|
Search: id:A132410
|
|
|
| A132410 |
|
Sequence allows us to find the solutions of the equation: X^3 - (X + 3)^2 + X + 6 = Y^2. |
|
+0
4
|
|
| 3, 4, 7, 12, 19, 28, 39, 52, 67, 84, 103, 124, 147, 172, 199, 228, 259, 292, 327, 364, 403, 444, 487, 532, 579, 628, 679, 732, 787, 844, 903, 964, 1027, 1092, 1159, 1228, 1299, 1372, 1447, 1524, 1603, 1684, 1767, 1852
(list; graph; listen)
|
|
|
OFFSET
|
0,1
|
|
|
COMMENT
|
To prove that X = n^2 + 3: Y^2 = X^3 - (X + 3)^2 + X + 6 = X^3 - X^2 - 5X - 3 = (X - 3)(X^2 + 2X + 1) = (X - 3)*(X + 1)^2 it means: (X - 3) must be a perfect square, so X = n^2 + 3 and Y = n(n^2 + 4).
An equivalent technique of integer factorization would work for example for the equation X^3-3*X^2-9*X-5=(X-5)(X+1)^2=Y^2, looking for perfect squares of the form X-5=n^2. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 20 2007
|
|
FORMULA
|
a(n) = n^2 + 3.
G.f.: -(3-5*x+4*x^2)/(-1+x)^3 = -2/(-1+x)^3-3/(-1+x)^2-4/(-1+x) . - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 20 2007
|
|
EXAMPLE
|
3^3 - 6^2 + 9 = 0^2, 4^3 - 7^2 + 10 = 5^2.
|
|
CROSSREFS
|
Cf. A028560, A005563.
Sequence in context: A034885 A130324 A020677 this_sequence A025047 A050342 A108700
Adjacent sequences: A132407 A132408 A132409 this_sequence A132411 A132412 A132413
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 12 2007
|
|
|
Search completed in 0.002 seconds
|
