A190577 a(n) = n*(n+2)*(n+4)*(n+6).

105, 384, 945, 1920, 3465, 5760, 9009, 13440, 19305, 26880, 36465, 48384, 62985, 80640, 101745, 126720, 156009, 190080, 229425, 274560, 326025, 384384, 450225, 524160, 606825, 698880, 801009, 913920, 1038345, 1175040, 1324785

Offset

1,1

Comments

Each term is the difference between a square and a fourth power:

n*(n+2)*(n+4)*(n+6) = ((n+2)*(n+4)-2^2)^2-2^4. More generally,

n*(n+k)*(n+2*k)*(n+3*k) = ((n+k)*(n+2*k)-k^2)^2-k^4 for any k; in this case, k=2.

References

Miguel de Guzmán Ozámiz, Para Pensar Mejor, Editions Pyramid, 2001, p. 294-295.

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Rafael Parra Machío, Propiedades del 2011: Un paseo a través de los números primos, section 7.3, p. 22.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = ((n+2)*(n+4)-2^2)^2-2^4.

G.f.: 3*x*(5*x^3-25*x^2+47*x-35)/(x-1)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Wesley Ivan Hurt, May 15 2023

Example

a(3) = 945 = 3*(3+2)*(3+4)*(3+6) = ((3+2)*(3+2*2)-2^2)^2-2^4 = 31^2-2^4.
a(13) = 62985 = 13*(13+2)*(13+4)*(13+6) = ((13+2)*(13+2*2)+2^2)^2-2^4 = 251^2-2^4.

Mathematica

Table[n (n + 2) (n + 4) (n + 6), {n, 1, 15}]
Table[((n + 2) (n + 4) - 2^2)^2 - 2^4, {n, 1, 15}]

Prog

(Magma) [((n+2)*(n+4)-2^2)^2-2^4: n in [1..40]]; // Vincenzo Librandi, May 23 2011
(Python)
def A190577(n): return n*(n*(n*(n + 12) + 44) + 48) # Chai Wah Wu, Mar 06 2024

Crossrefs

Sequence in context: A145752 A195266 A113480 * A102792 A013594 A160340

Adjacent sequences: A190574 A190575 A190576 * A190578 A190579 A190580

Keyword

nonn,easy

Author

Rafael Parra Machio, May 18 2011

Status

approved