984, 326700, 357204, 389550, 423792, 459984
OFFSET
0,2
COMMENTS
Parametric representation of the solution is (x,y,z) = (6*n^3, 3*n^3, 3*n^2), thus getting a(n) = 9*n^3 + 3*n^2.
FORMULA
a(n) = 9*n^3 + 3*n^2.
From Colin Barker, Oct 25 2019: (Start)
G.f.: 6*x*(2 + 6*x + x^2)/(1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. (End)
From Amiram Eldar, Jan 10 2023: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/18 + sqrt(3)*Pi/6 + 3*log(3)/2 - 3.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/36 - Pi/sqrt(3) - 2*log(2) + 3. (End)
From Elmo R. Oliveira, May 27 2026: (Start)
E.g.f.: 3*exp(x)*x*(4 + 10*x + 3*x^2).
a(n) = 3*A294315(n) = A087887(n)/2 = A033428(n)*A016777(n). (End)
MATHEMATICA
Table[9n^3 + 3n^2, {n, 0, 34}]
PROG
(PARI) concat(0, Vec(6*x*(2 + 6*x + x^2) /(1 - x)^4 + O(x^40))) \\ Colin Barker, Oct 25 2019
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( 6*x*(2 + 6*x + x^2) /(1 - x)^4)); // Marius A. Burtea, Oct 25 2019
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), Aug 13 2003
EXTENSIONS
More terms from Robert G. Wilson v, Aug 16 2003
STATUS
approved